Do (A implies B) and (A implies notB) contradict each other?
Do (A implies B) and (A implies notB) contradict each other?
Please give your reasoning, if you choose to answer, for why you think they do, or don't, contradict each other. And if you think they do contradict each other, does that mean they can't both be true at the same time?
Please give your reasoning, if you choose to answer, for why you think they do, or don't, contradict each other. And if you think they do contradict each other, does that mean they can't both be true at the same time?
Comments (815)
((A ? B) ? ~(A ? ~B)) ^ (~(A ? B) ? (A ? ~B))
...Which turns out to be false (link). Given material implication, they do not contradict when A is false, and therefore the two are not full-on contradictory. There would only arise a contradiction supposing A.
(NB: Given the way that common speech differs from material implication, in common speech the two speakers would generally be contradicting one another.)
* I think the second conjunct is redundant, but it illustrates the principle of contradiction.
"A implies B" =/= "A implies not-B", and so the conjunction is "P ^ Q" rather than "P ^ ~P" -- the contradiction would be "A implies B" and "A does not imply B" -- the "not" would have to apply to the logical connective "imply".
Implication (EDIT: i.e. "A implies B") is logically equivalent to "~A or B", so...
Let
P = A implies B
Q = A implies not-B
then...
P = not-A or B
Q = not-A or not B
[s]P =/= Q, and therefore you cannot derive the form "P ^ ~P", and so they do not contradict.[/s]
Combining P and Q what we instead obtain is: not A or B or not B (EDIT: or, we should say, not-A or B and not-A or not B), and since "B or not B" is a tautology we can simplify the expression to "not-A" -- the "B" is a sideshow because it doesn't matter if it's true or false when we put P and Q in conjunction, all that matters [s]if[/s]is the truth value of A.
Here is a JavaScript example:
In the example above the following occurs:
When we run this code, we see that both implications are true, demonstrating the logical conditions.
Presumably you could also run a truth table on this form. If neither conjunct is inherently fallacious and yet the truth table comes out fallacious (i.e. the composite proposition is always false), then presumably there would be a contradiction. The composite proposition would be:
(A ? B) ^ (A ? ~B)
And, yes, I agree with your rendition of the composite proposition. I just find it easier to think of material implication in terms of the equivalence between: A -> B <-> not A or B (Who said that philosophers never agree on things? As long as we stipulate the definitions... :D )
A true and B true means... The thing's false
A true and B false means... The thing's false
A false and B true means... The thing's true.
A false and B false means... The thing's true.
So no, it's not contradictory. As if A is false the statement is true, since both implications are true.
A & A=>B & A=> !B
That's, however, a contradiction.
A true and B true means... The thing's false.
A true and B false means... The thing's false.
A false and B true means... the thing's false.
A false and B false means... the thing's false.
I think where you're getting the contradiction from is the thought that stipulating A lets you derive B and !B, if you also stipulated A=>B and A=>!B. But in fact that's stipulating A and A=>B and A=>!B together.
Now, if you want to use English instead of logic, we can also examine that.
Lets use the word 'implies" which for many means, "could" or "not necessarily".
In which case A could, or could not lead to B. In other words, you're describing an induction. We cannot use A -> B here either, as that's not what it means. Once we had the result of the induction, there are two results we can conclude.
A always leads to (outcome) or
A sometimes leads to (outcome)
This can be determined through testing. For example, if I flip a coin, it sometimes comes out heads, and sometimes comes out tails. This can be written in logic as "Some A's lead to Bs, and some A's lead to not B's"
If of course The question was whether a coin always spins when flipped, we would see it would flip every time. In this case after the test it would be A -> B.
Does that answer your question?
https://en.wikipedia.org/wiki/Barbershop_paradox
Do you think Uncle Joe's argument is valid?
Quoting Tautologies and Contradictions
If (A implies B) and (A implies notB) is understood as (A ? B) ^ (A ? ¬B), it is not contradictory as it is true whenever A is false.
But OP is asking are the two contradictory with each other?
I think what is being asked here is whether one is the denial of the other. And the answer is no. Putting it in logical tables, denial would be whenever (A ? B) yields True (A ? ¬B) yields False. We know that there are values for A and B where both yield the same result.
The denial of (A ? B) would be ¬(A ? B), same for the other (p/¬p). A less obvious example is that ((A?¬B) ? (¬A?B)) is the denial of ((A?B) ? (¬A?¬B)). The first is the XOR gate, the second is XNOR. For any value of A and B, one will give False and other True.
The issue is, whenever A is true, the two do not give the same results. Meaning, if A is true (it is raining), it can't be true that A implies B (it is wet) and not-B (it is not wet) at the same time. As fdrake said.
But if we say it is not raining, the antecedent (A) is false, and the way material implication works in classical logic is that, if the antecedent is false, the implication is always true I remember we both took part in an excruciatingly long discussion about denying the antecedent.
Hoping my explanation was clear and free of errors.
"A -> not A"
mean that
"not (A -> A)."
But surely A -> A. Therefore, not "A -> not A."
Thus, A -> (notB or B) -> not-A. Or more succinctly, A -> not-A
A -> (notB or B)
It's more like
A -> (notB and B)
But yes, it makes sense why that would imply not a. I think your Comms in this thread have just been a little too succinct, you're not filling in enough blanks for people to follow your logic, what point you're making, or even if you answered yes or no to the op poll question
TL;DR: Denial (¬p) and contradiction (yields false in all cases) are not the same thing. Denial is a relationship between two propositions, contradiction is a feature of a single proposition. The two propositions given are not the denial of each other, so Prop1&Prop2 is not contradictory (it can yield True for some values). Material implication (?) is defined in such a way that when the antecedent is False, the result is always True. So, if the antecedent is False, both propositions are True.
I like this post because it's getting into why we might be tempted to say they contradict.
I think, at least in philosophy though maybe there's some other argument this stems from that I'm not aware of, that we should separate out implication from modality -- so when you introduce "possibility" and "necessity" those are entirely different operators from implication.
Though if there's some other argument going on I'm not aware of then you can link it -- I'm surprised to find so many people saying "Yes". lol
"A implies B and not-B, therefore clearly not-A."
Fuzzy logic/Many-valued logic...
What would you be trying to say? I am having trouble thinking of an example.
Thinking in terms of a Venn diagram, you could have many points of congruence or intersection between sets and be able to say something along the lines of what your previous post meant.
That's not the same as "(A implies B) and (A implies not-B)" -- that'd be "(A implies (B and not-B)).
A real world example is often hard to parse into material implication -- sometimes, yes, but sometimes it's hard -- the conjuncts of disjuncts, while they can be claimed, is even rarer :D
Though after we dismiss "B and not-B" as always false, we can see that the truth of the proposition will only rely upon A, since "implies" is logically equivalent to "not-A or (B and not-B)", and the truth of a disjunct is true if one of the propositions is true -- so if not-A is true then it is true, and if not then it is false -- since not all results in the truth-table are false it is not a contradiction.
In epistemic logic, such cases are known as known known's, known unknown's, and unknown unknowns. Yet, there are examples of making false inferences like unknown known's that would describe such a situation arising in epistemic logic...
Maybe the confusion comes from taking A as a proposition instead of a variable that may be 0 or 1.
I'm definitely coming at the question from the point of view of propositional logic -- something basic because it's already confusing enough as it is :D
Im not versed in logic at all but I'll give my thoughts.
If A is defined specifically (ie. no coin flipping or randomness involved in its definition) and we're assuming "implies" means "neccesitates" and not "can lead to", then it is contradictory.
If B and not B are different, then if A neccesary leads to one, it cannot lead to the other definitionally.
Ofc, if not B and B are the same, then theres no contradiction; and if A can be unspecific then it can lead to both (flipping a coin can get heads or not heads).
Are they not? First and last rows of the truth tables are all the same. Seems logically equivalent to me.
I'm using https://en.wikipedia.org/wiki/De_Morgan%27s_laws though I've made many mistakes so I could be wrong...
Since the last column is not all "F" it's not a contradiction, I believe. (though I see I confused myself earlier, looking at the T-table)
Yeah, it's not a contradiction, but the final row of the truth table for p?q ? p ? ¬q is the same p ? q ? ¬q for all values of p and q.
Maybe that's why it's confusing? [s]It's an implication of two implications, and material implication is already confusing . ..[/s] lol see I confused it even in response.
Awwww ... there are minds out there who think like this, and if I'm in the habit I can -- but it's not my normal way.
¬(g ? P(g) ? g ? ¬P(g))
i.e. ¬¬g, which is a classically acceptable proof of existence.
It was a bit of headox with a sentence like this:
"If Carr is out, then we know this: "If Allen is out, then Brown is in", because there has to be someone in "to mind the shop." Ugh.
So cleaning this up we get this:
They explain that there are three barbers who live and work in the shopAllen, Brown, and Carrand some or all of them may be in. We are given two pieces of information from which to draw conclusions. Firstly, the shop is definitely open, so at least one of the barbers must be in. Secondly, Allen is said to be very nervous, so that he never leaves the shop unless Brown goes with him.
So, we have A, B, and C
One must always be in.
If A is out, B is out
Therefore C is in
If B is out, A can be in, as B can leave the shop.
If A is out, B is also out, so C is in
So our combos are as follows:
1. C !A !B
2. !C A B
3. C A !B
4. !C A !B
5. C A B
"Suppose that Carr is out. We will show that this assumption produces a contradiction. If Carr is out, then we know this: "If Allen is out, then Brown is in", because there has to be someone in "to mind the shop." But, we also know that whenever Allen goes out he takes Brown with him, so as a general rule, "If Allen is out, then Brown is out". The two statements we have arrived at are incompatible, because if Allen is out then Brown cannot be both In (according to one) and Out (according to the other). There is a contradiction. So we must abandon our hypothesis that Carr is Out, and conclude that Carr must be in."
This doesn't make any sense. Clearly B can be out and A be in. Clearly C could be out and A and B, or A be in. I don't get it.
This is simply a language issue. What does 'impliciation' mean? That's why I went over all the different possibilities. In the end implication must mean necessary or not necessary, in which case the answer will be different.
If implications were coaches, then logicians would still ride.
Let A = "Unenlightened's testimony is unreliable"
Let B = "Unenlightened tells the truth"
not B ="Unenlightened does not tell the truth"
No, 0/1 is exactly the same as True/False for the purpose of logic tables. What I meant to say is that some people seem to think that A here implies not a variable (that may take 0 or 1) but a proposition that is being asserted as True (which is to say that it is 1). When A is 1, A?B and A?notB give opposite results.
Quoting Philosophim
We are not talking about modal logic.
Both are contradictory (thus False) if A is True. But as by the definition of material implication, both are True if A is False.
Often times in logic or arguments, the solution is clear. What often happens is we use language with poor definitions. Thus people make assumptions or conclusions that the user of the unclear language didn't intend, or the user themself is doing the same.
Notice how he used the word 'imply' and people immediately thought, Oh A -> B. But that's not imply. that's "Necessarily leads to." And its obvious that if A -> B, that A -> !B is a contradiction if you use the correct definition. "Imply" is vague enough that some thought 'necessarily', while others thought 'maybe'. And because you can get halfway there with the word, some might even thing A ->B means only sometimes.
A -> B
A -> ~B
are not together inconsistent, since they are both true when A is false, and the propositional calculus doesn't permit an inference of a contradiction from two formulas such that there is a model in which they are both true.
The digressions in this thread don't affect that very simple fact.
Yours is the best post in the thread imo. It is especially interesting that what I called fallacious your source explicitly calls a contradiction (). There are two different notions of contradiction occurring in the thread.
Quoting Lionino
A classical definition says that two propositions are contradictory if the denial of either entails the affirmation of the other, and vice versa. So there are materially four different relations, given that each of the two propositions can be denied or affirmed. This is what I was trying to get at in my first post.
What's interesting is that it is not possible to contradict a material implication even on the classical understanding of contradiction. This is why I think the more interesting question prescinds from material implication:
Quoting Leontiskos
You think the two propositions logically imply ~A? It seems rather that what they imply is that A cannot be asserted. When we talk about contradiction there is a cleavage, insofar as it cannot strictly speaking be captured by logic. It is a violation of logic.
Your presupposition here straddles the two definitions of a contradiction in an interesting way. Using the same example I gave privately:
"The car is wholly green." "No, the car is not wholly green."
This is a contradiction on all definitions.
"The car is wholly green." "No, the car is wholly red."
This is a contradiction classically but not according to symbolic logic, and your method would not find it to be a contradiction.
I think the third way to give a contradiction, besides the two already noted, is to use symbolic logic and say, "Assume P and suppose Q. If an absurdity results on Q, then P and Q are contradictory." This gets closer to the classical definition. It shows that they cannot both be true, but it does not show that they cannot both be false, and it does not show that the trueness or falseness of one results from the falseness or trueness of the other.
A contradiction is a formula of the form P & ~P, or in other contexts the pair {P ~P}.
We don't have to check four different things to see that formula, or in other contexts a pair of formulas, is a contradiction.
A set of formulas is inconsistent if and only if it proves a contradiction.
Quoting Leontiskos
Wrong. P -> Q is contradicted by ~(P -> Q).
Quoting Leontiskos
Yes, people say "contradictory", but as a terminological preference, I would say they are together inconsistent. A pair of formulas is a contradiction iff they are of the form P and ~P. A set of formulas is inconsistent if and only if it implies a contradiction.
Quoting Leontiskos
Wrong. If one is true then the other is false. If one is false then the other is true. [EDIT:] Correction: Obviously, it is not correct that it is always the case that there is at lease one true statement in an inconsistent set of statements.
Quoting Leontiskos
Wrong. If the members of a set of sentences that includes P are all true, and that set of sentences along with Q (that is not in the set) proves a contradiction, then Q is false. Put another way, we cannot derive a contradiction from a set of all true sentences. [EDIT:] Correction: Obviously, it is not correct that it is always the case that there is at lease one true statement in an inconsistent set of statements.
Quoting Leontiskos
Classical logic includes propositional logic such as treated in symbolic logic.
"The car is green" and "The car is red" is not a contradiction. But if we add the premise: "If the car is red then the car is not green," then the three statements together are inconsistent. That's for classical logic and for symbolic rendering for classical logic too.
They imply ~A.
Quoting Leontiskos
I don't know what you mean by 'cleavage' and 'captured' in this context. But in logic systems we can write contradictions. Indeed, we often intentionally prove contradictions from premises in order to refute the premises.
Wrong. Material implication does not require necessity.
Then give your proof.
If p is False, both propositions are true for any value of q.
Taking a look at (p?(q and ¬q)):
0 = False, 1 = True
(q and ¬q) is always 0 definition of contradiction
a?b is equivalent to (¬a or b) definition of material implication
our b in this case is (q and ¬q), thus b is always 0
if p is 0, ¬p is 1, so from (¬p or (q and ¬q)) we have (1 or 0)
The or operator will return 1 if any of the variables is 1 by definition
So (1 or 0) returns 1, so (¬p or (q and ¬q)) returns 1 if p is 0, so ((p?q)?(p?¬q)) returns 1 if p is 0
This kind of stuff is better to understand from the POV of electronics. The words "True" and "False" make things confusing. "1 or 0" is a logical gate with two inputs, returning 1, "True or False" is like "Duh, what else could it be?"
(A implies B) and (A implies notB) do not contradict one another.
It would be useful to have a page that generates an image of a given truth table.
Around a third of folk hereabouts who have an interest in logical issues cannot do basic logic.
might note that 's testimony is reliable.
Oh, and , (A?B)?(A?¬B)?¬A.
- very clear.
https://web.stanford.edu/class/cs103/tools/truth-table-tool/ ?
I am mostly complementing my own posts. I didn't know you were arguing the opposite, but alright :up:
Other methods are clumsy.
Are you serious? You don't know how to prove it yourself?
Proof:
(1) (A -> B) ... premise
(2) (A -> ~B) ... premise
(3) A ... toward a contradiction
(4) B ... from (1), (3) modus ponies
(5) ~B ... from (2), (3) modus ponens
(6) ~A ... from (4), (5) and discharge (3) by contradiction
Truth table:
A B......A->B......A->~B......~A
T T..........T............ F..............F
T F..........F.............T..............F
F T..........T.............T.............T
F F..........T.............T.............T
All rows in which both A->B and A->~B are true are rows in which ~A is true.
/
Do I get anything for doing your homework for you? Cash? Philosophy Forums poker chips? Internet Brownie points?
You misunderstand. A -> B is not implication as we may use it in English. In other words, "It being cloudy implies that it will rain soon," is not the same as "It being cloudy necessarily means that it will rain soon. Yes, in logic, its called an "implication", but in standard English its equivalent to "Necessitates". That notation is If A, then B. Or if A, necessarily B. Its not, If A, maybe B.
Thus, if one plugs the word 'necessitate' in, its clear that A necessitates B and A necessitates not B contradict each other.
What I see some people doing is declaring A -> A or B, which is perfectly fine. But that is not the same as stating A -> B, (which means it will only lead to B) then in the next statement A -> !B (A will lead only to not B).
Edit: we posted at the same time.
Well I screenshot, press control v on imgur.com, then copy image url address and use it on "image" button
Yes, this was my concern. Tones requires the assumption, as I thought he must.
I replied exactly to what you wrote. What you wrote is wrong.
A -> B
or
material implication
is not "A necessarily implies B".
Both propositions together imply ~A. The conjunction of the propositions implies ~A. The premise set {A->B, A->~B} implies ~A.
Quoting Lionino
Right.
https://people.cs.rutgers.edu/~elgammal/classes/cs205/implication.pdf
"Different forms for implications
p implies q is equivalent to
If p then q
q if p
q whenever p
q when p
q follows from p
p is sufficient for q
a sufficient condition for q is p
q is necessary for p
a necessary condition for p is q"
So what?
'->' is ordinarily regarded as standing for material implication that is not "P necessarily implies Q" nor "P necessarily leads to Q".
Please demonstrate your claim.
Look in any textbook on symbolic logic.
If you're not going to bother answering like I did, I'm going to hand wave you away. Don't be a troll. Show it.
Rather, I'm interested in you doing something more than making curt pronouncements from on high. This is a philosophy forum, after all.
Here is the alternative notion of contradiction that you are overlooking:
Quoting Aristotle on Non-contradiction | SEP
Tones is correct.
It is not trolling to point out an incorrect statement, and it not trolling nor handwaving to suggest that one can look in textbooks to see that the statement is incorrect.
Or look in any resource you like to see that '->' in logic is ordinarily understood as the material conditional and any textbook in symbolic logic will explicitly state that '->' is the sentential connective such that 'P -> Q' is interpreted as true if and only 'P' is interpreted as false or 'Q' is interpreted as true. And that is material implication.
If A (being True) is necessary for B (to be True), B?A. A is a necessary condition of B.
If A is sufficient for B, A?B. A is a sufficient condition of B.
Necessary and sufficient: A??B. A is a necessary and sufficient condition of B, so A and B are in constant conjunction (queue: Hume).
Another confusion that stems from 'p'/'a' meaning "[given proposition] is True" or meaning a variable that may take values 0/False or 1/True.
Actually, your imperative "Then give your proof" was curt, especially as spoken to someone giving you correct information.
The proof is so simple that one would think that someone posting claims about the matter would know the proof or would just look it up, or even run it in their mind for five seconds, rather than a challenging "Then give your proof". People don't need to give such utterly basic proofs. It's like saying, "Then give your proof that 1+1=2". Moreover, as good as proofs had already been given in this thread.
Run it in your mind for five seconds: If A implies both B and ~B, then A implies a contradiction, so we infer that A is not the case. You really need to challenge people to prove that for you?
Quoting Aristotle on Non-contradiction | SEP
That is the law of non-contradiction. What I said is a more formal way of saying the same.
Yes, we say 'necessary' and 'sufficient' conditions. But that is not "necessarily implies' or 'necessarily leads to'.
If P -> Q, then P is a sufficient condition for Q and Q is a necessary condition for P, and that is not to suggest any modal operator. But we don't say that P necessarily implies Q.
Yes, the first phrase means ?(p?q), the second means nothing to my knowledge.
That would be my leaning too.
The first page of the thread contains two basic ways of defining contradictions. You gave a third: mere negation and the attendant inverted truth table. That's a legitimate extrapolation, but still different from a non-formal assessment of contradiction.
I don't require such an assumption. "imply ~A" is not even a proposition.
I don't claim that one may not discuss all kinds of non-formal, formal, alternative formal, or philsophically formal or informal, or mystical New Age incantational informal senses.
'imply ~A' is not a proposition, and I didn't say that it is, so I don't require it as an assumption.
Anyway we clarified that here https://thephilosophyforum.com/discussion/comment/916729
Quoting Leontiskos
That is what I replied to.
If he is using the term of implication to mean, "could lead to", then that's fine. I've already written on that on the first page. I did not catch that 'material' conditional was anything different from the modal operator.
Quoting TonesInDeepFreeze
Your attitude is hostile and condescending without backing up your claim clearly. I had no idea that there was a specific logical term called a material conditional. You spoke so tersely and dismissively, I didn't take your reply seriously. Give detail and respect, and it will be given back by good people.
I will state again, you misunderstood what I was stating earlier. I'm replying to someone specifically in which I covered both types of meanings of the words 'imply', as the OP did not specify what they meant. One where "Imply" means "necessary" and one where imply means "Could lead to".
I already noted in the second case that A could lead to B and A could lead to not B are not contradictions. But they are contradictions in the sense of using the word 'implication' as necessary. Or to use Banno's link if we are going to use formal logical terminology, 'material conditional' vs a 'modal operator'.
I am not. I'm treating '->' as standing for material implication as is ordinary.
Quoting Philosophim
They are very different.
Quoting Philosophim
Actually, you insulted me. I hadn't written anything "hostile" or "condescending" but then insultingly you wrote:
Quoting Philosophim
Quoting Philosophim
I gave you the best advice anyone could ever give you regarding this subject: Look at a textbook. Then you could read a full explanation in full context, thus to inform yourself on the subject properly. And there is no better "proof" that '->' ordinarily means material implication than to read for yourself in an authoritative and widely referenced textbook.
The fact that '->' is ordinarily understood as the material conditional is ubiquitous. It's not my call to provide you references that you could easily find yourself by just looking at basic texts.
Quoting Philosophim
Terse in the sense of "devoid of superfluity" not in the sense of "brusque". To suggest looking in a textbook is the best advice I can give you; It should not needed be needed for me to further elaborate on that advice. However, if you asked me to recommend textbooks, then I would be happy to do that. And to be really dismissive I could have simply ignored you; instead I gave you the very best advice I can give.
Quoting Philosophim
You misunderstand. You said that '->' means 'necessarily leads to'. And that is false. And "necessary" and "could lead to" are not the two meanings of '->', as material implication is the ordinary meaning and does not mean "necessarily leads to" nor "could lead to".
You do me or no one else favors here. We're having a discussion, and if you want to make a point, link or note your point. "Look at a textbook" is dismissive and means you're removing yourself from the conversation.
Quoting TonesInDeepFreeze
Like this?
Quoting TonesInDeepFreeze
Not exactly the model of a sage and wise poster. You came on here with a chip on your shoulder to everyone. I gave you a chance to have a good conversation, but I didn't see a change in your attitude.
Quoting TonesInDeepFreeze
If I'm clearly using it as a strict conditional, as I noted in prior posts as I was talking to someone other than yourself, then I'm correct. Now if you had an issue with my use of -> or wanted to teach me the difference between a modal and material implication, something I did not know before today, you could have spent less then a minute citing a wiki post somewhere like Banno did. Instead, we have wasted time back and forth and your attitude didn't win you respect today.
This is not a place where we should banter back and forth for our egos. Its about spreading knowledge and wisdom with one another with good discussions. You seem to have knowledge, which is wonderful. Share it and teach. You'll earn respect. Don't bare it and preach. You'll get eyes rolled at you.
Material implication is often written in natural language as:
If A, then B
Or
A implies B
But they aren't perfect translations because all sorts of shit that sounds very dumb in natural language flies in symbolic logic. E.g. "if Trump won the 2020 election then we would have colonized Mars by now."
Anything follows from a false antecedent, so anything would be "true" following the claim that Trump won the 2020 election, since he didn't. But obviously in natural language the if/then here is talking about a counterfactual claim, and to say that if x had happened then y and not-y is (normally) nonsense talk.
There's no note or link needed. You can find out about material implication all over the place. I'm not your linking service.
Quoting Philosophim
You skipped what I said about "dismissiveness". You merely reiterate your claim without addressing my response to it. That is dismissive. And of course, by giving my best advice to look at a textbook, I am not thereby opting out of any conversation. Interesting that you continue to attack me rather than take my offer for some recommendations of books or other resources.
Quoting Philosophim
You leave out that I went on to give a proof in two versions. And it is appropriate to ask whether a poster is really serious asking for something that is, as far logic is concerned, as simple as showing that 4 is an even number. If in a thread about number theory someone happened to write "4 is even", and then another said "Prove it", you think that would not be remarkable enough to reply "Are you serious? You don't know how how to prove it?", let alone to then go on to prove it anyway.
Quoting Philosophim
Where is here? This thread? I came with no shoulder chip, not to anyone, let alone "everyone". If I permitted myself to do as you do - to posit a false claim about interior states - I would say that you do so from your own umbrage at having been corrected.
And my point stands that I did not insult you, whereupon you insulted me.Quoting Philosophim
By saying "don't be a troll".
You can converse as you please. I'm not stopping you. And I have read your subsequent posts, even after your insulting "don't be a troll" and have given you even more information and explanation. I have not shut down any conversation.
Quoting Philosophim
The first step is to at least point out that '->' does not mean "necessarily leads to". And, I'm glad that you do know that there is a difference now, and glad that my posting the correction has led to you knowing that there is a difference.
Quoting Philosophim
(1) I don't usually reference Wikipedia or similar sites. They often have misinformation and poor explanations. It is not my job to find a site for you, read it through to vet it for accuracy, then fashion a link for you.
(2) I gave you even better advice anyway.
(3) And if you had asked for more, then I would have recommended specific texts or suggest search terms for you, though you could fashion your own search.
Quoting Philosophim
I'll judge for myself what is or isn't my time wasted. But your time wasn't wasted since at least my correction led to you learning something about logic, and not just an incidental detail but rather a key fundamental aspect of logic.
Quoting Philosophim
I don't presume to be a teacher. But I have shared a tremendous amount of information and explanation over years in this forum alone.
The answer is 'yes' and they do not contradict each other. You can read the several differently arranged proofs of that in this thread.
Quoting Count Timothy von Icarus
I agree, I came across such things when studying logic as an undergraduate. I will just say that "if Trump won the 2020 election, then we would have colonized Mars by now" is neither true nor false (or at least cannot be determined to be true or false).
So I don't think it is true that "anything would be true following the claim that Trump won the 2020 election, since he didn't", because it could equally be said that "anything would be false following the claim that Trump won the 2020 election, since he didn't".
The way you would usually use it in any sort natural language statement would be to say: "Look, A implies both B and not-B, so clearly A cannot be true." You don't have a contradiction if you reject A, only if you affirm it.
This is a fairly common sort of argument. Something like: "if everything Tucker Carlson says about Joe Biden is true then it implies that Joe Biden is both demented/mentally incompetent and a criminal mastermind running a crime family (i.e., incompetent and competent, not-B and B) therefore he must be wrong somewhere."
In ordinary formal logic and classical mathematics, the material conditional obtains. But, of course, there are other natural language senses.
In everyday speech, one would not ordinarily say "If snow is green then Emmanuel Macron is an American, and if snow is green then Emmanuel Macron is not an American". But even then, the assertion is not inconsistent unless we also assert "snow is green".
A contradiction is a pair of statements of the form P and ~P.
Such as "Emmanuel Macron is an American" and "Emmanuel Macron is not an American".
Notice that "If snow is green then Emmanuel Macron is an American" and "If snow is green then Emmanuel Macron is not an American" is not of that form and together they don't imply the contradiction "Emmanuel Macron is an American" and "Emmanuel Macron is not an American". They only imply that contraction along with the statement "Snow is green".
An example where we do say both "A then B" and "A then not B":
Background: A lawyer is defending Ruth. We know that Ruth wore a blue dress on the night of the crime. And we know that the assailant did not wear a blue dress on the night of the crime. The lawyer says:
"If Ruth is the assailant, then the assailant wore a blue dress" and "If Ruth is the assailant then the assailant did not wear a blue dress. So, the assertion that Ruth is the assailant implies a contradiction, so Ruth is not the assailant."
That's an awkward and verbose way of saying "The assailant wore a blue dress, but Ruth did not, so Ruth is not the assailant". But despite it being awkard and verbose, it is correct English and logical.
/
As to what the purpose of formal logic is, there are different purposes. For logic that pertains to mathematics and computability, ordinarily the material conditional is used. For example, the computer you're using now is based on logic paths in which "if then" is the material conditional.
There's a mistake in the last row. The value of ~A v ~B is T. So there are two rows, not just one, where (A -> B & (A -> ~B) is true.
I'll add a link to your post here to the original post.
Thanks, you obviously know much more about formal logic than I do. However I was not thinking in terms of formal logic, since the original question contains no symbols from formal logic
Quoting flannel jesus
My point was that if you have two sentences 'if A then B' and 'if A then notB' they simply contradict one another regardless of whether A obtains. Of course we don't know what A is. If the two sentences were 'if monkeys had wings, then they could fly to the moon' and 'if monkeys had wings, they could not fly to the moon' the two sentences contradict one another regardless of whether it is true that monkeys have wings or whether it is true that if they had wings they either could or could not fly to the moon.
Quoting TonesInDeepFreeze
I think that is one way of interpreting it, separating the obviously false, or disconnected conditional "if snow if green" from the two contradicting statements about Macron, but assuming that there would be some logical connection between the conditional and the implications (and why would we even bother thinking about statements where there is no such logical connection) then the two statements do contradict one another.
Quoting TonesInDeepFreeze
Right, I do have some familiarity with logic gates. Are any of those useful logic paths nonsensical? Genuine question...
It seems to me that saying you don't have a contradiction is one way of interpreting it; that is, I don't think there is any fact of the matter. Consider "If lizards were all purple then they'd be a hell of a lot smarter" and "if lizards were all purple, they would not be a hell of a lot smarter": you don't see those two sentences as contradicting one another despite the fact that lizards are not all purple?
As someone pointed out, the use of variables suggest formality. But, of course, we may address the question in both formal and informal contexts. And I've done that.
Quoting Janus
That's incorrect, formally or informally. I explained why it's not correct.
Quoting Janus
I don't think so. Or I would like to know of a system or approach that supports it.
I sense that it is not logical connection you have in mind, but rather, what is called in logic, 'relevance'. If the relation between the antecedent and consequent is not relevant, then that does not accord with many everyday uses of 'if then'. But that doesn't entail that the two statements contradict each other. It only entails the statements don't accord with certain everyday uses. Same for green snow and Macron's nationality. But, of course, the material conditional also does not accord with relevance logic. Though I'd have to do a bit of reading to see whether even in relevance logic (A -> B) & (A -> ~B) is a contradiction when the implications are not relevant.
Quoting Janus
How are they nonsensical?
Why is it incorrect informally?
Quoting TonesInDeepFreeze
Yes, relevance is another way of saying logical connection in the context.
Quoting TonesInDeepFreeze
I asked you if any were nonsensical, I didn't say they were. In informal language if the antecendent has no relevance to the consequent then I would say that counts as nonsensicality.
Because even informally, the statements don't entail a both statement and its negation.
Quoting Janus
I wouldn't use the word 'logical' since that has a certain meaning in the study of logic that is not the same as 'relevance'.
Quoting Janus
Oh, I thought your question was rhetorical. I thought you meant that it is a genuine philosophical question about computing.
I don't know enough to answer your question. So I would turn it around to ask: Is there an argument to be made that they are nonsensical?
Quoting Janus
You asked if the logic paths are nonsensical. I thought 'logic paths' related to the logic gates you mentioned in your previous sentence.
Again, a contradiction is a statement and its negation. If there is a contradiction then you could show that both a statement and its negation are implied.
Again:
"if lizards are purple, then they would be smarter" and "if lizards are purple, then they would not be smarter" is not a contradiction.
"if lizards are purple, then they would be smarter" and "if lizards are purple, then they would not be smarter" and "lizards are purple" does imply a contradiction.
"if lizards are purple, then they would be smarter" and "if lizards are purple, then they would not be smarter" and "lizards are not purple" does not imply a contradiction.
/
I'm not sure, but maybe you should check whether you are conflating "not intuitive" with "contradictory".
I'm not sure what you mean: I was considering the two statements separately and it still seems to me, that regardless of the soundness or relevance of their content, that, taken informally as statements, they contradict one another.
Quoting TonesInDeepFreeze
I see those two statements as saying contradictory things about what lizards being purple would entail.
Quoting TonesInDeepFreeze
I see two of those statements, as above, as being contradictory and the third as being unsound. And I see the two contradictory statements as saying nothing about whether lizards are purple. I mean, I think it's fair to say that both of the conditional statements are untrue, because being or not being smarter has no logical connection with being purple. Or I could say that the two statements are nonsensical because the antecedent has no relevance to the consequent. However, I cannot but see them as contradictory.
What about these two statements: 'if I was more educated in logic, I would be able to see that those two statements are contradictory" and "if I was more educated in logic I would not be able to see that those two statements are contradictory"do those two statements contradict one another?
Or what about 'if I was more educated in logic, I would be able to see that those two statements are contradictory" and "if I was more educated in logic I would be able to see that those two statements are not contradictory"?
Quoting TonesInDeepFreeze
I don't believe I am conflating "not intuitive" with "contradictory", but of course I admit I could be wrong. Do I understand what 'contradictory' means? I think so.
What about "the present king of France is bald" and "the present king of France is not bald" do they contradict one another?
But the difficulties of material implication do not go away here. You are thinking of negation in terms of symbolic logic, in which case the contradictory proposition equates to, "Lizards are purple and they are not smarter." Yet in natural language when we contradict or negate such a claim, we are in fact saying, "If lizards were purple, they would not be smarter." We say, "No, they would not (be smarter in that case)." The negation must depend on the sense of the proposition, and in actuality the sense of real life propositions is never the sense given by material implication.
The reason we keep material implication is because we like truth functionality.
Note 's testimony.
An informal sense of 'contradict' is 'to imply the opposite or a denial of'; and an informal sense of 'denial' is 'a proposition so related to another that though both may be false they cannot both be true'
And an informal sense of 'contradiction' is 'a proposition, statement, or phrase that asserts or implies both the truth and falsity of something'.
Both of those accord with the formal sense.
Of course there are many other informal senses.
One informal sense of 'contradiction' is 'incongruity'. That might be what you have in mind.
It is not at issue that people may use different senses. It is senseless to argue with two incompatible senses both at work.
In context of the study of modern logic, in both philosophy and mathematics, 'contradiction' ordinarily means 'a statement that is the conjunction of a statement and its negation', from which follows 'a statement that asserts the both the truth and falsity of something'. And that is also an informal sense.
When someone says, "You contradicted yourself when you said you didn't visit the store", they mean "You said you you didn't visit the store but the day before you said that you did visit the store". Or, "You said you didn't visit the store, but you also said you saw Ted yesterday at 1:00. But at 1:00 Ted was at the store." That is the informal sense of 'contradiction' I refer to, and it accords with the formal sense.
No one can dispute that you find your example incongruous in some personal way, while what is incongruous to one person is not incongruous to another. But my point is that your example is not a contradiction in the ordinary sense in modern logic or in an everyday sense such as when someone says, "You contradicted yourself" to mean "You said you didn't visit the store but also the other day you said you did visit the store" they don't mean that it was merely odd or incongruous. Rather, they mean that you claimed both a statement and its negation.'
Of course, no one should deny you using whatever sense of 'contradiction' you like. Better yet, would be for you to define it. Meanwhile though, in logic, 'contradiction' has a precise definition and it accords with a natural everyday sense too.
Well... yes, kind of.
From falsehood, anything follows. Have you ever heard of this? This example before us is a great example of that.
You think if a then b, and if a then not b contradict each other.
Many other posters think they don't contradict each other, BUT with the caveat that if they're both valid statements, A must be false.
If a is false, and "from falsehood, anything follows", then (if a, then anything) fits. Replace anything with b, replace it with not b, replace it with a snail with a tophat, replace it with a???????b??????y????????????s??????????????s????????????a???????l?????? ???????d???????????e???????s??????????p???????????a??????i?????????r??????????...
As long as A is false, "if a then anything" obtains. You can verify this with the truth table. And this is where your sense of nonsense comes in. Do you see?
I don't think the principle of explosion is quite the same as material implication. It's kind of the opposite. We are running from a contradiction, not running on a contradiction. See and .
But I personally think it's not a coincidence that "from falsehood, anything follows" perfectly mirrors how, if you phrase "A -> B" as "from A follows B", then if A is false, you can say "A -> anything", from A anything follows.
I don't think that's a coincidence at all. I think the principle of explosion is actually really relevant here. But I understand that not everyone sees it that way.
In the same vein as above, 'true', 'sound' and 'valid have definitions in logic. Of course, it's your prerogative to use any sense you like. But it behooves us to be clear which definition is at play in a given context. If you are merely disagreeing based on a different definition, then my reply is, "Okay, then there are two different discussions: One based in the ordinary definitions in logic, and the other based in whatever other definitions you stipulate."
Quoting Janus
Again, that is based on your notion of 'if then'. It is not based on the ordinary notion in logic. So, again, two different discussions: One based on the ordinary notion in logic and the other based on your notion (or better yet, based on relevance logic). Also, you ignored my point about using the term 'logical connection'.
Quoting Janus
Ordinarily in logic, the expressions we study are not nonsensical. So the notion of contradiction would not apply. But, again, you'll use your own definitions, and that is a different discussion from a discussion in context of ordinary definitions in logic.
Quoting Janus
I don't know why you're asking me to comment on an example that is the same in form as the other examples.
Quoting Janus
I don't know what your point is, but to fulfill the exercise:
To save typing and copy/pasting:
L ... I know more logic
C ... I would see that the statements are contradictory
N ... I would see that the statements are not contradictory
-> ... if ___ then ___
~ ... it's not the case that
And I'll answer in the context of classical logic:
Note that in N, 'not' is in the scope of 'I would see that'. So, in mere sentential logic, N is an atomic statement, so I can't pull 'not' outside the scope.
(L -> C) & (L -> ~C) ... not a contradiction
(L -> C) & (L -> N) ... not a contradiction
I hope that's the only exercises I'll be doing here.
Quoting Janus
Perhaps you understand the sense you prefer to use. But it seems you don't understand the ordinary formal and informal sense I've explained.
I don't opine on what follows from your sense of 'contradiction', whatever your definition might be. But I do know what is the case with the ordinary formal sense that accords also with a primary informal sense. I hope that you don't hold that your use of your sense of 'contradiction' - however you might define it - trumps people talking about contradiction in the sense in the study of logic that accords with a primary everyday sense.
It does not follow; it is moot. According to material implication (A ? B) is true if A is false, but B does not follow given that A is false. We cannot derive B. That's why A is false in this case, because we cannot arrive at the contradiction of (B ^ ~B).
Of course I agree that we can't conclude B and notB. The fact that you're saying that makes me think you've misunderstood what I said.
I'm not affirming the antecedent, so I'm not affirming the consequent.
"Who are you, who are so wise in the ways of science?"
Put it together: ...therefore the consequent cannot be affirmed as true in this case. Therefore the consequent does not "follow."
You think when I use the word 'follow', and completely understandably, I mean "this thing is true". As in, I'm saying "B is true" period.
I'm not.
"follow" can also just be a synonym for implication. A -> B, From A follows B. If you assert A, B follows.
I can say "A -> B" without asserting B, and in the same vein, I can say "From A follows B" without asserting B. Because they're just different ways of phrasing the same thing.
I'm not asserting B. I'm asserting A -> B. You have to see the difference to understand. When you understand how I can assert A -> B without asserting B, you can understand how I can say "From A follows B", without me saying "B is true".
I've not claimed that anything I've said dissolves any difficulties with material implication.
Quoting Leontiskos
I'm thinking of it in context of symbolic logic, informal logic, and a primary everyday sense.
Quoting Leontiskos
I know of no context in which that sentence is a contradiction.
Quoting Leontiskos
What is your basis for that claim? Your observation of what people mean when they say such sentences. I'm not privy to those observations.
Quoting Leontiskos
There are two separate matters: negation and material implication. I've addressed both in this thread. It is not disputed that material implication often does not accord with everyday senses of 'if then'. But such everyday senses are not explicated for definiteness usually don't submit to rigorous logical, mathematical or philosophical treatment. And though logic, mathematics, computing, and philosophy are not everyday, they aren't thereby relegated irrelevance nor is their profound relevance diminshed.
The question at hand is, "What is the contradiction of, 'If lizards were purple then they would be smarter'?"
Quoting TonesInDeepFreeze
The negation of a material conditional will be different from the negation of an if-then statement in natural language, and my post was highlighting that difference.
Or as I said earlier:
Quoting Leontiskos
(1) You changed the sentence. Here is what you wrote:
Quoting Leontiskos
(2) "If lizards were purple then they would be smarter" is not a contradiction, a fortiori not a contradiction in symbolic logic especially. And "Lizards are purple and they are not smarter" is not a contradiction.
Quoting Leontiskos
There are many senses of 'if then' in natural language. But, of course, material implication does not accord with many of the natural language senses of 'if then'. That's never been at issue.
Explosion doesn't make the material conditional in Boolean logic used for computing nonsensical.
Right, it's not a coincidence. That doesn't entail anything about the material conditional in Boolean logic.
No, I was there giving an answer to the question at hand.
Quoting TonesInDeepFreeze
I give up. Go read Lionino's first post on the first page. He explains the two basic senses of contradiction operating in the thread.
Your answer is incorrect.
Quoting Leontiskos
You should give up, since Lionino's post is perfectly in accord with what I have said: the pair of statements are not a contradiction.
You don't even understand what is being said. :roll:
I suspect you don't understand what you wrote.
(1) I take 'the contradictory statement is P' to mean that P is a contradiction, as a contradictory statement is a contradiction.
(2) But maybe you mean it is a contradicting statement. Maybe you mean "Lizards are purple and they are smarter" contradicts some other statement? Well, it contradicts "It is not the case that lizards are purple and they are smarter". That doesn't vitiate anything I've said.
/
If you have some other definition of 'contradiction' then it would help to know it. Meanwhile, ordinarily, and not just in symbolic logic, a contradiction is a statement S and its negation. Or, less strictly but tantamount, a statement S and some other statement that implies the negation of S. At least along those lines. If that doesn't suit you, then so be it. But unless you provide a different definition, then I'll say that (A -> B) & (A -> ~B) is not a contradiction and does not imply one, whether regarding purple lizards and their intelligence or whatever sentential example you wish to mention.
And I already corrected your misinterpretation in <this post>.
Quoting TonesInDeepFreeze
I'm glad you finally figured this out and even came up with your own fun way of describing it in English. Now you should go back and reread the original post, using what you have learned about your misinterpretations.
To help you, @Janus' point about natural language is something like this:
Now Sue has contradicted Bob. The question is, "What has Sue claimed?"
Haven't read through all the posts so far. Still, to given one answer:
They'd contradict each other only if they are claimed to occur both at the same time and in the same respect. This irrespective of what "imply" might be taken to precisely specify.
For example, to the question "Is that song any good?" can validly be replied, "Yes and no," without any logical contradiction. Here, yes and no at the same time but in different respects: such that yes, the song is of substantial quality and thereby good and no, the song will be unable to make any revenue and is thereby bad.
Hence, here, song A implies (is taken to hold as a necessary consequence) quality B (here that of "goodness") and song A implies notB - this at the same time, but in different respects, and, hence, in perfectly non-contradictory manners.
Maybe this isn't the best use of the term "implies", but the same principle remains.
It's not a misinterpretation. To say that P is contradictory is to say that P is a contradiction. A statement is contradictory if it is a contradiction. A pair of statements are contradictory if they contradict each other. If you didn't mean that "lizard are purple and not smarter" is contradictory, then you would have done better to not write and not blame a reader for taking what you wrote in its plain meaning. And your other posts doesn't correct anyone.
Quoting Leontiskos
I gave you the courtesy of allowing that what you wrote is not what you meant. I explained previously why the less smart purple lizard sentence is not a contradiction. It's not my fault that what you wrote is not what you meant.
And I did reread your post. And I addressed it again. You skip what I wrote about your remark in the sense of 'contradicting'.
Quoting Leontiskos
You're not "helping" me. But I will help you, as I've been trying to help you from the start.
First, take out 'would' since subjunctives unnecessarily complicate.
So "Supposing A, does B follow?"
Sue claimed "It is not the case that B follows from the supposition A"
As I said, you're not "helping" me. You're just offering me a pointless exercise.
Now I'll ask you a question that is not pointless:
What is your definition of 'contradiction'?
Quoting Leontiskos
I gave two proofs. Your point in directing me to do that turned out to be ill-premised, as it was to say that I had overlooked an alternative notion:
Quoting Leontiskos
That is in accord with the standard definition in modern logic, the definition I gave.
Here's some help for you from the dictionary:
Merriam-Webster - Contradictory
(Adjective): involving, causing, or constituting a contradiction
| contradictory statements
| The witnesses gave contradictory accounts of the accident.
(Noun): a proposition so related to another that if either of the two is true the other is false and if either is false the other must be true
So you're wrong whether we interpret it as an adjective or a noun. You don't seem to have a great grasp of natural language.
Quoting TonesInDeepFreeze
It's like talking to a computer. "Get rid of that natural language, you're confusing our processes!"
Quoting TonesInDeepFreeze
You're still involved in ambiguity. In order to know what Sue denied we must know what Bob affirmed. As noted in my original post, your interpretation will involve Sue in the implausible claims that attend the material logic of ~(A ? B), such as the claim that A is true and B is false. Sue is obviously not claiming that (e.g. that lizards are purple). The negation (and contradictory) of Bob's assertion is not ~(A ? B), it is, "Supposing A, B would not follow."
It seems to me that that is true, and a very important point.
That's no help for me, since I already know that definition. Here's some help for you:
(1) 'constituting a contradiction' is tantamount to 'being a contradiction'. My own point.
(2) implying a contradiction involves either a statement alone being a contraction itself, or with other statments, implying a contradiction, in which case there are at least two statements, each contradicting the other, just as in the examples. But you referred to a contradictory statement, not to some pair of statements. In accord with my own point.
(3) you used the adjective, not the noun, so you can leave out the clause about the noun.
I took what you wrote at face value, and in accord with that dictionary definition.
Oh please, that is a really dumb remark and a lame attempt at putdown. You gratuitously seize on my mere offer to simplify for clarity so that I could better address your question.
Quoting Leontiskos
I'm not ambiguous. The question is ambiguous, given that 'follows' is not defined. Or what sense of 'follows' Bob and Sue are using.
No matter what Bob's sense of 'follows' is, Sue is negating his claim.
The rest of your paragraph boils down to reiterating that the material conditional is not usually operative in such natural language situations. As I've said, now verging on a hundred times, no one disputes that the material conditional does not suit a wide range of natural language senses.
It was a pointless question and exercise. All you had to say is "in everyday conversation, people don't adopt material implication", though that had been agreed upon many posts ago.
And notice that the matter of material implication does not entail that negation is not usually in the sense I've used it. No matter what sense of 'if then', Sue's claim is the negation of Bob's claim. The unpacking of negating an if-then is different depending on the sense of if then but doesn't require adjusting the sense of negation.
This is Kreeft and Dougherty's argument for the superiority of Aristotlean logic for many common uses (evaluating writing, rhetoric, scientific arguments, etc.)
See: https://thephilosophyforum.com/discussion/comment/916812
It seems to me though that the real benefit is that the "three acts of the mind," end up enforcing a sort of realism on the interpretation of syllogisms.
For what its worth, I think you're running into a mismatch between most people's general sense of seeing -> as a strict conditional. Perhaps in your field or life 'material conditional' is a common phrase, but for most people who use logic, this is never introduced. For them, its almost always seen as a strict conditional. Remember that this forum is populated by all types of people, and most of them are not logicians or philosophers themselves. Explaining and contrasting a strict conditional vs a material conditional should make the issue clear for most people.
"most people's general sense of seeing" -- I mean we all have places we come from and thoughts we start at, but if you walk into the chemistry department and start talking alchemy someone might correct you.
This is not a forum exclusive to collegiates, this is a general public forum. I have no issue with being corrected or told new things. While he may have responded well to you, he jumped into a conversation I was having with another poster without context, and when I asked him to clarify his issue he came across as dismissive. I encourage you not to do the same and jump into another conversation between two people.
We did have a conversation earlier though right? You asked my take on the barbershop paradox, and asked my clarification on what I meant by this being a language issue. Did you have any follow up on that?
@Janus' point applies to logic as well. Formal logic is parasitic on natural logic, and "logic" does not mean "formal logic," or some system of formal logic. A lot of folks around here get into trouble because they can't see beyond their own logical system. Tones even mistakes natural language for his own system, and normatively interprets natural language in terms of his system.
And to read Flannel Jesus' posts is to realize that he did not intend the OP in any special sense. I see no evidence that he was specifically speaking about material implication.
Yeah, looking at OP at least, I can see how there's ambiguity there: whether material implication, or some other meaning, was meant isn't specified in the OP and so whatever meaning was meant there's still ambiguity there (which may explain some of the divergence here that I'm surprised to find)
The part where "A" is used as a variable is what made me jump to propositional logic.
Your points about the difference between two versions of contradiction was interesting and I was thinking about it then got sidetracked in reading the back-and-forth.
Quoting Leontiskos
Yeah, we agree there. I think @TonesInDeepFreeze does too, given the various caveats they gave in their posts about different forms of logic.
And again we come back to: as long as the people doing philosophy stipulate definitions they agree :D
I keep thinking about my aversion to "? ~A" ().
The most basic objection is that an argument with two conditional premises should not be able to draw a simple or singular conclusion (because there is no simple claim among the premises).
I had not been following the line in the thread stemming from your claim, but 's proof fortuitously gave me an inroad, and his proof has nothing in particular to do with material implication.
So then we have (p?(q?¬q)), and I think your question immediately arises:
Quoting Count Timothy von Icarus
But I would press further and wonder whether we ever do say things along those lines, in a strict sense:
Quoting Count Timothy von Icarus
This actually runs head-on into the problem that I spelled out <here>. Your consequent is simply not a contradiction in the sense that gave (i.e. the second clear sense of "contradiction" operating in the thread). I don't think (p?(q?¬q)) ever occurs (in reality). This is obviously related to @Janus' critique. I want to say, "Yes, if that proposition were true, then ¬p would follow, but it is never true." Hence Lionino's point, which is elementary but essential:
Quoting Lionino
Which goes back to a central question. How are we interpreting the OP? In my sense or in Moliere's sense?
...I should also note that Tones gave an argument for ~A in which he attempted to prove it directly, without going through Lionino's equivalence proof. This is an acceptable argument by basic logical standards, but I have always had difficulty with argument by supposition. What does it mean to suppose A and then show that ~A follows? This gets into the nature of supposition, how it relates to assertion, and the LEM. It also gets into the difference between a reductio and a proof proper. The point is one I had already made in a post that Tones was responding to, "You think the two propositions logically imply ~A? It seems rather that what they imply is that A cannot be asserted" ().
Still, as I conceded to Lionino, I think his equivalence proof suffices to show that we can draw ~A if the proposition is true. It just seems that it is never true.
Right, and note also the way that Flannel confuses the conditions of a material implication with the principle of explosion beginning <here>.
Quoting Moliere
I gave an example of using "A" without material implication earlier, "Supposing A, would B follow?" ().
Quoting Moliere
I think that is a central point, which Lionino was the first to make explicit on the first page of the thread. It goes back to the uncertainty of the asterisk in my first post, as no one has set out the exact way that the two versions relate. The simple account, which I have set out, is that to contradict is to negate, and what it means to negate depends on one's logical context. Tones was assuming a truth-functional context where negation is the reversal of the truth table.
The original question was, "Do (A implies B) and (A implies notB) contradict each other?"
On natural language they contradict each other.
On the understanding of contradiction that I gave in the first post, they do not contradict each other, and their conjunction is not a contradiction.
On the understanding of contradiction that you gave in the second post, their conjunction is not a contradiction, but their conjunction does contain a contradiction (as showed).
It is that contradiction contained within the conjunction that bubbles up and creates all of the strangeness, and it is worth noting that this contradiction is a direct result of the idiosyncrasies of material implication; they are only logically consistent on account of material implication. It has been some time since I studied formal logic, but I would want to say something along the lines of this, "A proposition containing (p?¬p) is not well formed." Similar to what I said earlier, "When we talk about contradiction there is a cleavage, insofar as it cannot strictly speaking be captured by logic. It is a violation of logic" (). My idea would be that (p?¬p) is outside the domain of the logic at hand, and to try to use the logic at hand to manipulate it results in paradoxes.
But I'm sure others have said this better than I, and the principle of explosion is in fact relevant here insofar as it too relies on the incorporation of a contradiction into the interior logical flow of arguments.
Quoting Lionino
Perhaps. :lol:
((a?b)?(a?¬b))?¬a is valid
My point is that it is a vacuous instance of validity, more clearly seen in the form <((a?(b?¬b))?¬a>. It is formal logic pretending to say something. As I claimed above, there is no actual use case for such a proposition, and I want to say that propositions which contain (b?¬b) are not well formed. They lead to an exaggerated form of the problems that has referenced. We can argue about material implication, but it has its uses. I don't think propositions which contain contradictions have their uses.
This is perhaps a difference over what logic is. Is it the art of reasoning and an aid to thought, or just the manipulation of symbols? I would contend that one reason we know it is not merely the manipulation of symbols is because the rules are not arbitrary, and I am proposing the well-formed-formula rule as yet another non-arbitrary rule. Unless I am wrong and there is some good reason we need to allow for propositions to contain internal contradictions?
I am concerned that logicians too often let the tail wag the dog. The ones I have in mind are good at manipulating symbols, but they have no way of knowing when their logic machine is working and when it is not. They take it on faith that it is always working and they outsource their thinking to it without remainder.
Every time we make an inference on the basis of a contradiction a metabasis eis allo genos occurs (i.e. the sphere of discourse shifts in such a way that the demonstrative validity of the inference is precluded). Usually inferences made on the basis of a contradiction are not made on the basis of a contradiction contained within the interior logical flow of an argument. Or in other words, the metabasis is usually acknowledged to be a metabasis. As an example, when we posit some claim and then show that a contradiction would follow, we treat that contradiction as an outer bound on the logical system. We do not incorporate it into the inferential structure and continue arguing. Hence the fact that it is a special kind of move when we say, Contradiction; Reject the supposition. In a formal sense this move aims to ferret out an inconsistency, but however it is conceived, it ends up going beyond the internal workings of the inferential system (i.e. it is a form of metabasis).
Now suppose we draw out the argument for ¬A:
This is a covert metabasis. It is a metabasis that is not acknowledged to be a metabasis. This has to do with the contradiction, (B?¬B), which is interpreted equivocally as both a proposition and a truth value (false). The difference between a truth value and a proposition is flubbed because what is posited is purely formal, and can never exist in reality (i.e. a contradiction). In order to affirm such a proposition as being true, we must affirm something which could never actually be affirmed, and thus the formal logic here parts ways with reality in a drastic manner. Normal logical propositions do not contain contradictions, and therefore do not require us to do such strange things!
You could also put this a different way and say that while the propositions ((A?(B?¬B)) and (B?¬B) have truth tables, they have no meaning. They are not logically coherent in a way that goes beyond mere symbol manipulation. We have no idea what (B?¬B) could ever be expected to mean. We just think of it, and reify it as, "false" - a kind of falsity incarnate.* Is this then a critique of truth functionality? Maybe, but I want to say that truth functionality can have value where contradictions are not allowed.
* A parallel equivocation occurs here on 'false' and 'absurd' or 'contradictory'. Usually when we say 'false' we mean, "It could be true but it's not." In this case it could never be true. It is the opposite of a tautologyan absurdity or a contradiction.
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Edit:
We can apply Aristotelian syllogistic to diagnose the way that the modus tollens is being applied in the enthymeme:
Viz.:
In this case the middle term is not univocal. It is analogical (i.e. it posses analogical equivocity). Therefore a metabasis is occurring. As I said earlier:
Quoting Leontiskos
Now one could argue for the analogical middle term, but the point is that in this case we are taking modus tollens into new territory. Modus tollens is based on the more restricted sense of 'false', and this alternative sense is a unfamiliar to modus tollens. This is a bit like putting ethanol fuel in your gasoline engine and hoping that it still runs.
Note that the (analogical) equivocity of 'false' flows into the inferential structure, and we could connote this with scare quotes. (B?¬B) is "false" and therefore the conclusion is "implied." The argument is "valid."
That's a cute one. It seems to trade off the ambiguity of translating statements into logic. Obviously when we say someone's testimony is "unreliable" what we mean is that some of their statements are true and some are not true. But there is nothing contradictory about A implying some B are C and A also implying that some B are not C, and so we won't be forced to deny A.
I think one of the challenges inherit with formal logic is that even fairly straightforward arguments of the sort you might find on some science blog or op-ed end up requiring all sorts of stuff to formally model correctly: counterfactuals, modality, temporality, etc. It gets very complex very quickly.
It's even hard with comically bad arguments like the pic below. Premises like "solar panels reflect more heat than grass or trees," and "small changes in ground temperature can cascade into large changes in weather systems, including tornado formation," are all true, but the conclusion that solar farms are "tornado incubators" is still baseless.
B = Vampires are dead.
Not-B = Vampires are living.
As you can clearly judge, this truth table works with Ts straight across the top, since vampires are members of the "living dead." Fools who think logic forces them to affirm ~A are like to end up missing all their blood.
You might be interested in relevance logic, which tries to deal with the paradoxes of material implication: https://plato.stanford.edu/entries/logic-relevance/
There is a section in Tractatus where Wittgenstein declares that belief in a causal nexus is a "superstition" and holds up logical implication as sort of the "real deal." I think this is absolutely backwards. We come up with the idea of logical implication from experience, from the way the world works. This is mistaking an abstraction for reality, and this is ultimately where I think the discomfort with the paradoxes come from.
There is some interesting stuff on modeling relevance logic in terms of information theory I've seen but I forget where I found it.
But I think Kreeft is working with a caricature in the earlier parts, as he has a tendency to do:
He is falling into equivocation between validity and material implication. Modern logic agrees with Aristotelian logic in saying that, "It does not follow from the premises." That a material conditional is true does not mean that the consequent can be drawn, and if Kreeft tries to draw the consequent when the antecedent is false, the modern logician will rightly accuse him of failing to respect the conditions of modus ponens. I think Kreeft is involved in word games here, but in his defense he might say that the modern logician is involved in word games with his word "implies."
Oh, Kreeft knows full well that "material" is contrasted with "formal," and that content is formal, not material. :roll:
:grin:
I agree that material implication has problems, but if you want a tidy, "algorithmic" system, then these sorts of problems are inevitable.
The conservation part was the only reason I remembered it TBH.
Or one that isn't horrifically complex. I actually think that is what gets people more than the "paradoxes of implication." People can learn that sort of thing quite easily. What seems more confusing is the way in which fairly straightforward natural language arguments can end up requiring a dazzling amount of complexity to model.
Thanks, I may check this out in time. My sense, though, is that you can't fully formalize reasoning. In particular Aristotle's final condition for demonstrative knowledge at 71b22 of the Posterior Analytics is something that I think can never be fully formally modeled, "[demonstrative understanding depends on things that are...] explanatory ([i]aitia) of the conclusion[/i]." To understand why something is the way it is requires more than symbol-manipulation, and to understand why B follows from A requires understanding why B is the way it is.
Quoting Count Timothy von Icarus
That sounds like classic Humean backwards-reasoning. :lol:
In that very same paragraph of Posterior Analytics Aristotle points out that in order for an argument to produce knowledge the premises must be better-known than the conclusion, which strikes me as indicative Hume's significant error. Hume takes premises that are very implausible and leverages them to disprove beliefs that are highly plausible, and this is also why his arguments are rightly dismissed even by those who cannot offer a point-for-point refutation.
Quoting Count Timothy von Icarus
Right.
Some interesting points from both of you, so for me not "six pages too many".
Quoting flannel jesus
I woke in the middle of the night and realized there is an alternative interpretation of the above in natural language. I remain convinced that reading the two propositions as "B follows from A" and "B does not follow from A" means that they contradict one another.
However, reading (A implies notB) as "something other than B (caveat: also) follows from A". would be consistent with "B follows from A", because it would not deny that B also follows from A.
That's my kindergarten contribution for what it's worth.
That makes sense to me (even though symbolic logicians must interpret all such things as material implication, as they have no alternative). Related:
Quoting Leontiskos
Quoting Leontiskos
It seems to me that it is key to understand that, "The negation must depend on the sense of the proposition..." Material implication is the way it is for much the same reason that humans are the way they are given Epimetheus' mistake. When the logic gods got around to fashioning material implication they basically said, "Well if the antecedent is true and the consequent is true then obviously the implication is true, and if the antecedent is true and the consequent is false then obviously the implication is false, but what happens in the other cases?" "Shit! We only have 'true' and 'false' to work with! I guess we just call it 'true'...?" "Yeah, we certainly can't call it 'false'."
I haven't thought about this problem in some time, but last time I did I decided that calling the vacuous cases of the material conditional 'true' is like dross. In a tertiary logic perhaps they would be neither true nor false, but in a binary logic they must be either true or false, and given the nature of modus ponens and modus tollens 'true' works much better. It's a bit of a convenient fiction. This is not to say that there aren't inherent problems with trying to cast implication as truth-functional, but it seems to me that an additional problem is the bivalence of the paradigm.
Yet some here have interpreted the OP in such a way that this 'dross' gets repackaged and marketed as gold. The undesirable behavior of material implication is basically supposed to be ignored, not utilized for logic parlor tricks.
Quoting Janus
I think your basic intuition is correct. It resists the crucial methodological error of "trusting the logic machine to the extent that we have no way of knowing when it is working and when it is not" (). We need to be able and willing to question the logic tools that we have built. If the tools do not fit reality, that's a problem with the tools, not with reality ().
Yeah that's a good explanation for why it intuitively makes sense that they're a contradiction.
Consider this as an intuitive explanation for why they aren't a contradiction:
A implies B can be rephrased as (not A or B)
A implies not B can be rephrased as (not A or not B)
Do you think (not A or B) and (not A or not B) contradict?
Heres a possible real-world example (which I think is common knowledge in some circles, though I dont now recall where I picked up the example from):
A = a cat is sleeping outstretched on the threshold of the entry door to a house.
B = the cat is in the house
notB = the cat is not in the house
In this example, how does A not imply both B and notB in equal measure?
It so far seems to me that one can then affirm that, due to A, both B and notB are equally true. Furthermore, the just specified can then be affirmed with the same validity as affirming that A implies both B and notB to be equally false.
This, however, would still not be a contradiction, for while both B and notB occur (else dont occur) at the same time as a necessary consequence of A, they nevertheless both occur (else dont occur) in different respects.
As I so far see things, this addresses the principle of the excluded middle. But the fault would then not be with this principle of itself but, instead, with faulty conceptualizations regarding the collectively exhaustive possibilities in respect to what happens to in fact be the actual state of affairs.
This same type of reasoning can then be further deemed applicable to well enough known statements such as neither is there a self nor is there not a self. This latter proposition would be contradictory only were both the propositions clauses to simultaneously occur in the exact same respect. Otherwise, no contradiction is entailed by the affirmation.
Asking this thinking I (as a novice when it comes to formal modern logics) might have something to learn from any corrections to the just articulated.
Yes, this is similar to 's vampire argument.
Quoting javra
Right: another way of putting it is that you are expressing a phenomenon which is not able to be captured by the logic at hand.
Quoting javra
This reminds me of the reasoning and forms of denial that both Buddhists and Pyrrhonic Skeptics developed, for similar reasons.
Quoting javra
Right: if the terms of the argument are equivocal then the conclusion does not follow, and in this case the conclusion is the claim that a contradiction is occurring.
Quoting javra
It seems accurate to me. I see it as a heavier critique of bivalent logic than the one I was trying to give, but there are a lot of similarities.
Presumably the objection would be something like, "If you hear dogs barking outside and your roommate asks you if the cat is in the house, isn't he asking a question with only two mutually exclusive answers?" Or in other words, I think someone like Wittgenstein might say that the sense of "in the house" should be traced back to the purposes of the speaker, and that where the purposes are unclear the answer will also be unclear. If "in the house" means that the cat is locked inside and safe, then there is no ambiguity. I think there is some merit to this objection, but my guess is that there are nevertheless phenomena which bivalent logic struggles with. There are a lot of ways you could go with this sort of topic, and I am not well versed on polyvalent logic.
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Edit: I sort of forgot to address the way your post interacts with the thing you were quoting. When a symbolic logician writes out (B?¬B), they are not thinking of something like, "Yes and no," where the 'yes' applies to one aspect or consideration of the question and 'no' applies to a different aspect or consideration of the question. They are thinking of a formal contradiction, where something is and is not in precisely the same way. It is this formal contradiction which justifies their modus tollens and the affirmation of ¬A. If they are talking about something like your cat then as you rightly say the inferences that they depend on (and that I oppose) are no longer available for use. In that case I have no objection, for what I was objecting to has disappeared.
Note, though, that this problem does not seem to go away on polyvalent logic. It is still possible to syntactically represent an absurdity like (B?¬B) in more complex logics, such as Buddhist logic. It's just more difficult to do.
The other relevant matter is whether we are thinking about speakers or whether we are thinking about propositions in the abstract. There are strong arguments for the idea that a conclusion requires at least two premises, and although the argument I am taking issue with is arguably an enthymeme (with a hidden premise), there is still a strange way in which the conclusion has only one premise insofar as nothing additional needs to be independently affirmed. This moves quickly away from a speaker-conception idea of propositions and logic, as if arguments could be self-proving, with no middle term. What is at stake, then, is not a logical inference in the psychological sense but rather a formal identity between the truth tables of two propositions. It is a case where the "inference" is based on nothing more than the regrettable idiosyncrasies of the logical system, and that could never be made by those who are not privy to the arbitrary conventions of the system.
The obvious objection to this idea is to note that this restriction goes beyond the typical syntactical requirements for a formula being well formed. I do see this, and it is possible that in rejecting this I am doing irreparable damage to modern symbolic logic. Perhaps my idea is that if someone engages in these sorts of inferences then there should be added an asterisk to their conclusion on account of the fact that this form of metabasis is highly questionable. I mostly want attention to be paid to what we are doing, and to be aware of when we are doing strange things.
Yes, I forgot to give Count Timothy a mention. Thanks for pointing that out. I thought the example of the cat to be more real-world than that of vampires in that it is something that can physically happen in our world, and most likely has. But yes, its the same general issue in respect to logic.
As to polyvalent systems of logic, Im one to find the notion of partial truths to be quite applicable to the world we live in. One of the most immediate and commonplace examples of this can be that of what one is seeing at any moment M. We generally say, I am looking at to implicitly specify what we are willfully focusing on visually. But to specify what we are seeing is a quite different matter; such as, for example, were a judge to ask one what did you see that night? In such a context, it is literally impossible to give a full truth regarding what one saw or else sees: from the issue of needing to describe everything one is aware of occurring within ones peripheral vision to that of needing to describe all the minute details of what one witnesses in ones focal point of vision. A book would be required for this, and even then the telling would still be incomplete. Partial truths, such as that of what we are seeing, could then be contrasted with each other for degrees of fullness; such that what results, as one example among others, is the comparative attributes of some proposition being more true (or truthful), else most true, by comparison to some other proposition (and this without necessitating any falsehoods being expressed). In contrast, I find that falsehoods in the form of lies will always contain some (at least background) truths in order to be in any way believed by those to whom they're told. And in all this I find extensive interest. But maybe all this talk of partial truths is too far afield for the current thread. Although it certainly entwines quite well with most if not all systems of polyvalent logic.
All the same, thank you for the thoughtful reply, the "edit" portion of it included. Yes, my other post regarding "yes and no" was poorly formulated for the context; other such possible answers can include "it is and it isn't" and "they're the same but different" (which I do find interest in as pseudo-contradictions of sorts); but yes, they don't quite address implications.
Do (A entails B) and (A entails notB) contradict each other?
I think issues like the cat are simply mistranslations and over simplifications. The statement should be something like:
The cat is sitting across the threshold of the house, therefore some of the cat is in the house and some of the cat is in the house. This is not contradictory. The same issue is in play here: .
This seems more like a case of "user error." It's like you cannot blame a coding language for a programmer writing code that correctly instructs it to do the wrong thing. The paradoxes of material implication go beyond this sort of misuse though.
If you affirm A there is a straightforward contradiction implied (B and not-B). But the statement as a whole isn't contradictory. Consider the case where A is denied (indeed the statement implies not-A). So, think about mathematical proofs of the sort where we say something like "if A is true then B must be odd and not-odd." A number can't be odd and not odd so A must be false.
I don't find Quoting javra
to be physically impossible. With a quick whimsical online search, found this pic: https://www.instagram.com/atchoumthecat/p/C5ddrTwLzR6/
In this one pic, the 4-month-old cat's tail is partly inside and some of its whiskers are outside, and the threshold is to a sliding door rather than an entry door, but I think it amply illustrates my point: given a sufficiently wide threshold and a sufficiently small cat, A as I described it can in fact obtain. This as a real-world example.
Only if (A entails B) and (A entails notB) occur in the exact same respect (and, obviously, at the same time), which I find is most often the case.
I agree, and if formal logic contradicts the logic inherent in our ordinary ways of speaking and making claims about things, I can't see the fault laying with ordinary parlance.
Quoting flannel jesus
Actually I had thought that it was an explanation for why, on that interpretation, it makes sense that they are not a contradiction, while maintaining that on the other reading it seems to makes no sense to claim that they are not a contradiction.
I read 'A implies B' as (if A then B) or (not (A and notB)). It's a fair while since I studied predicate logic, though.
Quoting javra
Is it not a given that we should understand A and B to refer to the same things in both?
So many things wrong packed into just that one sentence. (1) My post was hardly that long. (2) It's length was a function of the explanation it contains. (2) I don't begrudge posters making posts at any length they want. (3) What is the purpose of mentioning length if not as wedge to discredit? (4) The post carries a lot more message than being a way to "feel better". (5) You did not even address the points I made, but instead you tried to dismiss it with innuendo (if not outright implication) that the post is just a bunch of me trying to fell better.
Quoting Philosophim
(1) I can't dispute that certain people feel certain ways. (2) I can dispute people's representations and characterizations of my posts and my interior, including feelings. (3) How you feel about me does not represent how all others feel about me. (4) And likewise, you can't argue against how you come across to me.
Quoting Philosophim
Hopefully so. And hopefully in this thread. The odds of that happening would be greatly improved by not sending your first post to me to include "Don't be a troll".
Quoting Philosophim
Don't need luck. I've been explaining my thoughts well. I am glad for your agreement.
Quoting Philosophim
(1) '->' is a symbol ordinarily used for material implication; while the strict conditional usually uses a different symbol or is written with '->' and a modal operator.
(2) Strict implication is what might be in mind in certain everyday contexts, but my guess is that relevance is crucial in average everyday contexts. If A is not relevant to B, then I bet most people would just take "if A then B" and "Necessarily if A then B" to be nonsense and, while some people would take it as false on account of being nonsense, others would just say it is plain nonsense.
(2) I have said at least a few times that I quite understand that there are many other notions of 'implies' other than material implication. There is relevance logic, strict implication, and I would bet there are other formal and/or philosophical approaches. And there ar everyday notions that may run from deliberate to not even having a worked out notion of what 'implies', especially to extent of untangling the original thread question (I would guess that you could ask many people walking around the original thread question and their best answer would be "huh?").
Quoting Philosophim
In a philosophy forum, in its 'Logic and Philosophy of Mathematics' section, a question posed that would hardly ever occur in everyday discourse, may be answered however one likes to answer it. Since in logic, the overwhelmingly most common sense is the material conditional, I answered in that context. I do not disallow anyone from answering in other contexts. (I did say that my answer is 'that simple" and regardless "digressions". But I also posted that I do not claim that material implication is the only context that can be countenanced).
Quoting Philosophim
That's fine. No one is stopping you from posting your explanation and contrast. But also, a hearty explanation would include other notions too, especially relevance.
Note that my corrections did not presuppose that only material implication can be countenanced.
Quoting Philosophim
Oh please, everyone enters a thread by "jumping in" in media res. Maybe you noticed that there aren't 'hand waving' icons to click to be recognized like in a video meeting. And there are no "having a conversation with another poster" that doesn't admit of anyone "jumping in" to comment. And "out of context" would mean that my remarks were misleading or not worthwhile on account of them needing to be modified by context. That was not the case with my remarks.
Quoting Philosophim
You didn't just ask me to clarify. You insulted personally with "Don't be a troll", and I did provide you with what you requested.
Quoting Philosophim
I encourage anyone to jump into any conversation, among any posters, as much as they like.
You don't have an exclusive right to be the only one commenting on what other posters say, including what they say to you.
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And as preemptory in case of complaints that I should just "let it go" with you, note that you are publicly making faulting me as a poster, not just as to what I've said on the topic but on a personal basis also. It is quite proper that I defend myself. I don't have to just let you run your posts running me down unanswered.
No. We presume this to be so most of the time for good reason, but it is not a universal given.
Consider: the metaphysical understanding of reality, R, entails both that a) there is a self and b) there is no self.
If the entailment here referred to in regard to both (a) and (b) occurs in the exact same respect (to include relative to the exact same metaphysical or else philosophical perspective concerning the exact same reality/actuality - this even if various levels of reality/actuality were to be implicitly endorsed, e.g. the mundane physical reality of maya and the ultimate reality of literal nonduality), then it would be a flagrant contradiction and thereby a necessarily false proposition. One can of course argue that this is in fact the case, but one cannot thereby establish that this is what the speaker in fact held in mind and thereby intended to express. (Example: the speaker might have intended that in the mundane physical reality of maya there of course occurs the reality of selfhood, this just as much as that in the ultimate reality of literal nonduality no such thing as selfhood is possible - with both these affirmed truths being equally entailed by the very same R at the same time but in the different respects just mentioned.)
Firstly, which metaphysical understanding of reality are you referring to? Those different entailments rely on different interpretations of what is meant by"self' so they are not speaking about the same things.
No, as per my previously given example, they are (or at least can be) speaking about, or else referencing, the exact same thing via the term "self" - but from two different perspectives and, hence, in two different respects (both of these nevertheless occurring at the same time). Again, one perspective being the mundane physical world of maya/illusion/magic-trick and the other being that of the ultimate, or else the only genuine, reality to be had: that of literal nondualistic being. The validity, or lack of, of this Buddhist metaphysics being here inconsequential. As logic goes, here A entails both B and notB at the same time but in different respects/ways.
That is a flat out lie.
I said a number of times that ordinary classical logic (which is not just "my" system, especially since overwhelmingly it is not exclusive to me, and especially since I study and have perused logics other than classical logic).
I said at least three or four times that it is not the case that in all respects formal logic represent everyday reasoning and discourse. I even said explicitly that material implication is not an everyday sense of "if then".
Leontiskos would have seen my posts saying those things, and, if I recall, he even replied to some of them. So either Leontiskos ignored what a wrote or read what I wrote but stated a flat out lie about me anyway.
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Quoting Leontiskos
That's a second flat out lie.
(1) The question in this thread did not specify whether it should be answered as to everyday senses of "if then" or the ordinary sense in the study of the most ordinary logic. And the question was posed in the 'Logic and Philosophy of Mathematics' section of a philsophy forum. So, I (and others) chose to answer as to material conditional. And I have stated explicitly that answers may be different in contexts of other formal logics and in everyday use.
(2) I have never stated a normative claim that classical logic trumps all other approaches.
Moreover:
(3) It is not "my" system, since it is not exclusive to me, and since it is not the only system I study or have perused.
Lenontiskos should not lie about my posts.
It wouldn't be unreasonable to glean that he meant it not just in everyday senses. But, of course, it is open enough that everyday sense may be in play. And just to be clear, he did not indicate that material implication should be excluded.
(1) There is no single everyday sense of "if then" or "implies". For that matter, I bet that in everyday discourse a lot of people would not even make sense of the original thread question.
(2) What senses people have in everyday conversation is an empirical question. It should not be presupposed that any particular sense is even the most common one without some basis. A reasonable conclusion about what people mean would have to take in natural languages world wide, not merely English.
(3) I think that language is not all that's in play, but rather also in play are the notions people have about things. We won't find in a dictionary what a person's response would be to "If snow is green then Winston Churchill was a Confederate general". Yes, it involves a person's meaning of "if then" but also that person's notions of what utterances are sensical, logical, or true.
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(4) In what context is the criteria above supposed to be in? Everyday discourse? An alternative formal logic? Other?
(5) What is the definition of 'simple sentence'? A sentence with only one clause?
(6) Does the criteria pertain only to conditionals? And what a the sentence is written as an equivalent non-conditional?
(7) If not only conditionals, the what of single clause sentences such as "Alex is a dirty rotten scoundrel"? From that single clause sentence I should not infer "Aex is a rotten scoundrel"?
(7) What is the basis for the criteria? Without knowing more about it, I don't know a basis for disallowing the inference of a simple statement from compound statements.
(8) It might be the case throwing out the supposed bath water might be throwing out the baby that is the class of certain valued everyday, mathematical and scientific reasoning.
No, "two different perspectives and, hence, in two different respects" just is two different interpretations of the concept or meaning of 'self'.
It's close enough for purposes of an informal illustration. Obviously, it is implicit in this particular example that 'incompetent' and 'mastermind' are to be regarded as mutually exclusive.
As in the concept/meaning of self as "that which is purple and square" vs. "that which is orange and circular" or any some such? And this in relation to "there both is and is not a self"?
I dont see how your answer can rationally follow. Nor have you put in any effort in justifying your contention via any reasoning or examples. Nor have you evidenced how the examples and reasoning I have repeatedly provided to support my own contention cannot feasibly, rationally, work.
But Ill leave you to it.
Without seeing a definition, I would take 'argument by supposition' to mean arguing from a premise or conditional:
Suppose A. Infer B. Infer If A then B.
Then, with the supposition A, and the inference If A then B, we infer B.
That is ubiquitous everyday reasoning. "Suppose Bob is an orchestra player. Every orchestra player can read music, so Bob in particular can read music. So if Bob is an orchestra player thenBob can read music." Then later, "We established yesterday that if Bob is an orchestra player then can read music, and today we confirmed that Bob is an orchestra player, so Bob can read music.''
Of course, everyday reasoning is not so stilted nor belabored. Such inferences occur virtually instantaneously, but such are the reasoning forms if we or the reasoner were to spell them explicit.
We suppose a proposition A and go through some more reasoning to get B, then we conclude that A implies B. Then we claim A and deploy that A implies B to claim B.
Also, we have two two forms of prove with negation:
(1) Suppose A. Infer a contradiction. Infer ~A
(2) Suppose ~P. Infer a contradiction. Infer A.
Those also are common in everyday reasoning.
What does it mean to do that? I don't know what is meant by "what does it mean to do it" other than the obvious:
There are no circumstances in which a contradiction is true. And if a statement P implies another statement Q, then any circumstances in which the P is true are circumstances in which Q is true. But if Q is a contradiction, then there are no circumstances in which P is true. So, since Q is a contradiction, there are no circumstances in which P is true.
I didn't merely attempt, I proved. And by reductio ad absurdum.
LEM is not needed for my proof.
Anyone is welcome to another context, but if another context is not stated, and since your remarks were related to my proof, I'll suppose that material implication is the context.
First, what is a definition of 'proof proper' in any context?
What is not "proper" about reductio ad absurdum?
How else would one prove a negation if not by reductio ad absurdum, or modus tollens (which is equivalent with reductio ad absurdum) or some other equivalent in either a natural deduction or logical axiom system?
And those are formalizations of everyday reasoning forms: Notice that vacuousness (which probably the main objection to material implication) is not involved
There are two forms:
(1)
Suppose P, infer a contradiction, infer ~P.
Or, suppose that P is true, infer a falsehood, infer that P is false.
That is intuitionistically (thus also classically) acceptable and common everyday reasoning.
"The plumber said he fixed the pipe. But, suppose the pipe was fixed. Then, since there are no other leaks around, we would't see leaking water. So it shouldn't be leaking, but it is leaking. So he didn't fix the pipe."
In everyday life, it may compressed:
"The plumber said he fixed the pipe. But if he fixed the pipe, then, since there no other leaks around, we wouldn't see leaking water. So he didn't fix the pipe."
(2)
Suppose ~P, infer a contradiction, infer P.
Of, suppose that ~P is true, infer a falsehood, infer that P is true. l
That is classical acceptable and common everyday reasoning, but not intuitionistically acceptable.
"Ralph said he didn't the candy bar. But, suppose he didn't eat it, then, since no one else was home, it would be where I left it on the table. But it's not there. So he did eat the peach."
In everyday life, it may compressed:
"Ralph said he didn't the candy bar. But, suppose he didn't eat it, then, since no one else was home, it would be where I left it on the table. But it's not there. So he did eat the peach."
What does "cannot be asserted" mean?
Anything utterable can be asserted. So do you mean "cannot truthfully be asserted"?
If both the propositions can be truthfully asserted then "A is not the case can be truthfully asserted".
Quoting Lionino
((a?b)?(a?¬b))?¬a
The ? operator means that everytime the left side is 1 (True), the right side is also 1, and same for 0 (False). So (a?b)?(a?¬b) does not mean that A is False, unless we say (a?b)?(a?¬b) is True. If (a?b)?(a?¬b) is False, ¬A is False, so A can be True or not¹. So (a?b)?(a?¬b) may be taken as a proposition p which can take the values of 0 or 1 as well.
More confusions stemming from whether 'A' means 'A is true' or a variable which may take the values True or False. It would be better if there were a way to easily distinguish the two. Perhaps a doctor thesis for someone out there. [hide]Or just ditch logic and use normal human language when talking of this stuff, Physics textbooks seem to be doing fine without them, and this thread is evidence of more harm than good.[/hide]
1 We know (a?b)?(a?¬b) is the same as a?(b?¬b), and (b?¬b) is a contradiction. So (a?b)?(a?¬b) just means A implies a contradiction. If (a?b)?(a?¬b) is True, A cannot be True, it has to be False. But let's say (a?b)?(a?¬b) is False, does that mean A is true? That is what the logical tables would say:
Woa, woa, easy on the draw there, pardner.
"On natural language they contradict each other" is pretty categorical.
You can't speak for all speakers of all natural languages. People have all kinds of notions of 'imply' and 'contradiction', ranging from thinking of implication as contexts as at least partially related to material implication and even in some contexts wholly related, through everyday senses of relevance, necessity, common sense in some unarticulated and not explicitly conceived way, to not even being able to make sense of the question. And what is 'everyday'? May it not include professions that concern circuits, models of weather and that kind of thing?
You don't have public surveys that would show how many people would regard those as contradictory, or nonsense, or to account for some who would be evidence against your universal claim.
Moreover, it's not just a question of language but of the kind of reasoning people recognize as correct or incorrect in everyday situations. The English language, for example, doesn't imply what people consider good reasoning. You can't look in a dictionary and grammar book to find out whether the propositions are to be regarded as contradictory, not even a description of colloquial use.
I answered that. I don't recall whether you addressed my answer.
But (a?b)?(a?¬b) being False simply means that A does not imply a contradiction, it should not mean A is True automatically.
Edit: I underestimated your post. Like your first post, this is far above and beyond anything else that is occurring in this thread.
I'd like to see what formation rules you come up with.
Quoting Leontiskos
I asked what you mean by 'cleavage' and 'capture by logic'. I don't recall that you replied.
Whatever 'capture' means, the meaning of [P & ~P' is specified.
What does "violation of logic" mean that entails that 'P & ~P' is well formed. It is well understood that 'P & ~P' is a violation of the law of non-contradiction. But it is not a violation of syntax to write 'P & ~P' nor "P is true and P is false". Indeed, a statement of the law of non-contradiction would itself involve stating a contradiction in order to deny it.
My idea would be that (p?¬p) is outside the domain of the logic at hand/quote]
(1) Which logic at hand?
(2) Even in everyday contexts, people say things like "You said Jack is to be trusted and you said he's not to be trusted. Which is it?" "You said to turn right at Elm and you said not to turn right at Elm. Would you please give me directions that don't contradict?'
Quoting Leontiskos
Did you mean "try not to"?
Anyway, I don't see the paradoxes arising from undo use of negation. Would you give an example?
Quoting Leontiskos
I would be interested to know who.
There are systems of negationless logic, but that is not what you have mind since you don't want to toss even negation. There might be systems that have negation but do not allow writing 'P & ~P', but I don't know of any.
And is your idea only for everyday reasoning, or do you want wipe out being able to state a contradiction in formal ogic, informal academic logic and rhetoric, mathematics, science, philosophy and other acdademic fields?
Moreover, even though formal logic and everyday reasoning often converge, wouldn't you want to allow formal reasoning to be expressed in everyday terms?
If I write a formal argument deriving a contraction from P, shouldn't I be taken as making sense when I instantiate it with natural language?
And what about an inconsistent statement that is not in the form of 'P & ~P'? I certain sentential logic and monadic predicate logic, it is checkable whether a formula is inconsistent, but not in dyadic or greater predicate logic. But, for excellent reasons, formation rules for formal language are checkable.
Does 'interior logic flow of arguments' just mean 'proof steps'?
It is not "pretending" anything. It has a precise meaning.
"The premise that Tom reneged on his library fines leads to a contradiction, therefore, Tom did not renege on his library fines".
I'd rather have a lawyer who understands contradiction than one who follows only Lenontiskos logic.
Yep. Worth noting that parsing this correctly shows that the original was incomplete - implied nothing.
More generally, parsing natural languages in formal languages, while not definitive, does occasionally provide such clarification. That's kinda why we do it.
Also worth noting that (A ? B) ^ (A ? ¬B), while not a contradiction, does imply one, given A:
(A ? B) ? (A ? ¬B)?(A?(B?¬B))
So in answer to the OP
Quoting flannel jesus
Taking "implies" as material implication, they are not contradictory but show that A implies a contradiction.
Quoting TonesInDeepFreeze
I had the same thought when I read that. It's wellformed. It is also invalid: A?¬A
This thread is bringing out some rather odd attitudes towards the relation between logic and natural languages.
Quoting Lionino
Isn't this a fairly big problem given that (¬¬A?A)? I take it that this is the same thing I have pointed out coming out in a different way? Namely the quasi-equivocation on falsity?
...Or is it simpler: ¬(a?(b?¬b)) ? a
Back to material implication:
Quoting Leontiskos
Given material implication there is no way to deny a conditional without affirming the antecedent, just as there is no way to deny the antecedent without affirming the conditional.
Symbolic logic definitely does not hold that that Lassie argument is valid.
That claim in the article is either sneaky sophistry or egregious ignorance. It is a ludicrous claim. It goes dramatically against what obtains in symbolic logic.
Let:
'Fx' stand for 'x has 4 legs'
's' stand for 'Lassie'
'Dx' stand for 'x is a dog'
The argument is:
Ax(Dx -> Fx)
Fs
Therefore Dx
[EDIT: There's a typo there. It should be:
Ax(Dx -> Fx)
Fs
Therefore Ds]
Symbolic not only does not say that that is valid, and not only does symbolic logic say it is invalid, but symbolic logic proves it is invalid.
Here is where the authors try to pull a fast one:
Correct: A valid formula is implied by any set of formulas.
Correct: If P is true, then, for any formula Q, we have that Q -> P is true.
Incorrect: If P is true, then Q -> P is valid.
Look what the authors did:
By saying "'Lassie is a dog; is true", they are adopting Dx [edit typo: should be Ds] as a premise. So, of course,
Ax(Dx -> Fx)
Fs
Dx
Therefore Dx
is valid.
[EDIT: There's a typo there. It should be:
Ax(Dx -> Fx)
Fs
Ds
Therefore Ds]
Or I invite the authors to show any symbolic logic system for ordinary predicate logic that provides a derivation of:
Ax(Dx -> Fx)
Fs
Therefore Dx
[EDIT: There's a typo there. It should be:
Ax(Dx -> Fx)
Fs
Therefore Ds]
Moreover, we prove that classical logic provides that its proof method ensures that the the premises indeed entail the conclusion. That is, if the conclusion is not entailed by the premises, then the conclusion is not proved by the premises. And that goes for both true and false conclusions. If the truth that Lassie is a dog is not entailed by the premises, then 'Lassie is a dog' is not provable from the premises.
That's a disgustingly specious and disinformational start of an article. And unfortunate that that speciousness and disinformation is propagated by another poster quoting it here.
But this is good:
Logicians have an answer to the above charge, and the answer is perfectly tight and logically consistent. That is part of the problem! Consistency is not enough.[/quote]
Indeed, that is why we prove both consistency and soundness.
So a proper answer is that the logic is consistent, sound and extaordinarily useful for many contexts. And it is useful even in everyday context where vacuousness doesn't even come up.
Quoting Leontiskos
The proof does not use vacuousness.
Quoting Leontiskos
(1) I'd like to see you forrmulate a grammar, formal or informal, that restricts to only locutions that "have use".
(2) Even though the form mentioned is not found in everyday discourse, it is equivalent with very basic everyday reasoning
Quoting Leontiskos
Not well formed in everyday language? Not well formed in formal languages?
And I would like to see your formation rules for either.
Quoting Leontiskos
Being able to write a contradiction is useful, as I explained.
Quoting Leontiskos
And the study of formal logic is not just a study of manipulating symbols.
Quoting Leontiskos
I take it that the machine Leontiskos was using when he wrote the paragraph above is working. That's a machine whose invention and development is steeped in formal logic, steeped in the use of sentential logic, steeped in syntax that has material implication, disjunction, conjunction and negation, steeped in truth functionality, steeped in 2-value Boolean algebra.
Whatever you call 'meta-logic, the subject of meta-logic is discussed plenty on this forum, by me and others.
The method of models formalizes the idea that a statement "is true", "could be true, but it's not" or "is false" or "could be false, but it's not".
People use terminology differently, but we have quite unequivocal terminology:
a sentence S is false in model M if and only if [fill in the the rigorous definition].
a sentence S is a contradiction if and only if S is of the form 'P & ~P'
a sentence S is inconsistent if and only if S proves a contradiction.
a set of sentences G is inconsistent if and only if G implies a contradiction.
"absurd" doesn't usually get a formal definition, but its use is usually along the lines such that:
"S is absurd" is equivalent with "S implies a contradiction".
Quoting Leontiskos
That is addressed also:
S is logically true if and only if S is true in all models.
S is logically false if and only if S is false in all models.
S is contingently true if and only if S is true in some models and false in other models.
I don't know what any of that means or a bunch of other stuff in a similar vein.
Quoting Leontiskos
Quoting Leontiskos
I understand them just fine, and not as mere symbols.
Quoting Leontiskos
You have it quite wrong.
No, the conditional with the negation of the consequent prove the negation of the antecedent.
A ... antecedent
B & ~B ... consequent
~(B & ~B) ... negation of the consequent
~A ... negation of the antecedent
Quoting Leontiskos
If that and bunch more in your post is not egregious doubletalk then I stand having it explained that it's not.
There's so much more in your post. Time is not enough.
That's just a matter of defining the words. If 'dead' and 'living' are defined so that they are not mutually exclusive, then of course we don't make the inference. It's silly to claim that sentential logic is impugned with the example.
Worse, his argument about Lassie and symbolic logic is specious, dishonest or stupid, and ludicrous.
A formal problem, philosophical problem, or problem in not adhering to everyday discourse.
It's not a formal problem. I don't know what philosophical problem you suggest. And I've discussed symbolic logic and everyday discourse.
Compare:
With:
Whether or not we affirm the negation of the consequent, the antecedent still ends up being false. In the first case the consequent is pre-false, in the second case our explicit negation makes it false. In the first case it is treated as "falsity incarnate," whereas in the second it is treated as a proposition. This demonstrates the analogical equivocity with respect to falsity that I pointed out earlier, with help from . I think Lionino was somehow seeing this through the overdetermination of the biconditional with respect to ¬¬A, but it's slippery to grab hold of.
Note that we could also do other things, such as treat the second premise as truth incarnate, but this is harder to see:
...that is, if we conceive of the consequent as a proposition and the second premise as truth incarnate, then ¬A does not follow from the second premise (or from the consequent, absent a premise that negates the consequent qua proposition).
-
Quoting Banno
And about two thirds of folk hereabouts who are good at manipulating symbols "have no way of knowing when their logic machine is working and when it is not" ().
Aristotle always wins in the long run, as he could both use the machine and understand when it was misbehaving. :wink:
This makes no sense to me.
Quoting javra
The notion of a self from the perspective of "the mundane physical world of maya/illusion/magic-trick" is not the same as the notion of a self from the perspective of "literal non-dualistic being", so you are not talking about one thing.
That said, the self has no definitive definition, so introducing such a thing in the context of discussing whether anything could be the same in different contexts or thought under different perspectives seems incoherent from the get-go.
Consider the following substitutions which do not suffer from such ambiguities: Render (A implies B) as "the presence of water implies the presences of oxygen" and (A implies notB) as " the presence of water implies the absence of oxygen": do the two statements not contradict one another?
What is the definition 'analogical equivocity'? Most helpfully it would be something like:
An thing has analogical equivoicity if and only if [fill in definiens here, and such that the definiens doesn't use terminology that hasn't itself been defined or common anyway]
It is the kind of equivocity present in analogical predication, where a middle term is not univocal (i.e. it is strictly speaking equivocal) but there is an analogical relation between the different senses. This is the basis for the most straightforward kind of metabasis eis allo genos. The two different senses of falsity alluded to above are an example of two senses with an analogical relation. I was using the term earlier because I believe @Count Timothy von Icarus has an understanding of it.
I know you want a strict definition, but the wonderful irony is that someone like yourself who requires the sort of precision reminiscent of truth-functional logic can't understand analogical equivocity or the subtle problems that attend your argument for ¬A. As we have seen in the thread, those who require such "precision" tend to have a distaste for natural language itself.
Anyway, a implies B and (not a or b) are synonyms in classic symbolic logic. They have the same truth table.
Yes, that's right in there. I'm not sure how they are "slipping" it in. That's exactly the point they are making and trying to highlight. You seem to be missing the point of the example, yet you've hit on it right here.
I think you and may be supposing that the example is being called in to show something it is not being called in to show. The point being made might be more obvious in context, IIRC there is a section on how the old logic presupposes a sort of epistemic realism right before it, and to "common sense" form without realism often seems silly.
...when it comes to vamps it's deadly serious. :death: :death: :death:
Yes, I am waiting for someone to clarify what is up with that.
I had written more under my own post but got myself confused also on whether the letter here is a statement or a variable.
Quoting Leontiskos
Yes. It is what I say here.
Quoting Lionino
But, as my second phrase in this post, I got confused as to whether ¬(a?(b?¬b)) means a variable that is in contradiction with (a?(b?¬b)) or simply that (a?(b?¬b)) is False.
¬(a?(b?¬b)) is only ever True (meaning A does not imply a contradiction) when A is True. But I think it might be we are putting the horse before the cart. It is not that ¬(a?(b?¬b)) being True makes A True, but that, due to the definition of material implication, ¬(a?(b?¬b)) can only be True if A is true. Likewise, what I say here in the first post:
Quoting Lionino
The denial of the implication can only be True if the antecedent is True. If the antecedent is False, the denial of the implication is False.
@TonesInDeepFreeze
Let a proposition P be A?(B?¬B)
Whenever P is 1, A is 0.
In natural language, we might say: when it is true that A implies a contradiction, we know A is false.
Now a proposition Q: ¬(A?(B?¬B))
Whenever Q is 1, A is 1.
Do you think it is correct to translate this as: when it is not true that A implies a contradiction, we know A is true?
Here is something I wrote when my internet was out, and before I was able to read this post of yours. I will read what you have written after posting this (and therefore there may be overlaps or redundancies):
Quoting Lionino
Note, too, how the LEM operates here.
On the LEM, (a?b)?(a?¬b) must be either true or false, and therefore a?(b?¬b) must be either true or false. If a?(b?¬b) is true then ¬a must be true and (b?¬b) must be indeterminate. It must be this way in order to avoid affirming the contradiction (and never mind the odd question of whether, in this case, saying ¬a must be true is the same as saying a must be false).
But on the other side of the LEM, if a?(b?¬b) is false then by material implication (a) must be true and (b?¬b) must be false. This is different from (b?¬b) being indeterminate. It is to affirm ¬(b?¬b), which is the same as (¬b?b) with an inclusive-or. The inclusive-or provides the second-order possibility of a true contradiction.
Another way this comes out is seen by comparing the truth table of (p?q) with the truth table of a?(b?¬b). If you run the table forward, by using the value of (b) to compute the value of (b?¬b), then (b?¬b) is always necessarily false. But if you run the table backwards, by using the value of (p?q) to compute the values of (p) and (q), then in that singular case where the entirety of the conditional is false we get (a ^ ¬(b?¬b)), which again results in the inclusive-or (¬b?b).
In one way, in saying that the inclusive-or is true we are admitting the possibility that both disjuncts could be true, which would be impossible. Put differently, negating this conjunction results in an exclusive-or rather than the usual inclusive-or, just as affirming the special conditional of (a?(b?¬b)) results in (a) automatically being false, even though there is obviously no rule that ((p?q)?¬p). Special cases and special exceptions are popping up.
What is happening, I think, is that the two different kinds of falsity are coming into play, for there are two different ways of calling (b?¬b) false. We can mean that it is false inherently, even without a negation (and this is reflective of the first argument I gave <here>).* Or else by saying it is false we can mean that we logically negate it, viz. ¬(b?¬b) (and this is reflective of the second argument I gave <here>). When are we supposed to reduce a contradiction to its functional truth value, and when are we supposed to let it remain in its proposition form? The mind can conceive of [s]propositions[/s] symbol-strings like (b?¬b) and ¬(b?¬b) in these two different ways, the manner in which we conceive of them results in different logical outcomes, and symbolic logic provides no way of adjudicating how the [s]propositions[/s] symbol-strings are to be conceived in any one situation. When we introduce contradictions into the logiceven in minor ways such as in the consequents of conditionalsthe logic becomes underdetermined. Different people are legitimately interpreting the [s]propositions[/s] symbol-strings in different ways.
* This is reflective of what I have been calling falsity incarnate
I am exactly on point. The paper says that symbolic logic permits a certain inference. But symbolic logic does not permit that inference. The paper's argument is specious exactly as I showed.
Quoting Count Timothy von Icarus
Death is serious. The example is silly, for the reason you quoted.
Quoting Leontiskos
A rigorous definition is best, for obvious reasons. And giving one is intellectual considerateness. But, again you put words in my mouth, thus a strawman. I said a certain form would be helpful. I did not say that I "require" it or any specific degree of rigor let alone "precision".
And I don't have "distaste" for language. You're just making that up. I love language and revel in its expressiveness, freedom, complexities, nuances and even its vagaries. You don't know jack when you presume to mischaracterize my sensibilities.
Your method is to characterize me in a certain way, to wedge in a prejudice that my arguments are tainted by a point of view, even though I don't have that point of view. You are arguing by characterization of an interlocutor, actually characterization.
You lied that I don't distinguish between formal language and natural language. I had said at least a few times that they don't agree in certain respects. Depending on the conversation, they don't agree on "if then".
You would do well to knock it off with trying to impugn me with lies about what I've said a sloppy mischaracterizations of my point of view and sensibilities.
No, you aren't, your post clearly demonstrated the point went right over your head. I even quoted you on the exact point where you say the same thing they are pointing out is counterintuitive. The fact that you think this is "slipping" something in or a "gotcha," demonstrates that you have missed the intent.
This is counterintuitive. That's what the example is there to illustrate; it's right their in the text. Then you go off "challenging" them to show something they've never asserted.
The intent in that part of the paper was to make symbolic logic look like it says that this argument is valid:
Dogs have four legs, and Lassie has four legs, therefore Lassie is a dog.
But symbolic logic does not say that argument is valid.
Ax(Dx -> Fx)
Fs
Therefore, Ds
is not valid. And the paper is correct in saying it is not valid. But the paper is incorrect in saying that symbolic logic says it is valid.
In other words:
All dogs have four legs.
Lassie has four legs.
Therefore, Lassie is a dog.
is not valid. And the paper is correct is saying it is not valid. But the paper is incorrect in saying that symbolic logic says it is valid.
The paper says that symbolic logic says it is valid on the basis that "Lassie is a dog" is true.
That is not correct, since if we are using "Lassie is a dog", then it has to be a premise in the argument, so the argument that the paper puts on symbolic logic would actually be:
Ax(Dx -> Fx)
Fs
Ds
Therefore Ds
which is valid.
/
If you think the paper is correct in saying that symbolic logic regards this as valid
Ax(Dx -> Fx)
Fs
Therefore, Ds
In other words:
All dogs have four legs.
Lassie has four legs
Therefore, Lassie is a dog.
then you may prove that symbolic logic says it's valid. That is, show a derivation in symbolic logic that starts with the premises
Ax(Dx -> Fx)
Fs
and ends with the conclusion
Ds
Hint: You won't be able to do it. Morever, using the method of counter-example, symbolic logic proves it is not valid.
You sure about that one?
No, that was also a typo. I corrected it all in the post as it stands now.
¬(A ? (B ? ¬B))?A is valid
¬(A ? (B ? ¬B))?A is valid
I'm recomposing.
Sure, but that's not really what the example is there to assert, as is clear from the rest of the paragraph. They mentioned replacing the fact about dogs 2+2 = 4 in the next line. It's "if a statement is true, then that statement is implied by any statement whatever," which is straightforwardly counter intuitive.
(2) ~(A?(B &~B)) implies A
"when it is not true that A implies a contradiction, we know A is true?"
No that is not a correct translation of (2). The translation universally quantifies over contradictions, which (2) does not.
The example is incorrect, no matter what its purpose is.
Quoting Count Timothy von Icarus
I would not say that.
P may be true in some interpretations and not in others. If in a given interpretation, P is true, then per that interpretation, for any statement Q, Q->P is true.
Translation:
It is not the case that if A then B & ~B
implies
A.
We can't say:
For all contradictions, if A does not imply the contradiction then A is true.
(1) True in what interpretations?
(2) "Winston Churchill was French" does not imply a contradiction. But that does not imply that "Winston Churchill was French" is true.
More precisely, "Winston Churchill was French" does not imply a contradiction. But that does not imply that "Winston Churchill was French" is true in all interpretations, but it does imply that "Winston Churchill was French" is true in at least one interpretation.
Is this supposed to be the translation of ~(A?(B &~B)) implies A?
I then edited it with better demarcations. Said another way:
"It is not the case that if A then both B and not-B" implies "A".
Isn't this the same as "if A then contradiction"?
I gave a direct translation, symbol by symbol to word by word.
The formula has a subformula that is a contradiction, but the formula doesn't itself say that it is a contradiction. The word 'contradiction' is not in the formula, and the formula doesn't quantify over contradictions. Again:
The formula does not say:
For all contradictions, if A does not imply the contradiction then A is true.
"Winston Churchill was French" does not imply a contradiction. But that does not imply that "Winston Churchill was French" is true.
More precisely, "Winston Churchill was French" does not imply a contradiction. But that does not imply that "Winston Churchill was French" is true in all interpretations, but it does imply that "Winston Churchill was French" is true in at least one interpretation.
With symbols:
If P does not imply a contradiction, then it is not implied that P is true in all interpretations, but it is implied that P is true in at least on interpretation.
No, your reading of it is incorrect because you seem to think it is saying:
All dogs have four legs
Lassie has four legs
Lassie is a dog
...is valid in symbolic logic. It doesn't say that. It says the exact opposite, that this is not valid.
What it turns to point out in the proceeding paragraph is different:
i.e., as you put it: if "P is true, then per that interpretation, for any statement Q, Q->P is true."
Perhaps it could be written clearer, but the point is not what you seem to think it is, which would be a silly mistake to make, but it's simply an example to draw out one variety of paradox of material implication.
Read it again. The paper says it is invalid, but that symbolic logic "disagrees".
Again, the paper is correct that it is invalid, but the paper is incorrect that symbolic logic disagrees.
Quoting TonesInDeepFreeze
That's perfectly clear.
Yes, disagrees that "a true conclusion logically follows from, or is proved by, or is "implied" by, or is validly inferred from, only some premises and not others."
This is perhaps ambiguous, but it's clarified in the following sentences.
It's not ambiguous. It's as plain as day, with a plain reading:
"Its premises do not prove its conclusion."
'it' refers to the Lassie argument.
"modern symbolic logic disagrees."
That is, modern symbolic disagrees that the Lassie argument's premises do not prove the conclusion.
Does that mean modern symbolic logic thinks the premises do prove the conclusion?
That is what the paper says. The paper is incorrect.
Not just incorrect, but incorrect due to egregious sophistry, ignorance or blatant lack of reasoning skills.
There's a paper that says the premises prove the conclusion of this argument?
Dogs have four legs, and Lassie has four legs, therefore Lassie is a dog
I want to see this paper, could you link me?
No, the paper says the argument is invalid, but that symbolic logic says it's valid.
The paper is a polemic against symbolic logic, and it argues (egregiously incorrectly) that symbolic logic regards the argument as valid because symbolic logic says that any true statement (such as "Lassie is a dog") is implied by any statements. In my first post about it, I explained exactly where the sophistry occurs. (There are some typos in my post, but I made edit notes to correct them.)
It's quoted in a post earlier in this thread.
There are many many posts in this thread. I don't have any means of efficiently searching for it, so that's why I'm asking you. If you would prefer not to link me up for whatever reason, I suppose I'll just have to accept that.
It's clearly discussing paradoxes of material implication, not arguing that "All A are B and C is B implies C is A," is valid anywhere. In fact it says it isn't valid tout court up above. If that was the point, it could have been stated much clearer and then everything following amounts to a giant non-sequitur. But it isn't a non-sequitur, it's the point of the entire section.
If they wanted to make the point you ascribe to them why wouldn't they use an example like:
All monkeys have tails.
Garfield the cat has a tail.
Therefore Garfield is a monkey.
It would show the absurdity much clearer. But the point is that "has four legs" doesn't imply "is a dog," in that argument, and yet Q ? P is always true when P is true, which is incongruous with how Q does not imply P in the other argument. I agree that it could be written clearer, but I think it's pretty uncharitable and ignoring all the context to assume they are claiming what you say they are.
Since there were so many typos in my reply, here it is corrected:
Symbolic logic definitely does not hold that that Lassie argument is valid.
That claim in the article is either sneaky sophistry or egregious ignorance. It is a ludicrous claim. It goes dramatically against what obtains in symbolic logic.
Let:
'Fx' stand for 'x has 4 legs'
's' stand for 'Lassie'
'Dx' stand for 'x is a dog'
The argument is:
Ax(Dx -> Fx)
Fs
Therefore Ds
Symbolic not only does not say that that is valid, and not only does symbolic logic say it is invalid, but symbolic logic proves it is invalid.
Here is where the authors try to pull a fast one:
Correct: A valid formula is implied by any set of formulas.
Correct: If P is true, then, for any formula Q, we have that Q -> P is true.
Incorrect: If P is true, then Q -> P is valid.
Look what the authors did:
By saying "'Lassie is a dog" is true", they are adopting Ds as a premise. So, of course,
Ax(Dx -> Fx)
Fs
Ds
Therefore Ds
Or I invite the authors to show any symbolic logic system for ordinary predicate logic that provides a derivation of:
Ax(Dx -> Fx)
Fs
Therefore Ds
Moreover, we prove that classical logic provides that its proof method ensures that the premises indeed entail the conclusion. That is, if the conclusion is not entailed by the premises, then the conclusion is not proved by the premises. And that goes for both true and false conclusions. If the truth that Lassie is a dog is not entailed by the premises, then 'Lassie is a dog' is not provable from the premises.
That's a disgustingly specious and disinformational start of an article. And unfortunate that that speciousness and disinformation is propagated by another poster quoting it here.
Again, look at the exact words in the paper:
"Its premises do not prove its conclusion."
'it' refers to the Lassie argument.
"modern symbolic logic disagrees."
That is, modern symbolic disagrees that the Lassie argument's premises do not prove the conclusion.
That is plain as day. All you need to do is read the exact words. It is amazing that you won't recognize it.
Discussing other aspects of the paper does not change that the paper claims that symbolic logic regards the Lassie argument is valid. Indeed, the paper goes on to give its incorrect explanation of why symbolic logic regards the Lassie argument to be valid.
I somehow find it more plausible that they were trying to highlight the incongruity between the fact "Lassie has four legs" does not imply Lassie is a dog in symbolic logic in the argument:
All dogs have four legs
Lassie has four legs
Therefore Lassie is a dog
And the fact that "Lassie has four legs" does imply Lassie is a dog if "Lassie is a dog" is true. That's the straightforward purpose given the context, particularly since the text is not particularly hostile towards symbolic logic aside from arguing that it isn't particularly helpful for most people's use cases.
This, rather than assuming they are trying to imply an falsehood to cast shade on symbolic logic in an extremely roundabout way using an example obfuscates their point (if that was the point they were making)doing all this to try to suggest something that is easily verifiable as false for ... what purpose?
IDK, maybe I am letting the principle of charity run amok.
"Its premises do not prove its conclusion," because they they do not stand in the right sort of relation to one another (this is the topic of the paragraph). This is true in a specific sense in Aristotlean logic, which denies as valid arguments in which the premises are inconsistent, arguments with conclusions that would follow from any premises whatsoever, or arguments with superfluous premises. Symbolic logic does indeed disagree with this. I assume this is what "it" refers to because that's what all the context suggests.
It correctly says it is invalid, but incorrectly says that symbolic logic "disagrees". It's right there in the paper. It is amazing that you ignore that plain fact.
Quoting Count Timothy von Icarus
It was stated in perfect clarity that symbolic logic "disagrees" that the premises don't prove the conclusion.
Quoting Count Timothy von Icarus
Because their point in that paragraph was that "Lassie is a dog" is true but not provable from the premises but that symbolic logic disagrees that is not provable from the premises.
Read the paragraph slowly, line by line, and you will see:
"Its premises do not prove its conclusion."
"modern symbolic logic disagrees."
Whatever you think was meant to be highlighted, or plausible, or your interpretation, the plain fact is that the paper says that symbolic logic regards the argument as valid.
That's enough right there. It's incontrovertible.
But, going on:
The reason they make that claim is to make a strawman against symbolic logic. It's a strawman because symbolic logic does not say the argument is valid. Then they incorrectly argue that the reason symbolic logic says the argument is valid is because "Lassie is a dog" is true and symbolic logic says that any true sentence is proven by any sentences.
Quoting Count Timothy von Icarus
Whatever you think the purpose is, it is utterly straightforward that the paper claims that symbolic regards the argument as valid. They incorrectly base that claim on the claim that symbolic logic regards the argument as valid because "Lassie is a dog" is true and symbolic logic says that any true sentence is proven by any sentences.
Quoting Count Timothy von Icarus
The article uses the Lassie example to argue that symbolic logic departs from common sense logic. But that is a specious argument, since symbolic logic is right with common sense regarding the Lassie argument.
Quoting Count Timothy von Icarus
(1) You are obfuscating. Your argument is that they couldn't have meant what they exactly wrote because of something else. It is in broad daylight that they claimed that symbolic logic disagrees that the argument is invalid. And what they say right after is premised on that. Again: They make the claim, then argue that the reason symbolic logic regards the argument as valid is that symbolic logic says "true proven by anything" (by the way, that is not what symbolic logic says, so another false claim in the paper).
(2) The example and argument is not roundabout. And it does not obfuscate their point. It makes their point blazingly clear as they themselves state it right after the example. They try to make symbolic look ridiculous for saying the argument is valid, then incorrectly say what is at root in symbolic logic that allows symbolic logic to say the argument is valid.
(3) The purpose is to present an argument about symbolic logic. I don't why they resorted to egregious sophistry to do that. Why does anyone resort to sophistry? Possibilities include: (a) They think they can get away with it, (b) They got caught up in themselves thinking they had a clever argument, (c) They don't understand symbolic logic ...
Quoting Count Timothy von Icarus
What is that supposed to mean?
Anyway, you continue to evade exactly what they wrote.
No, not tout court. There's an 'however'. They correctly say it is invalid but they also incorrectly say that, however, symbolic logic disagrees.
It's text in broad daylight.
Of course.
I don't know what to make of your use of "interpretation".
For sentential logic, an interpretation assigns to each sentence letter a value True of value False.
Most commonly this is represented as columns in a truth table.
This is ordinary semantics for sentential logic.
/
I hope my explanation regarding my answer to your question was satisfactory to you. What was your purpose in asking the question?
In other words, the first question addresses this very affirmation which you more recently added: Quoting Janus
While the second question I asked addresses this:
The very proposition of "there both a) is a self and b) is no self" has (a) and (b) addressing the exact same thing - irrespective of how the term "self" might be defined or understood as a concept, the exact same identity is addressed The proposition has nothing whatsoever to do with two different interpretations of what "self" means and has everything to do with "the self" both occurring and not occurring at the same time.
So your critique completely misses the issue addressed regarding contradictions and the possible lack of such in the proposition "(the understanding of reality R entails that) there both is and is not a self". "Self" remains identical, but the usage of "is" can in principle be interpreted in two different ways - this, at least, within the context of certain Indian philosophies - thereby potentially equating to the more verbose proposition that "selfhood occurs from the vantage of mundane reality which is illusory but selfhood in no way occurs from the vantage of ultimate/genuine/non-illusory/nondual reality - for there can be no selfhood in the the complete absence of any duality with other - and both these actualities (in semi-Kantian terms, that actuality of the phenomenal world of maya and that nondual actuality/reality which can only be utterly non-phenomanal and, hence, purely noumenal) co-occur at the same time (again, this within certain Indian philosophies)". And such an affirmation would not be logically contradictory.
------
Quoting Janus
Of course they do. Just as much as saying that "the car C is completely green and completely red at the same time and in the same respect* ". And this as I upheld in the post you initially replied to: again, most often, the proposition that A entails both B and notB will be logically contradictory.
* p.s., to be clearer, this with the understanding that any shade of brown is neither green nor red, but brown.
I haven't followed your posts, so there may be context I need. But at least at face value:
"the presence of water implies the presences of oxygen"
is not an "if then" statement, since 'the presence of water' and 'the presence of oxygen' are noun phrases, not propositions.
I haven't followed your posts, so there may be context I need. That said, (1) What is your context? Classical logic? Some other formal logic? Notions in everyday reasoning? (2) By 'logically contradictory' do you mean the proposition implies a proposition of the form P and ~P? Or something else?
The principle/law of thought which sometimes goes by the name of "the law of noncontradiction", this as articulated by Aristotle (with possible ambiguities as to whether the law applies only to epistemology or also to ontology at large here acknowledged).
What is your statement of the principle?
In short, A and not-A cannot both occur at the same time and in the same respect.
Would allow simplifying that to:
For any statement A, it is not the case that both A and not-A.
Is that a question or an affirmation?
Consider: The statement, "Water can be green", does not allow for "water can be green" and "water can be not-green (e.g., blue)".
So what I said does not simplify into what you've written.
If I understand, you take
It is not the case that both water can be green and water can be not-green.
as an instance of the law of contradiction. (?)
Do you also take
It is not the case that both water can be green and water cannot be green
as an instance of the law of contradiction?
No, the exact opposite. Hence the concluding sentence to you in my last post.
However, the statement "water can be green and water can be non-green (e.g., blue) at the same time and in the same respect" will be an instance of the law of noncontradiction.
I don't find that one can properly express the principle linguistically without "at the same time and in the same respect" - or this phrasing's semantic equivalent - being affirmed.
Ah.
Quoting TonesInDeepFreeze
Can we say the proposition "Winston Churchill was French" is A? If so, if A is True, what do you make of the following table?
In the interpretation where ¬(a ? (b ? ¬b)) is T, A is always T.
I meant 'non-contradiction', not 'contradiction'. I meant:
Do you take
"It is not the case that both water can be green and water can be not-green."
as an instance of the law of non-contradiction?
At least for my sake, you don't need to link me to a generator for such simple matters.
In the case that ~(A -> (B & ~B)) is true, A is true.
I asked what is your point in asking me these questions.
So when you say you claim the opposite, do you mean you claim the denial of:
"It is not the case that both water can be green and water can be not-green" is an instance of the law of non-contradiction.
If so, I don't understand. I would think that you are claiming:
"It is not the case that both water can be green and water can be not-green" is an instance of the law of non-contradiction.
Or maybe you're not seeing that the scope of 'it is not the case that' is only 'both water can be green and water can be not-green'.
On second thought, so we don't continue going around in endless circles:
E.g.: this water can be green today and blue tomorrow (not a contradiction). Or, this water here can be green and that water there can be blue (again, not a contradiction).
However, if it's affirmed that:
Quoting javra
Then logical contradiction does result.
Again, if A and notA do not occur at the same time and in the same respect, then no contradiction occurs. Only when A and notA [s]do[/s] are affirmed to occur at the same time and in the same respect does contradiction obtain.
I understand the proviso "in same time in all respects". But that proviso may be given more generally, upfront about all the statements under consideration:
(1) Caveat: We are considering only statements that are definite enough that they are unambiguous as to such things as time, aspects, etc. So we're covered in that regard.
Then we have:
(2) Law: For all statements A, it is not the case that both A and not-A.
Would (1) and (2) suffice for you as the law of non-contradiction?
Quoting javra
Is A a statement? If so, what do you mean for a statement to occur? If not, then what is A and what does it mean for it to occur?
That's true of classical logic, and more specifically it's fragment known as intuitionistic logic, due to the fact that the respective rule of implication is essentially the logic of functions (including those that ignore their arguments to produce a constant value), rather than the logic of causality - which is described by relevance logic and linear logic.
As for "Modern Symbolic Logic", it doesn't have a well-defined meaning since it refers to a plurality of logics that are separately described in terms of different mathematical categories.
To be clear, that is not my own claim.
Quoting Lionino
Not untrue to me.
Quoting Lionino
You skipped what I said about that.
Quoting Lionino
I didn't intend to enlighten you. You asked me for a translation, so I gave you the most direct translation.
Quoting Lionino
I talked about interpretations so that we are clear about what we claim about these things.
I think the context of the paper is classical logic, or a logic that has material implication.
I explained why "if a statement is true, then that statement is implied by any statement whatever" is a misleading characterization.
Usually we have to have LEM to have truth tables. For example, intutionistic sentential logic cannot be evaluated with truth tables with any finite number of truth values.
A contradiction is a contradiction. It is neither true nor false. It is the basis for both truth and falsity.
Thus one cant pretend to represent a contradiction in the form of a proposition and then apply the LEM to it as if it makes sense to call a contradiction true or false. As soon as a reified contradiction is allowed to enter logical discourse, a wrinkle is introduced which can never ultimately be ironed out. It is hazardous to try to answer the OP's question with a formal proof, because formal proofs cannot represent the notion of contradiction in its entirety. Granted, natural language also cannot do this, but it is at least capable of not-trying to do this. It is capable of apophaticism.
The tokens "(b?¬b)" and "¬(b?¬b)" are neither true nor false if by 'true' and 'false' we are talking about something that goes beyond validity and invalidity (i.e. something which the logical system presupposes and is transcended by). The first is invalid but not false. The second is valid but not true. Too often in logic we conflate validity with truth.
Quoting Lionino
Yes, exactly.
Quoting Lionino
Right, one could approach it from two different directions. This is what I was trying to get at earlier, "I think Lionino was somehow seeing this through the overdetermination of the biconditional with respect to ¬¬A" ().
Material implication is supposed to prescind from the reason why the implication is true or false, and therefore it is supposed to prescind from causal and temporal considerations. But in this case the temporal order in which one evaluates a complicated material conditional affects its outcome,* and I can't see how this would ever happen without having that contradiction in the "interior logical flow" of the argument. Similarly, in my last post:
Quoting Leontiskos
* Or perhaps it is not the temporal order, but rather the simple decision of whether to conceive of the different things as propositions or as truth values. A problem of ordering rather than temporal ordering, as it is the combinations that affect the outcome rather than the succession-order in which they are applied. If this is the nub then the problem is, as I said earlier, that truth-functional logic does not give us any instruction for how to handle contradictions. ...and again, there is no such thing as "handling contradictions," so this is to be expected. The problem is that we are trying to handle something that cannot be handled.
How does your newly provided caveat (1) added to your previously made statement (2) not fully equate semantically to what I initially explicitly defined the law of noncontradiction to be in full?
If (2) and the now explicitly stated (1) do fully equate semantically to what I initially stated explicitly, then you have your answer. Yes.
Quoting TonesInDeepFreeze
Quite obviously not when taken in proper given context. ("if a statement both does and does not occur [...]" ???)
Quoting TonesInDeepFreeze
Anything whatsoever that can be the object of ones awareness. For example, be this object of awareness mental (such as the concept of rock), physical (such as a rock), or otherwise conceived as a universal (were such to be real) that is neither specific to ones mind or to physical reality (such as the quantities specified by 1 and 0, as these can for example describe the number of rocks present or else addressed).
Quoting TonesInDeepFreeze
In all cases, it minimally means for it to be that logical identity, A=A, which one is at least momentarily aware of. Ranging from anything one might specify when saying, "it occurred to me that [...]" to anything that occurs physically which one is in any way aware of.
---
I get the sense you might now ask further trivial questions devoid of any context regarding why they might be asked. I don't have as much leisure time as many others hereabout apparently have. If further questions are asked, please provide a context to your questions. I will choose to not further reply without a sufficiently meaningful context being provided.
BTW, general questions about Aristotelian notions of the principle of non-contradiction can be answered in this SEP article.
This seems to be the source of your difficulties.
As has been explained, in classical logic a contradiction is false. Dialetheism considers what must be the case if some contradictions are considered to be true.The various paraconsistent logics consider what must be the case if A, ~A ? B; that is, if contradictions are not explosive.
All this to say that there are various ways to treat truth values, each with its own outcomes. (the list is not meant to be exhaustive - there are other options)
Each of these systems sets out different ways of dealing with truth values. How the truth value of a contradiction is treated depends on which of these systems is in play.
Asking, as you do, how to treat the truth value of a contradiction apart from the system that sets out how a truth value is to be dealt with makes little sense. It does not make much sense to speak of "the notion of contradiction in its entirety".
The point is that if there is no determinate entity that 'the self' refers to, if there is only the concept, and if there is no actual entity, then saying that we are speaking about the same thing is incoherent. On the other hand, if you stipulate that the self is, for example, the body, then what would A be in the proposition (A implies B) where B is 'there is a self' ? Let's say that A is 'the perception of the body': this would be 'the perception of the body implies that there is a self". 'The perception of the body implies that there is no self' would then be a contradiction to that.
Quoting TonesInDeepFreeze
An alternative way of putting it would be 'if water then oxygen'. 'If water then no oxygen' contradicts 'if water then oxygen' according to the logic of everyday parlance.
My point earlier with taking an alternative interpretation, that is with the 'notB' not being interpreted as 'not oxygen' but rather as signifying something other than oxygen, say hydrogen, then the two statements would not contradict one another.
See, for example:
Quoting Leontiskos
If you think that I am speaking about, "how to treat the truth value of a contradiction apart from the system that sets out how a truth value is to be dealt with," then pray tell how a contradiction is to be dealt with in classical propositional logic...? You seem to be implying that there are correct answers to the quandaries in this thread. For example, you seem to be implying that, according to the logic, one person is right and one person is wrong when they disagree about whether a given instance of (b?¬b) should be treated as a proposition/variable or as a simple truth value. What, then, is the right answer?
I freely admit that we could draw up additional rules to avoid the problems that are here arising. I have even proposed that we disallow "contradictions" from logical sentences. But I think the thread testifies to the fact that no such rules are generally acknowledged.
Edit:
Quoting Banno
I think the thread shows that this is not true. The problem here is that your answer lacks specificity, and contains the very ambiguity that is creating problems. Are we to consider a contradiction as if it were a variable that just happens to be false (e.g. ¬p), or are we to consider a contradiction as if it is a simple (non-complex) falsity (e.g. "If a conditional is false and its antecedent is true, then its consequent is by definition false")? The problem is that we can only be pretending to consult the truth table of a contradiction, and classical logic is premised upon the consultation of truth tables.
I'm puzzling over what this might be.
Quoting Leontiskos
As has been explained at length, in classical propositional logic contradictions are false.
Quoting Leontiskos
Another example of your practice of misattributing stuff to your interlocutors - as you did with . What I said is that the disagreement here is as to which system is in play. Hence there is no absolute answer as to which view is "right".
I've been ignoring Tones, as he is a pill and he inundates me with an absurd number of replies (15 in just the last 24 hours). Presumably he is the only one you believe has "explained this at length"?
Quoting Banno
I think the thread shows that this is not true. The problem here is that your answer lacks specificity, and contains the very ambiguity that is creating problems. Are we to consider a contradiction as if it were a variable that just happens to be false (e.g. ¬p), or are we to consider a contradiction as if it is a simple (non-complex) falsity (e.g. "If a conditional is false and its antecedent is true, then its consequent is by definition false")? The problem is that we can only be pretending to consult the truth table of a contradiction, and classical logic is premised upon the consultation of truth tables.
Edit:
I would suggest reading my post here:
How is it that both (B?¬B) and ¬(B?¬B) can have the exact same effect on the antecedent, allowing us to draw ¬A? How is it that something and its negation can both be false? This is key to understanding my claim that two different senses of falsity are at play in these cases.
It depends upon the values given to the variables.
They can contradict one another at times. At other times, they can both be true. They cannot do both at the same time.
Some A's have a plurality of implications. If A implies both, B and C, then "A implies B" and "A implies not B" is better understood as "A implies B and C". C is not B.
A implies B and C. C is not B. A implies both, B and not B. No contradiction.
QED
Your loss.
Quoting Leontiskos
Then the thread is in erorr. (p ^ ~p) is false in classical propositional logic.
Quoting Leontiskos
Not at all. A contradiction in first order predicate logic is an expression of the form (? ^ ~?). It is not an expression of the form ~?. The lack of specificity here is your attempt to make use of a notion of contradiction that is not found in classical propositional logic.
Quoting Leontiskos
Quoting Leontiskos
Nowhere in that post do you affirm (B?¬B).
10 pages later...
I read his responses to Lionino, but many of those posts are just completely blank. He deletes what he wrote. His ready-made approach doesn't answer the questions that are being asked, and you and Tones are two peas in a methodological pod.
Quoting Banno
This answer proves that you do not understand the questions that are being asked. If one wants to understand what is being discussed here they will be required to set aside their ready-made answers. They will be required to examine the logic machine itself instead of just assuming that it is working.
Quoting Leontiskos
Quoting Banno
I never said I did. Read again what you responded to. "Whether or not we affirm the negation of the consequent..."
You are telling me that (B?¬B) is false, and that this is presumably the reason why we can draw ¬A in the first argument. But in the second argument we can draw ¬A because of ¬(B?¬B). So I ask again: How is it that something and its negation can both [function as the second premise of a modus tollens]? By calling (B?¬B) is false you are presumably thinking of the first argument (and therefore both arguments) as a modus tollens.*
* You are thinking of the first argument as a modus tollens enthymeme, which is how I was conceiving of it earlier in the thread as well.
Now consider the way that is interpreting this first argument (and I think this is the same way that many others tend to think about this precritically):
This relates to what I said earlier, namely that a logical sentence that contains a contradiction should perhaps not be considered "well-formed" (I preempt an objection <here>). FALSE is not a term in classical logic, and we have no clear understanding of how it is supposed to be used. Banno seems to be wanting to use FALSE as the second premise in a modus tollens argument in order to draw ¬A. Is that permissible? Does this new term FALSE really work as a substitute for the second premise of a modus tollens? I don't think there is a clear answer.
Now suppose we have:
In this case can we also use ¬FALSE to draw the modus tollens? Is this new term that we have introduced into classical logic negatable? There is no clear answer.
Another way to read the first argument, and the one I prefer*, is as follows:
This is exactly what I said originally:
Quoting Leontiskos
I am trying not to repeat what I have said elsewhere, but the equivocation between what is false and what is absurd or contradictory was pointed out earlier:
Quoting Leontiskos
Speaking in natural language, the opposite of false is true, and yet the opposite of a contradiction is not true. Or rather, 'false' and 'contradictory' are opposites of 'true' in entirely different ways.** Thus when Banno says that a contradiction (b?¬b) is false, does he mean that it is false or that it is FALSE? It could be treated as false just as we treat ¬q as false, but in that case there is no possible modus tollens on the first argument (because in that case (p?¬q) would require a second premise that negates the consequent, i.e. ¬¬q or just q). In this case the opposite of false is true and everything carries on fine*** at least for the present moment, for as soon as we interact with (b?¬b) again we are again susceptible to treating it as FALSE instead of false.
But if Banno instead treats (b?¬b) as FALSE then the opposite of FALSE is ...? We don't know. This is really the opposite of ABSURD or the opposite of a real contradiction, and the opposite of such a thing is not simply 'true'.
Introducing ABSURD in the way I did above destroys the LEM of classical logic. Introducing FALSE significantly complicates the LEM of classical logic, and it is possible that it also destroys it.
* The one I prefer assuming we allow contradictions in our logical formulas, which I rather doubt that we should.
** I would want to say that the opposite of 'contradictory'/'absurd' is 'coherent' or 'consistent', not 'true'.
*** Fine except for the odd wrinkle that a cousin of an exclusive-or is produced where there would otherwise always be an inclusive-or (link).
Ah, so it's an esoteric mystery. :wink:
Quoting Leontiskos
The consequent is (B?¬B)
The negation of the consequent is ~(B?¬B)
Affirming the negation of the consequent is ?~(B?¬B)
if you don't affirm the negation of the consequent, you affirm (B?¬B).
Nowhere do you do this. Nowhere in these examples is it the case that "...something and its negation can both be false". That is, you do not show that somehow classical propositional logic affirms both
~(B?¬B) and (B?¬B).
Indeed, while your second example is a case of modus tollens, this is not clear for the first.
(The second is
1. A?(B?¬B) assumption
2. ¬(B?¬B) assumption
3. ¬A 1,2 modus tollens)
Modus Tollens tells us that "Given ???, together with ~?, we can infer ~?". In the first example you do not have ~?. It might as well be a Reductio, although even there it is incomplete. It should be something like:
1. A?(B?¬B) assumption
2. A assumption
3. B?¬B 1,2, conditional proof
4. ~A 2, 3 reductio
ans so A?(B?¬B)?~A
_________________
And yes, A?FALSE is not well-formed in classic propositional logic. So if your first example is to be understood as using MTT, it is not an example from classic propositional logic. Again, that is not something I have supposed, and you misattribute it.
Which takes us back to what I pointed out earlier - you are mixing various logical systems. The equivocation here is on your part. Don't put the blame for your poor notation on to me.
(Edit: Actually, Open Logic builds propositional logic from, amongst other things, ?. (Definition 7.1). And v(?) = F - the valuation of ? is "false" - in Definition 7.15. In this sort of build, ? ? ? could be well formed. @TonesInDeepFreeze might be able to clarify.)
I am attributing the modus tollens to you because you are the one arguing for ¬A. If you are not using modus tollens to draw ¬A then how are you doing it? By reductio?
Quoting Banno
As I noted earlier in response to Tones' reductio, a reductio is an indirect proof which is not valid in the same way that direct proofs are. You can see this by examining your conclusion. In your conclusion you rejected assumption (2) instead of assumption (1). Why did you do that? In fact it was mere whim on your part, and that is the weakness of a reductio.*
What if we reject (1) instead? Then A is made true, but it does not imply (B?¬B). Your proof for ¬A depends on an arbitrary preference for rejecting (2) rather than (1).
Quoting Banno
What is at stake is meaning, not notation. To draw the modus tollens without ¬(B?¬B) requires us to mean FALSE. You say that you are not using a modus tollens in the first argument. Fair enough: then you don't necessarily mean FALSE.
* A reductio requires special background conditions. In this case it would require the background condition that (1) is more plausible than (2).
A reductio is as much a proof in classical propositional logic as is modus tollens.
Quoting Leontiskos
Simply because I matched your example, which has
Quoting Leontiskos
and not ~A?A?(B?¬B).
Again, don't blame me for your problems.
edit: corrected A?A?(B?¬B)/~A?A?(B?¬B)
You think you get to arbitrarily reject (2) instead of (1) because I gave an example of the unaccountable inference that some in this thread are drawing? My whole point is that ¬A should not follow. What I gave is an example of the argument (claim?) that hypericin originally gave <here>.
So are you agreeing with me that the reductio does not prove ¬A?
The reductio shows that A?(B?¬B)?~A. As pointed out.
It could equally be used to show that A?~A?(B?¬B); but that was not the issue you raised.
Edit: correction
Again:
Quoting Leontiskos
Are you saying that if your logic professor asked you to justify an answer to my question you would tell him, "This guy on the internet set out an incomplete argument which he doesn't accept, in which the conclusion was ¬A. Therefore we must reject (2) instead of (1)"?
(i.e. "I matched [his] example" ())
I wonder if you will have trouble sleeping on such non-existent arguments?
Quoting Leontiskos
"Machine", singular. So back to my point, that Quoting Banno
and so
Quoting Banno
What I hope to have done over the last page is to show that you are mixing logics, resulting in your own confusion. You do not succeed in showing that "...something and its negation can both be false" in classical propositional logic.
You're reaching. :wink:
I have given my arguments, I have already responded to these charges.
At this point you either have an argument for "?¬A" or you don't. Do you have one? If not, why are you still saying that ¬A is implied?
Maybe not as much as you think.
Quoting Leontiskos
I'm not seeing a salient point here. Pretty demonstrably, you have made a series of claims that have been shown to be in error.
At this stage it is unclear what your general point concerning "metalogic" might be, beyond an "esoteric mystery".
I think calling them both assumptions has led to your confusion. Premise 1 is more of a GIVEN than an assumption.
We start out the scenario with it GIVEN that a -> (b and ~b). With that given, we say "let's see what happens when I assume a is the case". What happens is a contradiction, so we take a step back and realise, if it's given that a -> (b and ~b), a must be false.
"It is not the case that if A then B & ~B
implies
A"
I still can't make sense of it.
This is one of those funny places where symbolic logic seems to take a detour from what we mean in natural language.
It is what I ask here
Quoting Lionino
Tones replied that that is not true for all contradictions but for some interpretations. I couldn't make sense farther past it.
It seems as though, the right thing to say about basic classic symbolic logic is that EVERY statement is either true or false. So, you claim A, that's either true or false, period.
If A is false, then A implies anything. You can check the truth-table on implication: A -> B is always true if A is false. https://math.stackexchange.com/questions/1306280/implication-truth-table
So, no matter what B is, if A is false, A implies B - even if B ic (C and ~C).
So, if you KNOW that A doesn't imply (C and ~C), but you also know that if A was false, A has to imply (C and ~C) by the fact that anything follows from falsehood, then you must know that A must be true.
This makes sense in the universe of classic symbolic logic, where everything has explicit truth values and implication means what it means there.
Maybe that system of symbolic logic is just... English? Normal logical language? Idk.
I think there is a mystery why we can say it is false in this case. What rule of inference in classical logic are we appealing to? my point has been that the only legitimate rule of inference that we can appeal to itself turns out to be a metabasis. This is to say that such a rule of inference will never be valid in the same way that a direct proof is valid.
Some call it a "supposition," but they are fooling themselves if they think this answers my objection.
rule of noncontradiction, no?
Funnily enough, the rules of inference we're appealing to are in fact the very first ones listed on the Wikipedia page:
https://en.wikipedia.org/wiki/List_of_rules_of_inference
Reductio ad absurdum
Quoting Banno
I noticed that there is in fact a second problem with your reductio. You told me that in classical logic contradictions are to be treated as false, but in your reductio you do not treat the contradiction as false. You treat it as a contradiction, as an outer bound on the logic:
Quoting Leontiskos
You are appealing to the move of a reductio, Contradiction; Reject supposition. But there is no rule, False; Reject supposition. Therefore you are clearly not being consistent in treating the contradiction as false.
Now you could of course continue your proof and try to use a modus tollens to arrive at (A?¬A), But not only will this result in the same exact problem, but it would result in the additional problem of utilizing FALSE in the way I pointed out <here>.
(It would seem that you are wrong in claiming that classical logic treats contradictions as false. In fact it treats them as <ABSURD>. is correct that classical logic treats whatever "implies" a contradiction as false. Note carefully that "implies" here no longer means material implication.)
That is a good explanation. In a way the explanation is sort of a literal translation of the formula, but I was hiccupping at the "then you must know that A must be true" part. As you said:
Quoting flannel jesus
Here is something quaint. In modal logic, ¬?(a?(b?¬b)) entails ?a. But I guess that is simply a consequence of what we are talking about. Not only that, but ¬??x(A(x)?(B(x)?¬B(x))) entails ??xA(x). I am confident I am simply misunderstanding what "?(B(x)?¬B(x)" means, it can't be just "any contradiction", as Tones has pointed.
The main problem for me is, why can we read a?(b?¬b) as "a implies a contradiction" but not ¬(a?(b?¬b)) as "a does not imply a contradiction?
Well, if something results in a contradiction, we are able to rule it out, aren't we? At least we do it all the time in physics and mathematics.
No, you asked for the rule of inference from classical logic - it's right there, common knowledge in wikipedia. I don't see any good reason why my answer should be considered unacceptable, other than you just don't want to accept it. You asked for the rule, that's it.
@Banno
Check out the truth tables in Many-Valued Logic section https://plato.stanford.edu/entries/logic-paraconsistent/#ManyValuLogi
In this thread some users also use a third value for variables https://thephilosophyforum.com/discussion/15333/ambiguous-teller-riddle/p1 and, as we discussed, that is basically what my use of "(A or notA)" does as well.
It is also a way that people go around liar paradoxes https://plato.stanford.edu/entries/dialetheism/#SimpCaseStudLiar
Well, there we are going into epistemic territory. But it seems a bit related to what I say in
Quoting Lionino
Sure, so long as we are recognizing that "to rule it out" is a special move, unique and irreducible to any other move in classical logic. Specifically, the implication that we deny is in this case is no longer material implication.
Quoting Leontiskos
This all goes hand-in-hand with the fact that Banno's reductio has no power, strictly speaking, to draw the conclusion ¬A. A reductio is not a proof in the strict sense, and this is precisely what a metabasis is: a form of non-strict inference. Anyone can deny that reductio without being the worse for wear, logically speaking.
I'm sorry, but as someone who thinks that material implication is an example of the principle of explosion you are out of your depth here. Material implication is entirely different from the principle of explosion, and an argument from the authority of Wikipedia is not a sufficient answer to the question I am asking.
I don't think I claimed that. But as you're eager to reject basic reason, I'm not going to be one to stop you.
Where
These are just paradoxes of material implication, no?
The negation of a contradiction is always true, and being true it is implied by anything, true or false.
Same here, it seems to me to be a paradox of material implication that is the source of confusion.
In a normal conversation, we might ask "but what if A really only implies B and not B and not-B?" Or conversely: "what if A only implies not-B but does not actually imply B?" But the way implication works here it is not an additional premise we can reject, we don't assign a truth value to it except in virtue of the truth values of A and B themselves.
If A is true then B always implies it, whether B is true or false.
In a relevance logic or per Aristotle's comments, we might turn around and question if A truly implies either B or not-B regardless of the truth value of either A or B alone.
For example, we could set up something like:
Elvis is a man - A
Elvis is a man implies that Elvis is both mortal and not-mortal. - A ? (B and ~B)
Therefore Elvis is not a man.
Obviously, the common sense thing to do would be to reject "Elvis is a man implies Elvis is non-mortal," while affirming that "Elvis is a man does imply Elvis is mortal."
A truth table won't lay those possibilities out, it will only tell us how things work in terms of A or B being true.
However, there is a quite good reason not to do this in symbolic logic. Once you start getting into "what 'really' entails what," you get into judgement calls and a simple mechanical process won't be able to handle these.
But of course, you still need judgement to make sure your statements aren't nonsense, so you just kick that problem back a level. A proof from contradiction is only going to be convincing if we believe that A really does imply both B and not-B. I know plenty of skepticism has been raised against proofs from contradiction in general, outside of this issue, but for many uses they seem pretty unobjectionable to me.
I think that would be (A?¬(B?¬B)), which is True for any value of A and B. While we are talking about ¬(A?(B?¬B)), True only when A is True.
Quoting Count Timothy von Icarus
What about.
¬(A?(B and ¬B))?
When A is true ~A implies anything, including contradictions.
Here:
Quoting Banno
A reductio is not truth-functional. If we want to stick to strict truth functionality then we cannot accept reductios. In that case we can only think of them as indicating the inconsistency of a system, not as grounds for denying one thing or another. Technically a reductio is a form of special pleading (i.e. Please let me reject this premise rather than that one, for no particular reason). I think I can only be wrong about this if there is some principled difference between a supposition and a premise (or an assumption). A reductio is a kind of bridge between formal logic and the real world, and this is because of the background conditions it presupposes. In a purely formal sense no proposition is inherently more or less plausible than another, and therefore there is no reason to reject one premise rather than another in the event of a contradiction.
In general, the consistency of an axiomatic system isn't provable in an absolute sense due to Godel's second incompleteness theorem; the upshot being that consistency is a structural property of the entire system that isn't represented as a theorem by the system if it is sufficiently powerful.
Suppose that the logic concerned is weaker than Peano arithmetic, such that it can prove its own consistency. Then in this case, a proof of ¬¬a metalogically implies that ¬a isn't provable, i.e that a does not imply a contradiction.
But if the axiomatic system contains Peano arithmetic such that the second incompleteness theorem holds, then a proof of ¬¬a does not necessarily imply the absence of a proof of ¬a, since Peano arithmetic cannot prove its own consistency.
Elvis is a man A
Elvis is a man does not imply that Elvis is both mortal and immortal ¬(A ? (B and ¬B))
Therefore Elvis is a man. A
A, ¬(A ? (B? ¬B)) entails A. That makes sense.
Let's say now.
Elvis is not a man ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal ¬(A ? (B and ¬B))
Therefore Elvis is not a man ¬A
¬A,¬(A?(B?¬B)) entails ¬A. That makes sense.
Elvis is not a man ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal ¬(A ? (B and ¬B))
Therefore Elvis is a man A
¬A, ¬(A ? (B? ¬B)) entails A. That doesn't make sense to me.
But I guess it does make sense when we consider that ¬(A?(B?¬B))?(¬A?(B?¬B)) is valid. So, ¬A implies (B?¬B), from where we can say A.
But if ¬A?(B?¬B), it is a bit strange that we can derive ¬A above in the second schema. Perhaps because from a contradiction everything follows?
However, A,¬(A?(B?¬B)) does not entail ¬A. So from {Elvis is a man} and {Elvis is a man does not imply that Elvis is both mortal and immortal}, you can't derive Elvis is not a man, because there is no contradiction being states here from where we can affirm anything.
Therefore, if ¬(A?(B?¬B))?(¬A?(B?¬B)) is true, and ¬A?(B?¬B) can be read as not-A implies a contradiction, it must be that ¬(A?(B?¬B) cannot be read as A does not imply a contradiction.
Yes, good. :up: Kreeft's point comes back.
Quoting Count Timothy von Icarus
Right. As I have been saying it, "falsity incarnate" and "truth incarnate" are reifications. <FALSE> is a new idea, more or less foreign to classical logic.
Quoting Count Timothy von Icarus
Yep.
Quoting Count Timothy von Icarus
I think metabasis is useful, but I don't close my eyes to the fact that it is metabasis:
Quoting Leontiskos
Reductio ad absurdum is useful and important, but it is not formally valid in the same way that direct proofs are (and because of this it is a (useful) metabasis). We need to recognize that we are doing something special with a reductio, and that a reductio-inference to ¬A is quite different from a direct inference to ¬A. So if someone wants to say that ¬A is implied they must put an asterisk next to implied*.
Thanks. This is what I was trying to remember but could not find online (i.e. the complexities surrounding proofs of ¬¬a).
So I guess that, in order to say "A does not imply a contradiction", we would have to say instead (A?¬(B?¬B)). From there things start to make more sense.
Since ¬(A?(B?¬B)) does not translate to "A does not imply B and not-B". I have to fix my post above.
Quoting Lionino
to:
Elvis is a man A
Elvis is not a man implies that Elvis is both mortal and immortal ¬(A ? (B and ¬B))
Therefore Elvis is a man. A
A, ¬(A ? (B? ¬B)) entails A. A entails A.
Reminder that ¬(A?(B?¬B)) is the same as (¬A?(B?¬B))
Elvis is not a man ¬A
Elvis is not a man implies that Elvis is both mortal and immortal ¬(A ? (B and ¬B))
Therefore Elvis is not a man ¬A
¬A,¬(A?(B?¬B)) entails ¬A, from contradiction everything follows.
Elvis is not a man ¬A
Elvis is not a man implies that Elvis is both mortal and immortal ¬(A ? (B and ¬B))
Therefore Elvis is a man A
¬A, ¬(A ? (B? ¬B)) entails A, from contradiction everything follows.
Elvis is a man A
Elvis is not a man implies that Elvis is both mortal and immortal ¬(A ? (B and ¬B))
These two premises do not entail that Elvis is not a man, because there is no contradiction. A has to entail A.
I think, keeping explosion in mind, this makes much more sense in natural language.
So let's look at the cases with (A?¬(B?¬B)), which is finally translated properly as "A does not imply a contradiction".
Elvis is a man A
Elvis is a man does not imply that Elvis is both mortal and immortal (A ? ¬(B and ¬B))
Therefore Elvis is a man A
A, (A ? ¬(B? ¬B)) |= A
Elvis is not a man ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal (A ? ¬(B and ¬B))
Therefore Elvis is not a man ¬A
¬A, (A ? ¬(B? ¬B)) |= ¬A
Elvis is a man A
Elvis is a man does not imply that Elvis is both mortal and immortal (A ? ¬(B and ¬B))
These two do not entail that Elvis is not a man ¬A.
Elvis is not a man ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal (A ? ¬(B and ¬B))
These two do not entail that Elvis is a man.
I think I finally solved my own problem. When translating it to natural language, I was misplacing the associativity of the ? operator in this case.
So ¬(A ? (B? ¬B)) is the same as (¬A) ? (B? ¬B), which may be read as "Not-A implies a contradiction", it can't read as "A does not imply a contradiction". We would have to say something like A ¬? (B? ¬B), which most checkers will reject as improper formatting, so we just say A ? ¬(B? ¬B), which can be read as "A implies not-a-contradiction", more naturally as "A does not imply a contradiction".
@flannel jesus :cool: gigabrain has done it once again
Are you interpreting "a does not imply a contradiction" as the basis of a reductio (i.e. "Suppose a; a implies a contradiction; reject a")? If so, then I again think it is because a reductio is not reducible to a truth-functional move. A reductio requires more than negation and falsity.
Edit:
Quoting Lionino
Does this support my claim that what is at stake is something other than a material conditional? The negation does not distribute to a material conditional in the way you are now distributing it.
Do you believe for, for all statements (A -> B), you can do ¬(A -> B) and transform that into ¬A -> B?
In the meanwhile, I can finally go cook with peace of mind.
Quoting Leontiskos
But for the record I do accept this as a valid rhetorical move. However when it comes to propositional logic, from
P1: A
P2: A?contradict
The conclusion can be whatever we want, from explosion
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=A
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=A
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=C
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=D
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=P
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=Z
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=Z(P(G(F(x))))
Edited:
These two are different statements. By the way that the syntax of these operators is made, (¬(A ? (B?¬B))) is the same thing as (¬A?(B?¬B)). It is like 2(x*y)=2x*y, but 2(x*y)?x*2y
That's what I was asking, thank you.
I don't believe that's correct.
Yeah I messed that up.
Those two are not the same thing.
But ¬(A ? (B?¬B))) and (¬A?(B?¬B)) are the same thing.
Example:
(A?(B?¬B))?(B?¬B) is valid
But (A?C)?C is invalid.
Me? You understand that I think that?
But what just happened is that you did that, and I told you it's incorrect...
The answer is not for all statements. I never replied positively to the question, I copy pasted incorrectly in a post that is now edited.
Well you gave what certainly looked like an affirmation. If I ask you "is lemonade your favourite flavour", and you say "lemonade is the same as my favourite flavour", most people are gonna think that's pretty much a "yes" to the question.
Is * multiplication here? I don't think this is right either.
* is any operation. I know multiplication doesn't work like that obviously
You wrote
¬(A ? B) is the same thing as ¬A?B
Ill offer that the commonsense notion of self necessarily pivots around what we westerners commonly term consciousness - "the self" here always entailing the subject of ones own experience of phenomena (or, in for example the more philosophical jargon of Kant, the empirical ego (via which empirical knowledge is possible) - this as contrasted to what he specifies as the underlying transcendental ego). And, in this commonsense understanding of "the self", the body of which one is aware is then not the self which is in question - "the self" instead holding as referent that which is so aware. Nor does the notion you present in any way cohere with the descriptions of self as they are addressed in the Buddhist doctrine of anatta: this being the very metaphysical understanding of reality from which we obtain propositions along the lines of neither is there a self nor is there not a self. Which, on the surface, do at least at first appear to be contradictory (though, as I've previously argued, do not need to so necessarily be).
But, since, we each hold our own - sometimes more divergent than at other times - understandings of what terms signify, Ill here say that were the term self to be devoid of any referent outside of the occurrence of empirically observable physical bodies (maybe needless to add, that are living and so normally endowed with a subject of awareness, this rather than being dead and decaying), then I might find agreement with your general reasoning here.
Im however not one to find the term self - and, hence, terms such as I, you, us and them - devoid of referents, unempirical (imperceivable) though I take these referents as "subjects of experience" to be. All the same, this thread is not the place to engage in debates regarding what the subject of ones own experiences might specify or else be - although I do agree that it is not "an entity" in the sense of being a thing.
This doesn't make sense if * is "any operator" either. Replace * with + and 2(x*y)=2x*y is not true
I didn't, you are referring to this , which I already said was a copypaste mistake, it has been edited. I don't see what the issue is.
"Any operator" is not any mathematical operator you want. In that case, the ? operator is working like an operator ? where z(x?y)=zx?y, but z(x?y)?x?zy, it is a syntactic rule and nothing more.
The issue is you said you never wrote it, but you did write it. I understand it's a mistake. Therefore it's not correct to say you never wrote it, it's correct to say you wrote it by mistake.
Quoting Lionino
I don't know the rules of that game, my bad
since (a?b) is the same as (¬a?b),
(a?(b?¬b)) is the same as (¬a?(b?¬b)),
and ¬(a?(b?¬b)) the same as ¬(¬a?(b?¬b)), which is the same as (a?(b?¬b)).
So ¬(a?(b?¬b)) is the same as (a?(b?¬b)), so I think now it is a bit more clear why ¬(a?(b?¬b)) is True only when A is True, the second member is always False and the or-operator returns True when at least one variable is True.
(a?b) ? (¬a?b)
¬(a?b) ? ¬(¬a?b)
However (a?b) and ¬(¬a?b) aren't the same
So ¬(a?b) and (a?b) aren't the same
(a?(b?¬b)) ? (¬a?(b?¬b))
¬(a?(b?¬b)) ? ¬(¬a?(b?¬b))
(¬a?(b?¬b)) ? ¬(¬a?(b?¬b))
Since ¬(¬a?(b?¬b)) is the same as (a?(b?¬b))
(¬a?(b?¬b)) ? (a?(b?¬b))
Quoting Leontiskos
Or if you prefer: ??(?^~?)?~?
Or if you think it is only truth-functional if it fits in a truth-table,
At some point one has to realise that Leo has such an odd notion of logic that he is unreachable.
Again, no.
F's all the way down.
I gather you worked through this? Nice.
Leo does that sort of thing - claims you have said something you haven't, if it suits his purposes.
Yea, a?(b?¬b) can be read as "A implies a contradiction" but ¬(a?(b?¬b)) cannot be read as "A does not imply a contradiction", it is read instead as "not-A implies a contradiction". "A does not imply a contradiction" would rather be (a?¬(b?¬b)). So the opposite in natural language is not the same as the opposite in logical language, in this case.
¬?(a ? (b?¬b)) entails ?a
How should we read this in English? Because "{It is not possible that A implies a contradiction} entails A is necessary" is not obviously right.
Isn't it something like that "if it is not possible that A implies a contradiction, then A is necessarily true"?
Or "If in no possible world A implies a contradiction, then A is true in every possible world"?
Besides, ¬?(a?(b?¬b))??(¬a?(b?¬b)) is invalid, so the issue can't be solved like the original one was.
But I guess it can be solved in a similar way:
¬?(a?(b?¬b))??¬(a?(b?¬b)) is valid
¬?(a?(b?¬b))??(¬a?(b?¬b)) is also valid
Since ¬?(a?(b?¬b)) is the same as ?(¬a?(b?¬b)), it can be read as "It is necessary that not-A implies a contradiction". From that alone I think we can accept that it follows that necessarily A.
So, since ¬?(a?(b?¬b)) would be read by many as "It is not possible that A implies a contradiction", is that the same thing as "It is necessary that not-A implies a contradiction"? If not, "It is not possible that A implies a contradiction" is not a correct reading of ¬?(a?(b?¬b)).
?¬(a?(b?¬b)) would be "It is necessarily not the case that A implies a contradiction"
https://www.umsu.de/trees/#~8~3(a~5(b~1~3b))~4~8(~3a~5(b~1~3b))
So at least my reading is correct.
The issue with "It is necessarily not the case that A implies a contradiction" is that, if we remove the ? from ?¬(a?(b?¬b)), we end up with ¬(a?(b?¬b)), and this can't be read as "It is not the case that a implies a contradiction".
Something about this is that the more general ¬?(a ? (b?¬b)) |= ?a is also true. If we read ¬?(a?(b?¬b)) as "It is not possible that A is False", ¬?(a ? (b?¬b)) |= ?a starts to make a bit more sense.
Why not? I'm not seeing the issue here.
Quoting Lionino
Quoting Lionino
Quoting Lionino
A not implying a contradiction does not mean that A.
So ¬(a?(b?¬b)) can't be read as A not implying a contradiction
But (a?¬(b?¬b)) can be read as such, and it does not entail A. Thus "A does not imply a contradiction" is (a?¬(b?¬b)), not ¬(a?(b?¬b))
Same point I made earlier about alternative readings.
Hey Janus!
Evidently, normal parlance does not translate into logical notation so easily. I think there also may be a difference between "notB" and "not B". Given that no one paid much attention, I take it that my ignorance of formal logic was too obvious to mention.
:grin:
I've been reading the replies and trying to better understand, but with so little experience, and no time nor desire to practice, I'll remain an interested bystander.
Hello, creative. How are the fish hooks?
It exactly does not depend on the values given to the variables. That's kinda the point of using variables - you get to put different things in and get the same result.
So a+b = b+a regardless of what number you stick in to the formula, and a^(a?b)?b regardless of what statement you put in, too. Or so it is supposed to go.
One of the problems in this thread has been that the OP was not couched in formal logical terms, and just what was meant by 'notB' was not explained.
no ambiguity
Yeah, I get that much. As you said, that's kinda the point of using variables. I was just thinking that some statements implied a plurality of others, and hence, unless the others contradict one another, implying B and not B(C) does not imply a contradiction.
The international move has taken a year to get settled, but things are going well. Thanks for asking.
C is not B. Does that translate into notation the same way that notB does?
The hypocrisy there is astounding. The poster has written a whole lot of posts in this thread. Possibly a lot more characters I have written in this thread. Possibly more than anyone in this thread. I was away for a period while the poster had entered a lot posts, most of which are not short. When I got back, I replied to them. Only an arse would think that is not fair. And the poster, in a petty way, counts my posts, while some of them are individual for convenience. Moreover my posts have a lot of space in them due to my formatting. The poster writes stuff and I respond, often in detail. No one is a "pill' for that. In contrast with the poster's petty counting of posts, I don't begrudge the poster for filling up a lot of posting boxes. I think people should post as much as they want, explain as much as they want.
Quoting Leontiskos
A mere presumption indeed. And what is the point of the remark?
What in the world?! The poster takes issue with the fact that it happens sometimes that one needs to delete! I deleted some posts that were only started, because those posting boxes were out of sequence when I came back to finish them. Then l finished the posts in posting boxes that were in better sequence. Wow, how petty he is!
That might be because, for ease of clarity, the sentence needs parentheses.
"¬(A?(B?¬B)) entails A"
In order not to conflate with 'entails' to stand for semantic entailment, I'll simplify:
~(A -> (B & ~B)) -> A
I merely used the ordinary interrelations of the symbols.
(it is not the case that (A implies (B and it is not the case that B))) implies A
or
(not (A implies (B and not B))) implies A
or
if (not (A implies (B and not B))) then A
But you still have not told me what your point is in asking me this!
I am still looking at references to get a grasp of those terminologies. I think I have at least a picture of the notion of analogical equivocal and analogical univocal, mostly as I find the notions in certain philosophy of religion, but I guess found more generally in philosophy also. If I am not mistaken, the main idea, in greatest generality (not specifically regarding theological concerns) is that we have different kinds of subject to apply predicates to. When the predicate applied to the subjects means the same among the subjects, then that is univocal. But when the predicate means differently among the subjects then that is equivocal. That's the best come up with so far. But there is more terminology* and I don't follow what it means as applied to logic in this thread. *Especially it would help to know the translation of "metabasis eis allo genos".
I asked because I wanted to know whether you think they are equivalent, and if not, then knowing in what ways they are different would shed light on the differences in how the law of non-contradiction is taken. And, if they are equivalent then I could use my formulation in also your context, as I prefer my formation with which we don't have to give the caveat each time we talk about the kind of statements we're studying, including the law of non-contradiction and others.
Quoting javra
Then my formulation does not accord with your notion, since my formulation takes the law of non-contradiction to regard statements.
Quoting javra
I get you. A broad sense.
Quoting javra
They're not trivial. And I asked them to better understand your view.
Thank you for suggesting the SEP article. I am familiar with it even if I have not carefully studied all of it.
Does "apply" there refer to proofs of "(A -> (B & ~B)) -> ~A" or "~(B & ~B)"? LEM is not needed to prove those.
I don't think that way, except possibly as elliptic. Especially if the context demands analysis of the logical connections, I would says something like this instead:
'If water is present then oxygen is not present' contradicts 'if water is present then oxygen is present'.
As to to whether the above is true, of course, in many everyday contexts, it is regarded as true. In other contexts it is not regarded as true.
Quoting Janus
If I understand you correctly, I agree, and I touched on a similar example a while back.
This caveat will cover my posts in general: People may have different contexts with different meanings of terms, different fundamental conceptions, different philosophical positions, and especially different formal systems that may diverge from classical logic, contract it, or extend it. If a context is not made explicit, then it is reasonable to respond in the context of classical logic, which I will do. And doing so is especially apt when responding to criticisms or examination of classical logic.
Reductio ad absurdum is proven valid as it is proven that inferences using it are truth preserving, just as with the other rules of classical sentential logic. An argument is valid if and only if there are no models in which the premises are all true and the conclusion is false. Reductio ad absurdum is proven to provide only valid arguments.
Quoting Leontiskos
The argument may be better written:
(1) A -> (B & ~B) ... premise
(2) A ... assumption toward a contradiction
(3) B & ~B ... 1, 2 modus ponens
(4) ~ A ... 2, 3 RAA with 2 discharged
There we see that the premise A -> (B & ~B) proves ~A, while the temporary assumption A was discharged.
And it is valid, as every model in which "A -> (B & ~B)" is true is a model in which "A" is true.
Quoting Leontiskos
Modus tollens and reductio are two sides of the same coin.
A Hilbert system may have modus tollens (or to have classical, the non-intuitionistic version) in axiom form, and a natural deduction system may use RAA (and, to have classical, the non-intuitionistic version) in rule form.
To prove a negation, there have to be axioms or rules to do it. Modus tollens and RAA are in a sense versions of each other and they both do the job.
That's not what I said.
If I recall correctly, you said that "A -> (B & ~B)"* may be translated as "A implies a contradiction". (*Or it might have been a related formula; not crucial since my point pertains to all such examples.)
That is not the case as follows:
(1) The sentence has a sub-sentence that is a contradiction, but the sentence itself does not mention the notion of 'contradiction'.
(2) To say "a contradiction" is to implicitly quantify: "There exists a contradiction such that A implies it". And that quantifies over sentences. If we unpack, we get: "There exists a sentence Q such that Q is a contradiction and A implies it".
A translation of "A -> (B & ~B)" is:
If A, then both B and it is not the case that B.
and not
"A implies a contradiction".
(3) "B & ~B" is a particular contradiction, not just "a contradiction". Even though all contradictions are equivalent, a translation should not throw away the particular sentences that happened to be mentioned.
(4) If we have that A implies B & ~B, then of course, we correctly say "A implies a contradiction". But that is a statement about A, not part of a translation.
"If A implies B & ~B, then A implies a contradiction" is true, but it is a statement about the sentences, not a translation of them.
We need to be careful to recognize that "A is false" is not simpliciter. Rather, "A is false in model M" (with sentential logic, a model may be represented as a row in a truth table).
Quoting flannel jesus
But I would add (and this was my point to @Leonino) it is not the case that
"Winston Churchill was French" implies every sentence, simpliciter.
Rather, it is only in interpretations in which "Winston Churchill is French" is false is it the case that for every sentence "P", ""Winston Churchill is French" implies P" is true.
It would be rare that any one poster can answer all questions. I have given corrections, explanations about the logic that is being critiqued, answers to specific questions, and perspective on the subject. And describing my approach as "ready made" is empty polemics.
A person may have all kinds of independent, self-fashioned, not ready-made ideas. And one may critique "ready-made" classical logic, but when a critique misconstrues or misrepesents what it is critiquing, then it is quite appropriate to point out the errors and even explain why they are errors. Moreover, it is appropriate to point it out when what is critiqued is subjected to criteria not pertinent. Of course, classical logic doesn't accord with much of everyday discourse. And a pencil doesn't accord with the colors in the world. That doesn't entail that, for certain purposes, a pencil is not preferable to a box of crayons. Black and white photography doesn't accord with the colors in the world. That doesn't entail that black and white photography is wrong. An x-ray doesn't accord with the way we see things in everyday life. That doesn't make an x-ray wrong. And it is apt to explain how x-rays work, what they represent, and their purpose, just as it is apt to explain how classical logic works, what it represents and its purpose, especially in context of ill-premised remarks about it.
Logicians and philosophers of mathematics examine logics with intense scrutiny and may be interested in all kinds of formal and philosophical alternatives. And critiques of classical logic should be welcomed; indeed they have been vital in the history of logic and mathematics. But when critiques are ill-premised and describe the logic in inaccurate ways, then, of course, how can that be dealt with other than indeed explaining how the "machine" does operate?
But in what sense is the "machine" working?
Well, it does not work to express much of everyday discourse. Is there anyone who has denied that? And it does not accord with certain philosophies and alternative formal systems. And it does not provide answers to all the philosophical questions. But it does work as an axiomatization for mathematics for the sciences (indeed, the P vs NP problem is commonly described as the most important mathematical problems for practical implications for purposes of business, with a one million dollar prize for its solution). It does work as the basis for the computers we're using when we post. It does work in the sense that it gives rigorous definitions and unambiguous formulations of mathematics and for other fields of study. It does work in the sense that, in principle, we can most objectively (that is algorithmically) check that any given purported proof is indeed a proof in the system. It does work in the sense that the soundness and completeness of the logic is rigorously proven. And it does work in the sense that it provides beauty to its students and experts, and such that it induces great creativity and insight and questions that themselves lead to not only flourishing of mathematical inquiry but discoveries applicable in other fields. And it works in the sense that it provides the starting point and the basic rubrics for the invention of alternatives to it.
That is a bad misconception. It's not how it works.
Proof is from a set of premises. A natural deduction proof lists premises. A reductio premise however is discharged at the end of the proof. It is not an ordinary premise.
Let G be a set of premises and a sentence P is not a member of G. And we want to show that G proves ~P. Then we may use any of the members of G in our argument. But, along with members of G, we also may suppose P to derive a contradiction, thus to show that G proves ~P.
"plausibility" is not invoked.
Modus tollens is an inference rule or axiom, depending on the system. That's syntactical.
The notion of falsity is semantical.
Syntax and semantics work in synch in classical logic. But to use modus tollens in and of itself does not require mentioning falsity.
The inference rules don't opine as to falsity. Rather, syntactically, when a contradiction results from a conditional premise, then the rules allows ending the proof with the negation of the conditional premise and discharging the conditional premise. (Same for subproofs within a proof.)
Truth and falsehood are handled in the semantics. And we show that the syntax and semantics are in accord as:
For any set of sentences G and any sentence P:
P is provable from G if and only if there is no model in which all the members of G are true and P is false.
Intuitively, a 'model' represents a possible set of circumstances.
In other words, if P is provable from G, there is no set of circumstances in which all the sentences in G are true but P is false.
And we prove that rigorously.
To prove a negation, we must have a rule to use to do that. And any alternative (that adheres to soundness and provides for the completeness of the calculus) to modus tollens or RAA would be equivalent with them in the sense that the resulting systems would provide the same inferences as each other.
What I say is that "P implies a contradiction" is not a translation of "P -> (Q & ~Q).
I would not accept that reading, for the reasons I mentioned several posts ago.
Most briefly: Yes, if P implies Q & ~Q, then P implies a contradiction. But that is a remark about "P -> (Q & ~Q)" and not a translation of it.
RAA is an inference rule.
A sentence Phi is truth functional if and only if the truth or falsity of the sentence is a function of the assignment of truth or falsity to the atomic sentences occurring in Phi.
I don't know what you mean by an inference rule being truth functional.
But an inference rule may be truth preserving, as RAA is:
RAA is among the natural deduction rules. And regarding those rules:
If the rules provide that P is provable from a set of premises G, then any model in which the sentences in G is true is a model in which P is true.
What does "absolute sense" mean?
Godel-Rosser is that system of a certain kind don't prove their own consistency. That doesn't entail that there are not other systems proven to be consistent by secure means.
The system S could be inconsistent, in which case, if "S is consistent" is expressible in the language of S, then S proves "S is consistent" even though S is inconsistent.
I don't know what role A is suggested to have. Some formula A does not prove a contradiction, I guess. I don't know how that is supposed to bear upon the consistency of a system with other axioms. ~A and ~~A. Don't know why choosing that pair rather then A an ~A, or maybe this has to do with intuitionism.
Yes, granted. I used the word "translation" wrong in basically all of my posts. I meant "is a true statement about..." instead.
Now, the conclusion that I arrived at is that "A does not imply a contradiction" is not an accurate statement about ¬(A?(B and ¬B)), it would be a true statement about (A?¬(B and ¬B)) instead. When it comes to ¬(A?(B and ¬B)), as it is the same as (¬A?(B and ¬B)), "not-A implies a contradiction" is a true statement about it.
Regarding reductio ad absurdum, last night I was having a dream. I was walking a trail I know well and I noticed that the topography was inaccurate. I am usually semi-lucid when I dream, and so I decided to try to change the topography to make it more like it is in real life. As soon as I did this I noticed that this is similar to a reductio. In both cases an additional, uncharacteristic level of will emerges.
The reductio is an uncharacteristically teleological move for truth-functional logic:
Quoting TonesInDeepFreeze
The one who performs the reductio sees an opportunity to produce a contradiction and then decides to pursue it in order to achieve the inference desired (which inference is, again, a metabasis).
The Medievals would have called the truth-functionality of classical logic something which pertains to the intellect (as opposed to the will). It is supposed to be purely formal, purely intellectual, and in no way willful. As you and I know, this is not entirely true since any human act involves the will, and therefore in any logic the teleological end of the acts at hand must involve the will. Nevertheless, a reductio involves the will over and above the way that direct inferences involve the will. The reductio attempts to leverage a contingency about the problem at hand in order to wield the contradiction and draw a conclusion. Hence the difference between a supposition and a mere assumption is that the supposition acknowledges the teleological motive of the will in a way that an assumption does not (Tones called his move a supposition whereas Banno called the same move an assumption).
So we could say that the metabasis eis allo genos and the essence of a reductio ad absurdum is found not only in the unique inference that concludes a reductio, but also in its starting point: the supposition. This is what separates the supposition from the other assumptions, even though this difference is mental and not formal. And if we pay very close attention we will see that the formal conclusion is different from the teleological conclusion. The formal conclusion is that the system which includes the supposition is inconsistent. The teleological conclusion is that we should reject the supposition rather than a different premise. There is a miniature inference from the formal conclusion to the teleological conclusion, and this tends to be ignored by most students of classical logic. Put differently, the reductio strictly speaking only tells us that something cannot be supposed. It is a second step to say that that which cannot be supposed is in fact false.
Note too that if someone is a strict univocalist with respect to inference then the reductio and all metabasis is disallowed. A reductio is an inference in a sense that is analogous to the way that, say, a modus tollens is an inference. If -inference- cannot function analogically, the reductio cannot succeed. These are fun little wrinkles in the purported truth-functionality of classical logic. Of course some might in fact disallow reductio and prefer a stricter logically system, but this system will be less powerful vis-a-vis achieving natural inferences.
Quoting Lionino
I don't know if you saw my edit, which may now be redundant:
Quoting Leontiskos
Quoting Lionino
Interesting, thanks for digging into this. Actually thanks for digging into all of the stuff you have dug into in this thread. It has helped me piggyback onto a lot of other ideas.
Quoting Lionino
Very good. This is another instance of the wrinkle that is created when the contradiction is allowed.
Quoting Lionino
Yes, but as @Count Timothy von Icarus and I have noted, it seems simpler to say that (¬(p?q)?p). The antecedent of a negated material conditional is always true, and this goes back to my point in the edit you may have missed above.
Quoting Lionino
Good. This back to 's point about ¬¬a.
Thank you for recognizing my point.
/
For any A, we have these possibilities:
(1) There is a contradiction Q such that A implies Q.
(2) There is a contradiction Q such that A does not imply Q.
(3) For all contradictions Q, A implies Q.
(4) For all contradictions Q, A does not imply Q.
If there is a contradiction that A implies, then A implies all contradictions.
If there is a contradiction that A does not imply, then there are no contradictions that A implies.
So we could state equivalences among (1), (2), (3), (4).
/
Theorems:
(5) (A -> (B & ~B)) <-> ~A
(6) ~(A -> (B & ~B)) <-> A
(7) (~A -> (B & ~B)) <-> A
(8) ~(~A -> (B & ~B)) <-> ~A
From (A -> (B & ~B)) we infer that A implies all contradictions.
From ~(A -> (B & ~B)) we infer that A implies no contradictions.
From (~A -> (B & ~B)) we infer that A implies no contradictions.
From ~(~A -> (B & ~B)) we infer that A implies all contradictions.
There's no metabasis (change).
Again, to show a derivation that a set of premises G proves a sentence ~P, we may use any of the members of G as premises in the derivation, then enter P as a conditional assumption, then derive a contradiction and infer ~P with the conditional assumption P discharged.
There is no switching or "metabasis". Rather, at the very start, we state our premises and stick with them. The conditional assumption P is not one of our premises, but rather it is conditional assumption that is then discharged.
You would do well to look at a specific natural deduction system to see the exact way its rules are formulated. And also to look at a proof of the deduction theorem that is the basis of natural deduction.
This is part of the difficulty. If (b?¬b) is a particular contradiction, then what is a non-particular contradiction? That is what you must ask yourself. When I said things like:
Quoting Leontiskos
...or when @Lionino distinguished proposition-qua-variable from proposition-qua-truth-value, we were both pointing to this same valence where a material symbol (b?¬b) has two legitimately different mental conceptions associated with it. In your language we would say that it can be conceived as a particular contradiction or a non-particular contradiction (non-particular being, in my terms, "falsity incarnate," or FALSE, or ABSURD, and in Lionino's earlier phrasing, contradiction-proposition-qua-truth-value, which truth value is necessarily false as opposed to contingently false).
The analogical equivocity of the particular metabasis under consideration incorporates both particularity and non-particularity simultaneously. This equivocity is precisely what the reductio runs on, for when the reductio is begun the implicit contradiction is considered in particular terms, but by the end of the reductio it has been isolated and re-conceived in its non-particular sense. Or, at the beginning of a reductio when we only suspect a contradiction, that contradiction lives within the system in a particular way and not in a non-particular way. Once it is "ferreted out" it becomes non-particular, a contradiction qua contradiction, and this non-particular sense is what is required for the special reductio inference.
Again: We derive that a set of premises G implies a sentence ~P by using any of the members of G as lines, then entering P on a line, then deriving a contradiction, then inferring ~P with the line for P discharged.
The rule is the same whether we call the line entry of P an 'assumption', 'conditional assumption', 'assumption toward a contradiction', 'assumption to be discharged', 'supposition', 'conditional supposition', 'supposition toward a contradiction', 'supposition to be discharged', 'conditional premise', 'premise toward a contradiction', or 'premise to be discharged'.
Those are merely ways of referring to the line; they handles we use; the logical basis of the rule does not depend and is not affected by what handle we happen to use to describe the line entry.
'non-particular' is your word. It's up to you to say what you mean by it.
There are particular contradictions and we can generalize about them. One such generalization is that all contradictions are equivalent.
If you know what you mean by 'particular', then surely you know what you mean by 'non-particular'? If you can identify a particular contradiction, surely you can identify a contradiction that is not particular?
Perhaps now you are beginning to see the point?
There are particular apples and we can generalize about them. There is no apple that is not a particular apple. But we do say things like "If x is an apple, then x has a core". That is not claiming that there is an apple that is not a particular apple, but rather we can make generalizations about apples.
Quoting Leontiskos
Perhaps now you're beginning to see the point that a poster has no fault in posting several posts, some of them fairly long, in reply to several posts, some of them fairly long. Moreover that it is not unreasonable to post a long post in reply to a short one. And that it is a lot better not to lie about a poster by claiming he said things he did not say and indeed as he said the opposite of what he said. And that it is ridiculous to fault a posters for sometimes needing to delete posts, such as when needing to get rid of an otiose posting box left from a post that was started deferred to another post or not even posted.
/
Back to RAA: You've not shown any fault in RAA, but rather your fault in not understanding it.
Banno may speak for himself, but I don't know what difference in reference you mean by spelling 'false' without caps and with all caps.
Nothing is "reconceived" in natural deduction. It seems you don't know how a natural deduction system is formulated.
Quoting Leontiskos
What in the world? That is not my "language"; it's a bunch of your own verbiage.
I can't make sense of almost all of the rest of your post and similar posts. Probably, I'll focus on the parts that are most blatantly false or ill-premised about the logic you're talking about.
((a?(b?¬b)) ? ¬a) is truth-functionally valid, but the implication in the first half of the biconditional is not the same implication that is used in a reductio ad absurdum.
The easiest way to see this is to note that a reductio ad absurdum is not formally valid, and we can see this by noting that the reductio in 's post (now highly edited) does not prove his conclusion:
Quoting Leontiskos
That the conclusion of a reductio is not formally provable is the first hint that the implication in Banno's formula is not the implication of reductio ad absurdum. If it were then a reductio would be formally provable.
This is related to Lionino's point about the associativity of the ? operator in the case of a contradiction:
Quoting Lionino
And it is also related to Tone's point
Quoting TonesInDeepFreeze
If "A implies a contradiction" were a translation of the sentences, then it would mean ¬A. That is, it would be formally equivalent to ¬A. This is what Banno thinks his truth table has shown, and such is in line with 's claim that "(p ^ ~p) is false in classical propositional logic," as if we could formally translate a contradiction as "false" (whatever that is supposed to mean). Banno's response to Tones should therefore be, "
Quoting Banno
That was my interpretation of Banno, not Banno himself. See the post just above, posted a few seconds ago, for more detail on Banno's view. For instance:
Quoting Leontiskos
My point in response to you is that by limiting yourself to truth-functional logic you don't know what you mean by a "particular contradiction." Unless you think you can use the word "particular" without having any idea what it would mean for something to be non-particular?
It's not a translation of the sentences discussed. That point was recognized by @Lionios who originally claimed it to be a translation.
You said:
Quoting TonesInDeepFreeze
In saying this are you saying, among other things, that these two claims are not equivalent?
(My claim here is that (1) represents a reductio whereas (2) does not, even though thinks his truth table has shown that (2) translates a reductio.)
You lie again. You lie in the face of my having said the exact opposite. And you again make coherent discussion impossible.
I said that I study different logics; I don't limit myself to only a truth functional framework.
But we happen to be discussing your criticisms of a truth functional logic, so it is crucial to be clear what that logic is. Saying that what the logic actually is does not imply that that is the only logic that we may consider.
You repeated your tactic from earlier: Falsely painting my view about logic in general, then using that false painting to incorrectly impugn my statements about the logic we happen to be discussing.
Nonsense, and you entirely failed to answer the question:
Quoting Leontiskos
I think your charges of "misrepresentation" are all bosh, but if you want to prove that you are not limiting yourself to a truth-functional context, then you will have to answer my question. Objecting without answering the question does nothing at all. The rhetorical question is my proof for my claim.
One is a statement in the meta-language and the other in the object language. They are different levels of statement.
Yes, exactly right. :up:
And the point is presumably that the statement in the object-language does not translate the statement in the meta-language, except imperfectly.
From my edit:
Quoting Leontiskos
@TonesInDeepFreeze - The problem here is not so much that you do not know what you mean by 'particular.' No one in this thread has been able to understand what that concept means, even though four of us have now pointed to it (@Lionino, @Leontiskos, @Count Timothy von Icarus, and @TonesInDeepFreeze). Or more precisely, the first three have pointed to the duality between two different conceptions of (b?¬b), and when you used the word "particular" you pointed to one side of that duality. My post was an attempt to get you to see the other side, the "non-particular" side. The problem is only that you refuse to acknowledge the duality, and that the word "contradiction" in the meta-language does not capture the implied ¬a in the object language. That word "contradiction" is the non-particular contradiction.
My own meta-theory about why we can't pin down the exact sense of the duality is that it is because we are trying to specify something that cannot be specified, namely contradiction per se (or in your terms, "generalized contradiction" or non-particular contradiction). We can sort of gesture towards what we mean by a the word "contradiction" in the meta-language, but we cannot pin it down. I would say that this is simply because contradictions lack intelligibility. We are attempting to speak about something that cannot be, and what cannot be in this particular sense in fact cannot be said, at least with any high degree of precision. Nevertheless, we do use what cannot be said when we carry out a reductio ad absurdum.
You claimed that I don't distinguish between material implication and everyday discourse. But I had explicitly said, about three times, that I do recognize that material implication does not accord with many of the everyday senses of 'implies' and even that there are other formal logics with which material implication does not accord. You lied and then lied again.
You claim that I adhere only to truth functional logic. But I had said, at least twice, that I am interested in the study of many kinds of logic. You lied and then lied again And here, you are making claims about classical logic, and so we need to at least be clear how classical logic is actually formulated, and saying how classical logic is actually formulated is not a claim that one must only view logic in terms of classical logic.
There's a lot more I want to comment on in the posts today, but I'm out of time. I hope to have time later.
I didn't. Does that mean you are lying here? Instructively, it does not, because lying is not the same as saying something that is false. Even if, arguendo, I said something false, it does not follow that I was lying. This sort of nonsense is why I have been ignoring your posts, and have only read a handful of them. As Philosophim said:
Quoting Philosophim
I don't have time for silly spats and allegations. If you can't answer simple questions without telling me that I am lying a dozen times then I will just put you back on ignore.
Quoting TonesInDeepFreeze
The conclusion of a reductio is like, "This is an apple." Namely, "This [particular] instantiates the general or universal nature of appleness." The reductio says, "This is a contradiction." So again, to talk about a particular contradiction without a sense of a non-particular contradiction does not make sense. The reductio makes use of both, and this is one of the ways it goes beyond merely formal considerations. The general/universal sense of contradiction is not defined within classical logic. It is a metalogical construct. Specifically:
Quoting Tautologies and Contradictions
This quote that Lionino gave on the first page is accurate as far as it goes, but the trick here is that if (b?¬b) is considered a particular contradiction, then what is the non-metalogical concept of universalized contradiction supposed to be? It can only be (b?¬b), or perhaps (p?¬p). With apples we can point to apples that are obviously different and yet which are all commonly apples. With contradictions there is nothing obviously different about any of the "particulars," and therefore a problem arises. On a formal, non-metalogical reading of 'contradiction,' the reductio which says, "This is a contradiction," would be saying, "(b?¬b) ? (p?¬p)," which is a vacuous tautology given that there is no formal difference at all between the two sides of the biconditional. The move in the reductio of, "This is a contradiction," is opaque to the internal logic.
(The source that Lionino quotes does give a generalized formal note, but I will leave it to another post to ask whether that generalized formal note belongs to the language or the metalanguage, and this also pertains to the interpretation of truth tables.)
Quoting TonesInDeepFreeze
The fiction in the reductio for the formalist is that there is some formal difference between an assumption or premise and a supposition. I say that there is not.
The fiction in the case that you presented is that there is some formal aspect of membership. There is not. If we must consider G as a single entity, then what you say holds. But why do we have to consider G as a single entity? To do so begs the question at hand. For example, in 's reductio we could stipulate that (1) is part of G and (2) is not, and therefore we must reject (2) instead of (1), but this is pure stipulation.
The point here is that when a logician tests some sentence for consistency with a preconceived set of sentences, he is imposing a special order that is not part of the formal structure of the system. The system does not care one whit whether (1) is part of G or (2) is part of G. "G" means nothing to the system.
(Then, regarding plausibility, if the logician deems that P is highly plausible he will destroy and reconfigure G in order to accommodate P. If he is concerned only with formal considerations then he will say, "P is inconsistent with G," but as I said earlier, this is not a reductio. A reductio is not only a claim for inconsistency, but also a justification to reject some proposition.)
My conclusion was that, if we infer that, the logical sentence is at odds with common sense. Because ¬(A ? (B ? ¬B)) (is true) entails A (is true). Well, if from ¬(A ? (B ? ¬B)) we infer that A implies no contradiction, from the fact that A implies no contradiction we may conclude that A is true. When we think about it, prima facie just because something does not imply a contradiction it doesn't mean that it is true. So this inference is at odds with reason.
Because of that, and from the fact that ¬(A ? (B ? ¬B)) ? (¬A ? (B ? ¬B)), I would rather read both as "not-A implies a contradiction". From there, it is much more clear how either of them entails A is true. If the contrary of something is a contradiction, surely that something must be true (from LNC) which is exactly along those lines that proofs by contradiction work in mathematics.
Therefore, I think a more intuitive reading of ¬(A ? (B ? ¬B)) is "not-A implies a contradiction". While it is from A ? ¬(B ? ¬B) that we may infer that A does not imply a contradiction. Truly, as A ? ¬(B ? ¬B) does not entail A. Naturally, the natural language understanding of logical formulas does not change their validity or anything that matters, but an intuitive reading is essential to know in what contexts the formula may be applied.
A full breakdown of these relationships between the logical formulas and the natural language statements, with examples, is here in the quoted post:
Quoting Lionino
Well I forgot what was the stake of the discussion you all are having, but
Quoting Lionino
The idea here is that ¬(A?(B?¬B)) can't become (A¬?(B?¬B)), if such a thing were proper writing, it can only become (¬A?(B?¬B)). When we want to deny the implication of the contradiction, we have to write (A?¬(B?¬B)). However that is what happens when you have a contradiction as the consequent, I will see what happens when you have P (whatever) as a consequent.
Formula: A?B
Reading: A implies B
Checking: A?B, A |= B is valid
The reading seems fine.
Formula: ¬(A?B)
Attempted reading: A does not imply B.
Checking:
¬(A?B), A |= B is invalid.
¬(A?B), B |= A is valid.
¬(A?B), B |= ¬A is also valid.
That is because ¬(A?B) is always False when B is True, so taking ¬(A?B) as a true premise then taking B as a second true premise gives a contradiction, from where everything follows.
Same thing for ¬(A?B), ¬A |= ¬B and ¬(A?B), ¬A |= B, the formula ¬(A?B) is only ever True when A is True, so ¬A gives a contradiction here.
¬(A?B), ¬B |= A is valid.
A does not imply B, B is false, therefore A?
And that is not due to a contradiction, as ¬(A?B),¬B|=¬A is not valid. The argument ¬(A?B), ¬B |= A is valid by itself.
Therefore it seems to me that "A does not imply B" is also not a natural language interpretation of ¬(A?B), and I can't think of any interpretation for it. Likewise, I can't come up with any formula for "A does not imply B".
For ¬(A?B), Claude 3.5 gave me 'A more precise reading would be: "There's a case where A is true and B is false, and B is indeed false, therefore A must be true."', which makes sense as it is the same as A ? ¬B.
I think that "A does not imply B" can't even be put in terms of logic, because "A does not imply B" conveys no information.
Difference when the consequent is a contradiction:
(¬A?B)?¬(A?B) is not valid.
(¬A?(B?¬B))?¬(A?(B?¬B)) is valid.
So when the consequent is a contradiction, the ¬ may be pushed in. But when the consequent is a normal statement, you can't.
Quoting Leontiskos
How can we start inching towards the difference between false and FALSE? First I should say that the proposition (b?¬b) can be either. It can be interpreted as false or as FALSE each time we touch it with our mind. What this means is that terms like (b?¬b) or false are metalogically equivocal or ambiguous given the question we are considering (and because of this 'false' is a bad choice on my part).
It seems to me that in classical logic false in the simple sense is purely relational and context-dependent. To falsify a proposition is to negate it. We can stipulate the falsity of p with the term ¬p, and all this means is that we are affirming the negation of p. This is what I have earlier called contingent falsity, for p is not necessarily false. It is only made false by stipulation or by a contingent inferential consequence (e.g. modus tollens). This is the normal sense of falsity in classical propositional logic.
But what then is FALSE? This is harder to say, just as it is perhaps harder to say what the non-particular sense of contradiction is than to say what the particular is. First I want to note that it differs from 'false' in that it is in no way relational or context-dependent. Second, it has transcended falsity-as-negation. Why? Because, considered as a simple, it is just false, period; and second, because its own negation is ¬FALSE, and this negation means something that is also not context-dependent or relational.
To illustrate, let p = (b?¬b). Usually we can cast a non-simple proposition in terms of a simple proposition, but in this case that fails. The inferential moves that hold for most p's do not hold for this p. For example:
Quoting Lionino
So at the very least FALSE is a "proposition" which is simple and always-false.
My hunch here is that the classical logic system treats everything in this purely relational and context-dependent way and assumes that every non-simple proposition can be cast as a simple 'p' while preserving all of the validity relations. Because of this (b?¬b) is de facto treated in the same manner, even though it does not work to treat it in this manner. When we are handling the "proposition" (b?¬b) we are constantly making exceptions to the normal rules of logic.
Thus if we really wanted to allow sentences to contain (b?¬b), then we would have to add exceptions to all of the rules of classical logic. For example:
The revised textbook would need to add an asterisk: "This is a fallacy*."
" *Except in that case where Q = (b?¬b)"
Right, so it's another case of abnormal behavior occasioned by the contradiction.
Obviously the same thing arises:
((A?B)?¬A) is not valid.
((A?(B?¬B))?¬A) is valid.
And in each of the invalid cases if "B" could be made necessarily false they would presumably hold.
A?(B?¬B), ¬B does not entail anything besides the statements themselves.
(A?B)?¬A, ¬B does entail however (A?B)?¬A, even though (A?B)?¬A is not True for any B, only when B is False.
Quoting Lionino
Right. And I think this would always hold with a contradiction. A contradiction could be replaced by B if a second premise stipulates ¬B. By "each of the invalid cases" I meant your invalid case and my invalid case.
More important than that
A negated material condition is only True when the antecedent is True and the consequent is False. When we know that ¬(A?B) is just (A?¬B), it becomes obvious.
[s]When we state "A does not imply B" as the first premise, we can't conclude B from {A as a second premise}, or A from {B as a second premise}, but we can conclude A from A, and B from B; we also can't conclude ¬A from ¬B, ¬B from ¬A, A from ¬B, B from ¬A, or anything at all besides the second premise. To be sure we are not getting into a contradiction, we must make sure we can't conclude ¬B from B and vice-versa either.
So, with that in mind, I think that "A does not imply B" could simply be put logically as: C. A dummy formula.
C, A does not entail B.
A does not imply B; A; B is not a conclusion.
C, A does not entail ¬A.
A does not imply B; A; ¬A is not a conclusion.
C, A does not entail ¬B.
A does not imply B; A; ¬B is not a conclusion.
C, A entails A.
A does not imply B; A; A is a conclusion.
C, ¬A does not entail B.
C, ¬A does not entail A.
C, ¬A does not entail ¬B.
C, ¬A entails ¬A.
A does not imply B; ¬A; ¬A is a conclusion.[/s]
Retracted. It is dumb.
This is why I think it is more interesting to compare the sense of a reductio ad absurdum to ((a?(b?¬b)) ? ¬a). Common language is equivocal in a way that the sense of reductio ad absurdum is not.
My thesis is that the internal contradiction causes some rules of classical logic to come into conflict, and that ¬a is implied only given some rules and not others. I think the equivalence is stemming from the pseudo modus tollens I noted <here>, and not from a reductio. We still have no proof for ((a?(b?¬b)) ? ¬a). Folks are assuming that the implication is based on a reductio, but it seems more and more clear to me that it is not. FALSE can be represented by replacing the (b?¬b) by C and then adding a second premise where C is false (i.e. modus tollens).
If this is right then (b?¬b) introduces instances of formal equivalence that are not provable.
So similar to this:
Quoting Leontiskos
-
Quoting Lionino
Perhaps it is right to say that the contradiction introduces exceptions to invalidity but not to validity.
Such as?
Quoting Leontiskos
I think that is a valid way to frame it. The thing about (B?¬B) is that, differently from other formulas, it is always False.
Quoting Leontiskos
I don't think that is logically rigorous. As you say, it is not a term in classical logic, and for good reason.
If you want to say A always implies False, A?(B?¬B) is good for that. While A?¬(B?¬B) is "always implies True".
Quoting Leontiskos
If A implies a contradiction, not-A can be stated from LNC.
Dogs are fish. Fish, among other things, is defined as not-mammals. Dog is defined, among other things, as mammal. So we end up with "A mammal is not a mammal". Thus, "dogs are fish" has to be false, so "dogs are not fish" has to be true from LNC.
"... cannot be affirmed" does not stand to me as useful, as the LNC + LEM don't accept a third value.
Neither do I. This distinction between false and FALSE is not my doing. It seems to be another case of Leontiskos confabulating arguments on the part of those who disagree with him. Quoting Leontiskos
Presenting a statement that someone has not made is not presenting a translation.
Quoting Leontiskos I agree with Tones that you habitually misrepresent positions that are counter to your own, here and elsewhere.
_________________
Quoting Leontiskos
I'll agree with that. It is incomplete. As Tones pointed out RAA is an inference rule, not a sequent within classical propositional logic. The inference allows one to infer ~? given a proof of (? ^ ~?) with ? as assumption, a form displayed in the truth table.
Quoting LeontiskosThis is rubbish. Given a proof of B and ~B from A as assumption, we may derive ~A as conclusion. This is the form of reductio inferences and is quite valid.
I think that what has Leo worried is the notion that in an informal reductio with multiple assumptions, we have to have grounds for choosing which assumption we deny. So for example if we have assumption A and assumption B and assumption C, and from these we infer some contradiction, we then have the option of rejecting any or all of the assumptions, and a choice to make.
This is not the case in formal uses of reductio.
Given ?,? ??^~?, we can write that ??~? or we can write that ??~?. (Tree proof 1; Tree proof 2)
Leo seems to think that choosing between ??~? and ??~? somehow involves an act of will that is outside formal logic. He concludes that somehow reductio is invalid. His is a mistaken view. Either inference, ??~? or ??~?, is valid.
Indeed, the "problem" is not with reduction, but with and-elimination. And-elimination has this form
?^? ??, or ?^? ??. We can choose which inference to use, but both are quite valid.
We can write RAA as inferring an and-sentence, a conjunct:
[s]?,? ??^~?? (??~?) ^ (??~?)[/s]
(?^?) ??^~?? (??~?) ^ (??~?)
(fixed error)
...and see that the choice is not in the reductio but in choosing between the conjuncts.
Leo is quite wrong to assert that Reductio Ad Absurdum is invalid.
_________________
For the folks following along at home, the greek letters allow us to write about the sentences of classical propositional logic. We can substitute for any greek letter, consistently, a well formed formula from that logic. "?" is read as "infer". or "we can write" So we can set out modus ponens as
?,??? ??
Read this as "given rho and rho implies mu, infer mu". Substitute any WFF from classical logic into this form, consistently, and you will have a valid inference.
More often folk will use capital letters instead of greek, but here I thought it useful to seperate these out from the use of capital letters in the OP
Such as your example indicates here: . Does classical logic not presuppose that such substitution is truth-preserving?
Quoting Lionino
Yes, indeed.
Quoting Lionino
The problem is that, as I have been trying to explain, (B?¬B) is ambiguous, and can be interpreted as p or as FALSE (i.e. always-false).
Quoting Lionino
You are giving a reductio, so this all goes back to my points about reductios:
Why did you reject (1) and not (2) or (3)? The reductio is not formally valid in that tight sense. So I would maintain my point that what is at stake is not a reductio but rather the modus tollens where (B?¬B) is taken in the sense of FALSE.
Is this just obvious, ?
Yes, it is truth preserving. That
(A?(B?¬B))?(B?¬B) is valid
But (A?C)?C is invalid
does not make me think rules of logic are conflicting, because the equivalence or not with the second term of an «and-operator» is not a rule of logic.
Quoting Leontiskos
As I replied to sime, interpreting (B?¬B) as P is not a good move, for P can be True or False, (B?¬B) cannot be True ever.
Quoting Leontiskos
What about
1 A
2 A?¬B&B
3 ¬A
You would object why I rejected 1 instead of 2? I guess I see your point that it is not valid in a tight sense. After all, from A, A?¬B&B, everything follows, not just ¬A.
Edit: I would say that it is valid, but why choose one rather than the other, then I am not convinced that that is something that follows from logic.
I literally said it was an interpretation, not a translation. I still see it as the better option.
Quoting Banno
This does not contradict what I have been saying. Many in this thread have been concluding ~?, not (??~?) ? (??~?).
Rather than:
This is on point.
Quoting Banno
Tomato tomato. A reductio without choosing between them is not yet a reductio.
Ok, Presenting a statement that someone has not made is not presenting an interpretation. :roll:
Quoting Leontiskos
Quite so. So what? It remains that RAA is a valid inference in classical propositional logic.
Right. As I said, "Perhaps it is right to say that the contradiction introduces exceptions to invalidity but not to validity" (). Still, it seems to me like a general rule that I should be able to denote a complex proposition with a simple proposition - that the second formula should not be able to be made valid by substituting a particular C.
Quoting Lionino
My question then is whether we ever utilize (B?¬B) without conceiving of it as a kind of P.
Quoting Lionino
Yes, good.
So do we have a proof for ((a?(b?¬b)) ? ¬a)? Or is this an instance where the two sides are formally equivalent and yet we cannot utilize logical inference to move from the left side to the right side (or vice versa, I suppose)? I think the only way we can utilize logical inference is by using the modus tollens and conceiving of (b?¬b) as FALSE.
Put differently, is there any other circumstance where we apply a reductio to a single implication and draw a conclusion from it?
If P can only be False, yes; otherwise, no.
Quoting Leontiskos
Uh
Quoting Banno
I think Leontiskos is talking about choosing between the conjuncts, while Banno is correctly stating that reduction ad absurdum is formally valid.
Quoting Leontiskos
Modus tollens is p?q is True, q is False, therefore p is False. «1»
While reductio would be:
p
p?absurd/contradictory
therefore not-p «2»
So I think that a?(b?¬b)) ? ¬a can indeed be proven by modus tollens:
a?(b?¬b)) is True, (b?¬b)) is False, therefore a is False (from «1»).
Proving a?(b?¬b)) ? ¬a by reductio would be:
a
a?absurd/contradictory
therefore not a (from «2»)
I don't see a meaningful difference.
Here's the quote:
Quoting Leontiskos
Am I not allowed to inquire and apply my disjunction as to what you might mean when you say that "in classical logic a contradiction is false"? :yikes: What's your deal!?
Quoting Banno
Hmm? See:
Quoting Leontiskos
It's your term.
This conversation is increasingly inane. Again, I seem to have reduced you to reciting gobbledygook.
Right.
Quoting Lionino
It's the thing I've been going on about the whole time.
Quoting Lionino
One involves a supposition and one does not. The indirect proof (reductio) strictly speaking arrives at the conclusion of a disjunction, whereas a bona fide modus tollens does not:
What I claim is that what is at stake is not a bona fide modus tollens:
Quoting Lionino
Note how you require natural English to give this argument, namely with the premise, "(b?¬b)) is False." That was part of my point in <this post>. Compare the modus tollens you gave to this one:
Are they both valid? As I said:
Quoting Leontiskos
Quoting Leontiskos
Edit:
Earlier quote in context:
Quoting Leontiskos
i.e. It is unclear that "false" means the same thing in both premises. One is necessarily/always false, one is not. Does modus tollens care?
I'm glad you followed this.
It doesn't. Explained last time you made this claim...
Oh, and what is "FALSE"?
Yeah, you said you preferred the reductio to the modus tollens. Clearly @Lionino is following the conversation I am having with you much better than you are following the conversation I am having with him. I am discussing the modus tollens with him (alongside the reductio).
Quoting Banno
I tried to look at this here:
Rubbish.
Leo, what is "FALSE"?
The proof still exists from your heavily-edited post. Why are you editing posts long after they have been responded to? See:
Quoting Banno
Quoting Banno
I literally just gave you a link to a post where I examine that. Did you even read past the first sentence? You're acting like Michael. :roll:
Second argument:
These are both modus tollens arguments. One could construct a reductio ad absurdum similar to either one, but these arguments contain no supposition. They are meant to be direct proofs, not indirect proofs.
The first argument is treating (B?¬B) one way, and the second is treating it a different way. Both arrive at the same conclusion from a diametrically opposed second premise. What I have been saying from the beginning of this discussion is that there are two different senses of falsity at play, which are not being properly differentiated. Surely the diametrically opposed second premises here demonstrate those two senses.
The second premise of the second argument is straightforward, as it treats (B?¬B) like any other negatable proposition and then draws the standard modus tollens conclusion. In this case the falsity of the consequent is based on negation.
The second premise of the first argument is not straightforward, as it treats (B?¬B) unlike any other proposition and then draws an exceptional kind of modus tollens conclusion with but a single premise (it is thought that the second premise is redundant, and in any case it is a premise written in English). In this case the consequent is thought to be false even though not negated. This is what I have been calling FALSE (link).
* This is an instance of FALSE
(@Lionino)
When we do a reductio
A, A?¬B?B ? ¬A is valid
But A, A?¬B?B ? A is also valid
So the question is: how do we choose between either? Isn't it by modus tollens?
p?q is True, q is False ? p is False
a?b?¬b is True, b?¬b is False, ? a is False
And what I am after is a straight forward explanation of what "FALSE" is. Referring back to old posts that do not give a clear explanation anyway is in no way helpful.
Thank you. As I put it:
Quoting Leontiskos
Quoting Lionino
Yes, or rather I would want to say that a reductio is not involved at all. The modus tollens is what is really operative.
---
Quoting Banno
The irony here, Banno, is the post where you pivoted from the modus tollens to the reductio (you literally rejected my modus tollens interpretation of your desire to draw ¬A). In the post you were responding to I attributed the modus tollens argument to you, and you then claimed that I had misattributed that argument. Whatever the case, <that post> still stands. Here is what I told you when you rejected my interpretation:
Quoting Leontiskos
That post is meant to be a tu quoque. I have been arguing against drawing ¬A. You have been arguing for drawing ¬A. That post is meant to say, "If you want to draw ¬A, you will have to tell us what you mean by FALSE." And recall:
Quoting Leontiskos
What you have said is that, "in classical logic a contradiction is false," () and this is the basis of my questions about your position. "A contradiction is false; therefore we can draw ¬A." How so??
Edit: More precisely:
Quoting Leontiskos
Quoting Leontiskos
If ¬(B?¬B) is true, as it must be, then this is not a valid use of modus tollens.
Again, as i pointed out previously, you are comparing two very different arguments. But further your case is not helped by your not setting out the inferential steps in each case. Indeed, your new first example is not well-formed in classical logic. Why is the second line a quote? Was that in error (if so, then I suggest you edit it...). And if you were to modify it so as to be well-formed, what would it look like, if not ¬(B?¬B)?
And how is "(B?¬B) is false" and example of FALSE? Because of the quotes? What do they do?
What you write here is just a muddle.
Because Lionino's second premise was also a quote. It is no coincidence that we are using quotes to express this special kind of modus tollens. Additionally, I pointed to this very fact in my own post. You aren't following along very carefully.
I think you have here managed to set out Leo's confusion far more clearly than has Leo.
Can you see the answer?
Can I? Yes. Do I? No.
Well, first, ?,? ??^~?? (??~?) ^ (??~?) is nonsense. I stuffed up. Am I allowed to edit it? :wink:
It should be
(?^?) ??^~?? (??~?) ^ (??~?)
See the original post where I already explained this interpretation:
Quoting Leontiskos
:grin:
Presumably that's TRUE? Or is it 'TRUE'?
It might help if you explained what FALSE is.
As it stands, I can't take much more of this gobble. It's clear that reductio is valid as used in classical logic.
And since we said ? earlier, we choose ??~??
Even if that solves the problem between the conjunct, A, A?¬B?B can entail quite literally anything. So I wonder why we choose ¬A as an entailment instead of quite literally anything, if not by modus tollens?
The image I posted doesn't help me much.
P?(Q?R), ¬¬P, ¬Q, ¬R entails ¬P
But it also entails anything one wants, including Q, P, ¬J.
So here's the apparent problem:
A, A?¬B?B ? ¬A
A, A?¬B?B ? A
It seems we can infer both A and ~A from the same thing. But that's because the two assumptions, A and A?¬B?B, are inconsistent.
2. If b is true (b ? ~b) is false. If b is false (b ? ~b) is false, so (b ? ~b) is false.
3.~a ? ~(b ? ~b) - contraposition (1)
4. ~a - modus ponens (2,3)
No modus tollens required. You can do it as a disjunctive syllogism too. For the second premise you can assume the b is true or false. Doesn't matter because the conjunction will come out false.
But what I am more interested in would be a system that takes:
a?b
a?~b
a? (b ~b)
As premises with their own truth value, judging whether or not the implication is actually valid (i.e. that the two are even related). I suppose this is what relevance logic is for. It would make tables very long, but it doesn't seem that complicated.
Yes, this seems right to me. I would add that if we have to choose between A and A?¬B?B, I will choose A every time. That is, if for some reason we must take the path of the reductio, why would we choose to affirm A?¬B?B?
This faces the same problems that the modus tollens faces, as your second premise would function just as well for the second premise of the modus tollens.
What is at stake here is premise (2) (which I have called FALSE) along with using FALSE in a standard inference. As @Lionino has shown, it is not safe to assume that an inference which works with P will also work where P=(b?~b).*
* From the start I have argued that this is because (b ? ~b) is a metabasis on falsity (when interpreted as a truth value), and any inference that uses it in this way is involved in this metabasis.
See:
Quoting Leontiskos
This has a place in many philosophical critiques. For example, it underpins the DuhemQuine thesis, and greatly complicated Popper's fablsificationism.
Not quite so in formal systems. In classical logic if assumptions a,b,c,d lead to a contradiction we can only conclude that their conjunct is false.
But this allows is also to conclude certain implications. We might infer for example that a^b^c?~d. That is, if a, b, and c are true, it must be d that is false.
Worth keeping in mind
Well, yes, explosion.
The problem is that this A, A?¬B?B ? ¬A was given as reductio ad absurdum. But this entails anything. However, you gave the RAA as ??(?^~?)?~?, in which case there is no explosion. But then the question is: how do we prove ??(?^~?)?~?? If this is considered the proof of ??(?^~?)?~?, RAA depends on both MP and MT. The issue is that the proof is also explosive:
Tones said "Modus tollens and RAA are in a sense versions of each other and they both do the job.". To me, RAA depends on modus tollens.
Quoting Count Timothy von Icarus
I don't think that follows.
a ? (b ? ¬b), ¬a ? ¬(b ? ¬b) entails ¬a because of the contradiction.
But if we change (b ? ~b) to c: a ? c, ¬a ? ¬c does not entail ¬a .
(b ? ~b) is False, so ~(b ? ~b) is True, ~a implies ~(b ? ~b) but not the converse (3), so we can't derive ¬a from that.
They have two possible routes which could be used to reach their destination: Mt. Caradhras or the Mines of Moria. Gimli suggest that they take the Mines, but Banno knows that "the dwarves dug too deep in their greed, awakening horrors in the depths." So they try the pass of Caradhras, but it turns out to be unworkable, smothered by snow and storm. Do they dare tempt the Mines of Moria? Viewers must wait and see... :grin:
Note: For those considering the mines, two posts may be especially useful: first, second.
Quoting Banno
But my fuller position is that any inference utilizing strange senses of would-be familiar logical concepts must be used with care. I am not opposed to the Mines of Moria, but I don't think people are taking enough care in traversing them.
But couldn't we just assume B here and get ~A just the same? "If B then ~A," seems to work fine here because the conjunct is still going to come up false.
And then we can do the same thing assuming ~B. That covers all our options assuming LEM.
I guess I'm not seeing the trouble here. I can see the trouble with proofs by contradiction in mathematics that prove things for which no constructive proof exists. That makes sense because, on some philosophies of mathematics, an entity doesn't exist until the constructive proof does (and perhaps it can't exist).
To my mind the explosion only occurs if you don't reject either of the two premises. If you reject either of the two premises via reductio, explosion is avoided, no? When we reject (A?¬B?B) and accept A, explosion no longer follows. The same is true if we accept (A?¬B?B) and reject A.
Quoting Leontiskos
And yet your claim that Reductio is invalid is just wrong.
What I have consistently said is that reductio is not valid in the same way that a direct proof is. Perhaps I slipped at some point and called it invalid. In any case, I don't see this as a big mistake. As I said earlier, what is entailed is the disjunction, not some subset of the disjuncts. That is, in this case if we are to avoid the contradiction as a reductio requires that we do, then we must reject either (A?¬B?B) or else A.
RAA is ((S?¬P)?(B?¬B)) ? (S?P) this is valid and not explosive.
((S?¬P)?(B?¬B)), S entails P (valid).
((S?¬P)?(B?¬B)), S is not explosive.
Without S as a true premise, the argument doesn't follow.
S is a statement (or collection of them) that are know to be true, P is the statement to be proven.
(S?¬P)?(B?¬B)
S
? P
The criticism that applies here:
Quoting Lionino
does not apply to:
(S?¬P)?(B?¬B)
S
? P
because it is not explosive.
When arguments are given as A, B ? C, that is "A is true, B is true, therefore C is true".
What does "depends on" do here?
Modus Tollens: ???, ~? ? ~?
RAA: ??(?^~?) ? ~?
Both are equally useable rules of inference.
So, what is a "direct proof"? I gather you think using MT is direct, but RAA isn't? What's the distinction here?
While you are there, what does "FALSE" mean?
You're basically preaching to the choir. <This> is the third time I presented that idea. But a proof that requires an additional assumption is different from one that does not.
Edit: Sorry, misunderstood - I guess I'm wondering how a proof by exhaustive cases fits in. That's what you're introducing, and it's a good introduction.
Quoting Count Timothy von Icarus
I am wondering if allowing contradictions in this way fiddles with the LEM, but this is just conjecture.
Quoting Count Timothy von Icarus
The trouble is simply that (b?¬b) has been consistently creating unexpected behavior in this thread, but I find your approach interesting. Do continue.
The problem is that modus tollens can be proven syllogistically quite easily, but how do you prove that you may derive ~? from ??(?^~?)?
If a ? c it does. Contraposition, flip em and switch em (reverse the order and negate both).
Brad always wears his hat (a) on Mondays (c).
If Brad is not wearing his hat (~c) it cannot be Monday (~a).
a ? c, ¬a ? ¬c does not entail ¬a
Modus Ponens is ???, ?, ? ?.
Not seeing it in 4.
Modus tollens requires no "and-elimination" step. Is that a good way to put it in your language?
Straight RAA does not require the "and elimination". It's an additional step when there are multiple assumptions.
Quoting Lionino
If you don't want to read the posts where I quite sincerely tried to get at this, we could just say that FALSE is what is necessary to get the modus tollens to run with only one premise. It is the sense of the contradiction required for the valid inference.
It is what is supposed to answer this question:
Quoting Lionino
I consider it an open question as to whether this question is answerable.
I have never seen a reductio that does not have multiple assumptions.
Edit: this is what I think a one-premise reductio would look like:
Quoting Leontiskos
I didn't contrapose them when I copied and pasted it, I inverted it.
1. a ? (b ? ~b)
2. If b is true (b ? ~b) is false. If b is false (b ? ~b) is false, so (b ? ~b) is false.
3.~(b ? ~b) ? ~a - contraposition (1)
4. ~a - modus ponens (2,3)
~(b ? ~b) is given in 2
~(b ? ~b) ? ~a
That's not contraposed, I do see now that I didn't contrapose them in the post despite referencing it.
It's ~c ? ~a.
This is of course assuming ~c because c would be affirming a contradiction.
And the disjunctive syllogism
1. a ? (b ? ~b)
2. ~(b ? ~b)
3.~a V (b ? ~b) - material implication (1)
4. ~a - disjunctive syllogism (2,3)
(2) amounts to
2. ~(b^~b)
:lol:
I give in.
Well, I am taking it I have solved the thread. I will also take credit for extending it for 10+ pages by asking this:
Quoting Lionino
If Claude 3.5 is right (I feel it is), you are basically doing modus tollens there.
The problem as I see it is that those who will not move into an analysis of the language are trying to solve a metalogical problem with the logic itself, and this cannot be done. We must move into metalogical discourse, and because of this I would propose analyzing the nature of (b?¬b) and the attendant inferences using English rather than (redundant) logical translations. The logical translations involving that term seem by this point to be clearly underdetermined.
Here is the thing about this. Something not implying a contradiction does not mean that something is true. Of course. We have established that "A does not imply a contradiction" is not a good reading of ¬(a?b?¬b). Yet "A implies a contradiction" is a good reading of a?b?¬b. So if we say "A implies a contradiction" is false, it is the same as saying "A does not imply a contradiction", so saying (a?b?¬b) is false is the same as saying ¬(a?b?¬b) is true (see truth tables).
So, because "¬(a?b?¬b) is true" is the same as "(a?b?¬b) is false", ¬(a?b?¬b) should indeed be read as "A does not imply a contradiction" when it is true.
I can't take this anymore.
Prove RAA without MT?
Interesting problem.
I was already convinced that RAA is insufficient. That as you say:
Quoting Lionino
RAA will not prove ¬A.
Quoting Lionino
The logic of the RAA proves (¬S v P), and the RAA choses one or the other.
Quoting Lionino
The two might not be the same in natural language, perhaps there is the mistake, hence ¬(a?b?¬b) still shall not be read as "A does not imply a contradiction" when it is true.
Thread solved again.
Quoting Leontiskos
I don't understand.
(S?¬P)?(B?¬B)
S
? P is supposed to be the definition of RAA according to where I got it from.
Seems right, you can prove MT with contraposition and MP.
But you can do a disjunctive syllogism too.
Then again, this probably also amounts to the same thing.
What's at all wrong with this?:
(S?¬P)?(B?¬B)
¬P
? ¬S
We're going in circles. Time to go <meta>.
We don't know whether P is true or not. We know S is true. S being true and P being false leads to a contradiction. Therefore we have ascertained that P is true. No assumption is needed or allowed.
Oh, by the way, S does not follow from (S?¬P)?(B?¬B) and ¬P, because that would be a contradiction. Perhaps you meant.
(S?¬P)?(B?¬B)
¬P
? ¬S
But the issue is that we already know S is true.
And: (S?¬P) ? (B?¬B), ¬P does not entail ¬((S?¬P) ? (B?¬B)), so you can't deny the first premise either.
Quoting Count Timothy von Icarus
Mayhaps.
That's not how contraposition works.
Edit. I see you corrected yourself already, nevermind.
??(?^~?) (premise)
~(?^~?) (law of non contradiction)
:. ~? (modus tollens)
Ah. Thanks. So, B and notB are negations of one another while "not B" in normal parlance means different than B but not necessarily contradictory to B.
As I see it, the problem is that this is a misunderstanding of a reductio. A zero-premise reductio makes no sense, and a one-premise reductio misunderstands what a reductio is doing. As pointed out, a reductio is meant to show the inconsistency of some assumption given a set of premises (i.e. more than 1!). My response was here.
When you say "we know S is true" you are stipulating. What does it really mean to "suppose" that P is true? The supposition move is not a purely formal move. The LEM says that everything is either true or false, not that some things are supposedly true. A reductio is doing something over and above a formal move. A system with two inconsistent assumptions results in the dichotomy between the two assumptions, and there is no formal difference between an assumption and a supposition.
Here's another way to put it:
We can say:
But we could equally say:
This is different from the principle of explosion, but it results in the same formal indifference between the two conclusions, given the four assumptions (two of which are called 'suppositions').
When you run a reductio with only one premise you are basically acting out the same principle as, "One man's modus ponens is another's modus tollens." The second "argument" is no more silly than the first. We just think the first is less silly because the premise of the first argument seems to contain more complexity, and therefore it seems to mimic Tones' system of premises. But it arguably does not contain more complexity (and even if it did, it is only a single premise). It is just a simple conditional with a contradiction in the consequent. My following post tries to draw out the way that a contradiction is complex/compound only in a curious and unique way.
This is the formal conclusion, before the and-elimination step of the reductio takes hold:
...which is the same as, "? (A ? (A?(B?¬B)))" We are merely picking an assumption to be true or false, for no reason.
Whether we call (1) a supposition or (2) a supposition is arbitrary, and purely stipulative. There is no formal reason to draw one of the disjuncts of (3) rather than another.
Quoting Lionino
The issue is that we don't know S is true any more than we know that ¬P is true. All we really know is that S and ¬P are inconsistent. Given their inconsistency, one must be false. Picking one to be false without any other information is an arbitrary move.
Now there is something special about (B?¬B) which makes it seem plausible that we could successfully draw the modus tollens inference in this case, even though there is no other case where a one-premise modus tollens is possible. There is a way in which we can conceive (B?¬B) such that the modus tollens goes through. What is this conception of (B?¬B)? Let us call it FALSE, in order to be able to talk about it. FALSE is (B?¬B) conceived in the manner which allows us to draw the modus tollens inference.
FALSE is a kind of emergent property of (B?¬B) which no other conjunction of the form (P?Q) possesses. It is unique, and its uniqueness is what ostensibly allows it to uniquely draw a one-premise modus tollens.
(B?¬B) itself is also unique, insofar as it is both simple and compound. It is compound because it is a conjunction, and conjunctions are compound. It is simple because it can be seen to be false as a whole, without any variable assignment (hence the simplicity of FALSE). Both are necessary given the fact that a modus tollens requires the consequent as a whole to be false; and that there is no non-compound basis for something which is false without any variable assignment (and when we combine this with the fact that (B?¬B) is a unique compound proposition,* we see why every other modus tollens requires a second premise).
Now a modus tollens requires that the consequent be false:
Quoting Leontiskos
The modus tollens inference will be valid if two conditions are met. First, it must be the case that (B?¬B) can be legitimately interpreted as FALSE. Second, the modus tollens must be able to support FALSE as a consequent:**
Quoting Leontiskos
If both conditions hold then the argument cited at the beginning of this post is valid, and ¬A can be drawn by modus tollens. If at least one condition fails then the argument is not valid and ¬A cannot be drawn by modus tollens.
The more general point here is that the uniqueness of (B?¬B) presents us with a difficulty: can it be used in an exceptional way within standard inferences, or not? Can it be treated as FALSE in order to produce strange inferences, such as one-premise modus tollens, or not? And is there a principled way to decide this question? Is there a way to know the ways it can be used and the ways it cannot be used? I dont know, and my point throughout the thread is that I am wary of this whole approach, even though it seems intuitive to many.
* I take it that formally equivalent propositions such as ¬(B?¬B) do not count as separate bearers of FALSE.
** I realize I am mucking up the original definition of FALSE a bit here, but that is hard to avoid. The point is only that the interpretation must be both supportable, and it must suffice for the inference.
(@Lionino, @Banno, @Count Timothy von Icarus)
(I am probably going to need to begin moving away from this thread.)
This is perhaps my favorite proof for the modus tollens thus far. The question is whether that second step justifies the modus tollens. Does the "law of non contradiction" in step two allow us to think of the contradiction as a simple kind of falsity, which requires no truth-assignment? And if so, does that thing (whatever it is), allow us to draw the modus tollens? These are the questions I have been asking for 12 pages.
See my posts <here> and <here> for some of the curious differences between (?^~?) and ¬(?^~?).
This isn't a proof of Modus tollens. This is a use of Modus tollens.
You've been asking for 12 pages for a proof of Modus tollens?
From within a theory T, the statement S is known to be true. We don't know whether P is true or not.
We verify that S?P implies a contradiction.
Thus P cannot be the case within that theory T.
That is paramount for proofs from contradiction in mathematics.
Quoting flannel jesus
That is modus tollens. My question is why do we accept ~? from ??(?^~?) if not by modus tollens? I don't think we do. The RAA was given here as ??(?^~?)?~?, this statement seems to be justified by MT. The RAA I found elsewhere is (??¬?)?(?^~?),???, which is different [hide="Reveal"](whether it depends on MT or MP or whatever I don't know, but it doesn't even matter in the end as me and Count discussed)[/hide].
I edited that post a bit, perhaps after you read it. For example:
Quoting Leontiskos
Quoting flannel jesus
I was not trying to say it was.
We do because S fully follows from the axioms of a theory.
Ok so we're playing a game (I don't mean that pejoratively, I like games) where we have to prove the conclusion without using modus tollens, is that right?
What are the rules of the game? Are we allowed to use the rule of non contradiction?
You said before that the proof for modus tollens is easy - does that proof obey the rules of this game? If so, which proof of Modus tollens do you like? There are multiple, I want to make sure I'm using the right one.
I will play this game, if you answer my questions then we can have a solution.
Which premise do you think provides us with such information?
(S?¬P) does not favor S over ¬P in the case of a contradiction. If a contradiction follows from (S?¬P) it is no more rational to reject ¬P than to reject S.
I think you are referring to background conditions that are not formally present, hence my point. If we really had a set of axioms and a theory instead of a single assumption, then you could say that it follows from a theory. In this case we do not have that.
But if RAA is given as (??¬?)?(?^~?),???, it is a valid move on its own, distinct from MT.
That is all.
S.
Quoting Leontiskos
(S?¬P), S does.
Edit: (S?¬P)?contradict, S does.
Yours wasn't a proof of MT, it was MT itself. MT can be derived from MP and contraposition.
And note that you supposed ¬P (which is the same as preferring S). Either way its a random pick for the second assumption. The point here is as I have said:
Quoting Leontiskos
Or disjunctive syllogism. I did verify that this is a way to prove MT.
I didn't suppose ¬P.
I find out that (S?¬P) leads to a contradiction within that theory. Not an assumption.
I know that S follows from the axioms of the theory. Not an assumption.
Conclusion: P.
Socrates is a man. Not an assumption.
Men are macroscopic. Not an assumption.
Socrates is macroscopic. Conclusion.
Ok, I'm going to assume you mean this proof (the one wikipedia lists as "Via contraposition"):
1
P? Q (Given)
2
¬ Q (Given)
3
¬Q ?¬P (Contraposition (1))
4
¬ P (Modus ponens (2,3))
This is the proof of modus tollens that you like - it proves modus tollens without assuming it, correct?
So it's pretty straight forward to use the same format, I'll take my previous argument which assumes Modus Tollens:
??(?^~?) (premise)
~(?^~?) (law of non contradiction)
:. ~? (modus tollens)
And reformat it to be in the style above, the proof that you like of modus tollens that doesn't assume modus tollens:
??(?^~?) (premise)
~(?^~?) (law of non contradiction)
~(?^~?)?? (contraposition)
:. ~? (modus ponens)
"All men are mortal; Socrates is a man; therefore, Socrates is mortal" is of no value, since we could not know that the premise, "All men are mortal" is true unless we already knew that Socrates is mortal. So we learn nothing from the syllogism.
https://philosophy.stackexchange.com/questions/104220/do-you-gain-further-truth-from-syllogisms
I didn't really talk about proving MT at any point, so I don't know why you are asking that.
You don't know why I'm asking what?
You presumably have a proof of Modus tollens that you like, that doesn't itself assume modus tollens.
Whatever format of argument that proof takes - that argument that doesn't assume modus tollens, but proves Modus tollens - I can use that exact same format of argument to prove Y similarly without assuming Modus tollens.
Give me a proof of mt you like that isn't circular.
If you believe that's the same as assuming mt, then that means the proof of mt that uses those two assumptions is circular.
Assuming contraposition and MP is the same as assuming MT.
Sorry I misread a quote from above. You are right. You supposed S.
Quoting Lionino
You know equally well that ¬P follows. Conclusion: ¬S.
You are importing "the axioms of the theory." They are nowhere to be found. They are background conditions, absent from your proof.
Rule 1. Don't assume mt.
Rule 2. Don't simultaneously assume contraposition and MP
Can I assume MP if I don't also assume contraposition? Can I assume contraposition if I don't also assume MP? Are there any more rules you haven't explicitly stated yet?
Are there any proofs of MT that obey the rules of the game we're playing? Obviously the one you've been talking about doesn't obey .
Quoting Banno
(i.e. What is the difference between a direct proof like modus tollens and an indirect proof like reductio ad absurdum?)
I said:
Quoting Leontiskos
Put differently:
Quoting TonesInDeepFreeze
A direct proof requires no recourse to the meta-language. When the reductio identifies a contradiction it is dipping into the meta-language. That exchange earlier with Tones was about whether the reductio is truth-functional. It turns out that you cannot represent a reductio in the object language.
Another way to put it is that in modus tollens we have two premises whereas in reductio ad absurdum we have a premise and a supposition, and the difference between a premise and a supposition only exists at the level of the meta-language.
Edit: Indeed, this is instructive given that the unique <modus tollens> we are considering also <uniquely requires recourse to the meta-language>. No other modus tollens requires recourse to the meta-language. Nevertheless, the recourse that it requires is different from the recourse that a reductio requires. <If we avoid the meta-language we will only continue banging our heads against the wall>.
(@Lionino)
This of course makes the argument you brought up for MT circular. That's fine, we can move past that and find one that obeys the rules presumably.
Another interesting point goes to natural language. "A?(B?¬B) means ¬A."
Compare:
Without recourse to the meta-language, there is no way to adjudicate. I think this goes back to 's point.
This sounds like the "Scandal of Deduction," and it actually holds not just for syllogisms but for all deterministic computation and deduction. From an information theoretic perspective, because the results/outputs of computation and deduction always occur with a probability equal to 100% it follows that they are not informative. Everything contained in the conclusion must be contained in the premise; we learn nothing from deduction. The premises must always assume the conclusion.
For some reason, philosophy has generally taken Hume's Problem of Induction more seriously than this problem, but they are equally intractable from a formal perspective. The entire early modern move to prefer deduction and "analyticity," and it's continuation in analytic philosophy, essentially just ignores that deduction is as undermined as induction.
I wrote an introduction on this a while back for 1,000 Word Philosophy, but they weren't interested in the topic.
https://medium.com/@tkbrown413/introducing-the-scandal-of-deduction-7ea893757f09
I do believe I have a solution here that goes beyond psychologism and grounds an answer for why deduction is indeed informative in physics and biology: https://medium.com/@tkbrown413/does-this-post-contain-any-information-3374612c1feb
The problem shows up because logicians, who tend to be the folks most interested in this problem, only look for formal solutions. But the issue is that "eternal implication," or "implication occuring outside time" is assumed. We can think of computation abstractly, but it remains defined by step-wise actions. Yet these abstractions are taken to be "real" as opposed to merely tools.
However, in the brain or in digital computers two things hold:
1. Computation always occurs over time.
2. Computation involves communication and can be thought of in terms of communication models (some very good work on this has been done and the two end up being almost the same thing, "information processing" indeed.)
Hence, deduction is informative because it involves communication. A message cannot be received before it is sent.
Floridi and others have tried very complex formal explanations and I think these just miss the point. What could be more "surface level" information than what color font a word is written in? We recognize color automatically, seemingly in "no time at all." The color "eternally" implies the word.
Except there is the Stroop Test. If you spell out the names of colors and put the font in a different color, people have a hard time reading off the color of the font. They take much longer. Their error rate goes from virtually always 0 to something fairly high if they are trying to do it quickly.
Why? Because deduction/computation, be it in computers or humans, always involves communication and must occur over some region of space-time, not "all at once and all in one place." Aristotle gets at this in his essentially processual conception of demonstration in the Posterior Analytics.
Again, the criticism I had of Wittgenstein for assuming implication is the "real deal," and causality is "superstition" applies here.
It seems plausible that:
It's as if (?^~?) can be whatever we need it to be for our current purposes, and this should not be surprising.
Note:
Yes, I was thinking about this as well.
Some might have it that 2+2 is just another name for 4. 2+2 is 4.
And in some sense it might be right to think of it this way, but not in terms of computation. Why?
Consider a program P that solves Hamiltonian path problems like the "traveling salesman," through brute force. Then consider an input I for that program with an absolutely massive number of nodes. Then consider O as the output of the program, which figures out the shortest path between all the nodes.
In a certain sense, we might be inclined to say that P(I) = O in the same way we would like to say 2+2 just is 4. However, if I includes enough nodes then all of the world's super computers running P(I) until the heat death of the universe still won't have been able to actually compute O yet.
So then, in a very important functional sense P(I) is not "the same thing as O." If we let abstraction get taken for reality we end up with some weird "problems."
This shows up with non-constructive descriptions as well. "The first number that violates the Goldbach Conjecture," is a rigid designator (if such a thing exists). However, there is loads it doesn't tell us, like what digit the number would start with. Discovering this would seemingly require some sort of Herculean computational effort (given a simple search, and given it exists) even though we have a rigid designator description of what we are looking for.
Well, what to make of this? Perhaps mathematics is better thought of in terms of signs and relations instead of identity. This will all be explained in my forthcoming magisterial book introducing Hegelian-Semiotic-Process-Thomism, also to be known as "The Correct Philosophy." :cool:
Quoting Leontiskos
Note that this is a sufficient condition and not a necessary condition. The same thing can be expressed in terms of the "meta-language":
Quoting TonesInDeepFreeze
Whenever some logical move requires recourse to the meta-language, we are involved in metabasis. <The three senses> of interpreting a contradiction that I set out are all utilized in the service of a metabasis. This sort of ambiguity always attends metabasis. Sorting out the ambiguity requires us to go beyond the object language at hand.
Weren't you mocking me earlier in the thread for relating this side of logic to the principle of explosion? Or was that someone else?
This is really the problem of knowledge as expressed in places like the Meno:
Aristotle applies his notions of act and potency to basically say that in knowing something partially we can come to know it more fully. When the mind engages in argument this is what it is doing, according to Aristotle. We are unfolding implications previously unseen.
Right, and this is related to my claim:
Quoting Leontiskos
I believe that given the way formalized logic works, there can be sentences which are formally equivalent and yet underivable from one another. According to Sime one implication of this can be seen in terms of Peano arithmetic (link).
Quoting Count Timothy von Icarus
Going back to Meno, if argument was not temporal then we could presumably never gain new knowledge. The other interesting question is how to account for forms of non-temporal knowledge.
Quoting Count Timothy von Icarus
But probably only because it is NP-complete. When P is not NP-complete it is a more difficult question whether P(I) is the same thing as O. P(I)=O and 2+2=4 are very different in that sense.
I'm not sure it is. Consider zip bombs, which were a way to overwhelm PCs and make them crash or ton overwhelm anti-virus software. The zip bomb is just a highly compressible file, something like a a text block of nothing but a the same few letters repeated a tremendous number of times. This is extremely compressible because it can be summarized as something like "print XYZ 10¹?? times."
What makes the Hamiltonian Path problem intractable is precisely the extremely large number of operations and this can be true for any program provided it has enough steps.
E.g.:
Excel used to be terrible at this sort of thing in terms of using up a ton of resources to try to visually represent (relatively) small amounts of data.
Not sure what you have in mind here. Non-temporal knowledge makes me think of God's self knowledge in Plotinus or something like that.
Some posts ago, I shrugged the issue aside by saying the two don't mean the same thing. Well, when is "A implies a contradiction" False? When A is True. And before, we decided that "A does not imply a contradiction" is a "translation" of A?¬(B?¬B).
Here is the problem: A entails (A?¬(B?¬B)), which means that A entails {«A implies a contradiction» is false}; but (A?¬(B?¬B)) does not entail A, which means that {«A implies a contradiction» is false} does not entail A. That agrees with common sense.
Since one entails the other but other does not entail one, we may say that everytime «A implies a contradiction» is false, «A does not imply a contradiction» is true; but it is not everytime «A does not imply a contradiction» is true that «A implies a contradiction» will be false. Therefore there is an assymetrical relationship between the two statements quoted.
The prover confirms my intuition:
(a?¬(b?¬b)) does not entail ¬(a?(b?¬b))
¬(a?(b?¬b)) entails (a?¬(b?¬b))
You might want to ship it with lubricants, it sounds unpenetrable.
Quoting Leontiskos
They are found in "S". Or you can just replace "S" with axioms of the theory. Axioms are naturally assumed.
Quoting Leontiskos
I don't. I know that S and ¬P can't coexist. I know that S, so ¬P can't be the case. ¬¬P is P.
Tidbit: my notifications:
Which one of the above phrases are you saying is the english translation of (a?¬(b?¬b))?
It's either «A implies a contradiction» is false
or «A does not imply a contradiction» is true, right?
But consider this:
((a ? ¬(b?¬b)) ? ¬a) ? (a ? (b?¬b)) is valid
(a?¬(b?¬b)) doesn't actually stop a from implying a contradiction - it can be assumed true, and still be the case that a implies a contradiction.
¬(b?¬b) just means (b v ¬b)
«A does not imply a contradiction»
(a?¬(b?¬b)),¬a entails (a?(b?¬b)). That is a problem. Now I don't know what «A does not imply a contradiction» is in logical language, as a?¬(b?¬b) and¬(b?¬b)?a also don't work . Perhaps the question is spurious.
https://www.umsu.de/trees/#~3a~5(a~5~3(b~1~3b))
https://www.umsu.de/trees/#~3a~5(a~5(b~1~3b))
I would reword it from "a does not imply a contradiction" to "a implies this particular non-contradiction".
And when a is false, it implies everything, contradictions and non contradictions.
How not? There are several algorithms to solve mathematical problems in this context being symbol manipulation from one form to another. For example, solving square matrices of dimension NxN through reduction of order requires n^3 operations. If we have a 300x300 matrix, aren't the matrix and the determinant related to each other intrinsically anymore?
Twenty years ago, many things that are computable now would not be computable, does it make X?Y then but X=Y now?
¬A,J entail (A?(B?¬B))
If ¬A is a premise, it will always entail A?(B?¬B), no matter what the second premise is. Why? Because, if the second premise is the denial of ¬A (¬¬A which is A), it will be explosive therefore everything goes; if the second premise is not the denial of ¬A, A is given as False (because ¬A is given as True) and so A?(B?¬B) is True because B?¬B just means False in all cases.
Quoting flannel jesus
The issue is that if you say "non-contradiction" it can mean pretty much mean anything, while ¬(B?¬B) is not just any non-contradiction but something that is always True no matter what. That is a detail, the goal anyway is to translate "A does not imply a contradiction", not any other phrase.
The phrase «A does not imply a contradiction» really means specifically «A being true, it does not imply a contradiction». I think this meaning is indeed encapsulated in A?¬(B?¬B), especially when it can be translated as «A implies True».
The more global meaning of «A does not imply a contradiction» is «A being true or false, it does not imply a contradiction», which is not something we would say in everyday language, after all, one of the conditions of «A being true or false, it does not imply a contradiction» is that «A being false, it does not imply a contradiction».
Does the phrase «A being false, it does not imply a contradiction» make any sense whatsoever?
This might seem crazy to you, but I would translate that as just A, or in other words, A is true. If you are stating, for a fact, in classical logic, A definitely doesn't imply any contradictions, and we also know that false statements necessarily do imply contradictions, then the only way to say "A definitely implies no contradiction" is to say "A is true"
Quoting Lionino
For clarity, "both mortal and immortal" here is just meant to stand for any contradiction, in an "educational" manner.
The advantage of (A ? ¬(B and ¬B)) is that when you say ¬A (that A is False), it does not entail A.
But if your "translation" is just "A", we have a contradiction and therefore everything follows. Thus, I think that just A is not a good way to put "A does not imply a contradiction".
Furthermore, (A?¬(B?¬B))?A is invalid, so my suggestion and your suggestion are not agreeable between each other.
Is that the right English translation of that?
Keep in mind that ¬(B and ¬B) is equivalent to B or ¬B - would you say A ? (B or ¬B) can be worded as "Elvis is a man does not imply that Elvis is both mortal and immortal"?
For example:
'Beautiful', everybody knows what that mean; 'not beautiful', does that necessarily mean ugly? No. If "beautiful" is P, converting "not beautiful" as ¬P will give rise to things that violate common sense.
All in all, as you said, classical logic is useful for constrained cases, we are abusing its good-will here.
Take Javi's thread https://thephilosophyforum.com/discussion/15333/ambiguous-teller-riddle/p1 , where a third value is needed.
That was my proposal at least, as I proved, even before that post, that it cannot be ¬(A ? (B and ¬B)).
Quoting flannel jesus
I am obliged to say yes here. So yes.
It positively implies something, rather than "does not" imply something.
Thinking of the two as timelessly equivalent leads to the Scandal of Deduction. P(I) should allow us to know O in "no time at all," on the view that they are the same exact thing. In reality, no computation occurs in "no time at all."
Consider a second program Q that simply takes the output of P as its input and prints it to tell a user what the shortest path is.
Well, if we have Q(P(I)) it is clear that Q cannot print until P is finished computing. But if the output of Q is equivalent with Q, this is a strange thing indeed.
A helpful way to think about this is by looking at the way physical computation always involves communication and the ways in which it can be explained/moddled in terms of communication theories: https://link.springer.com/article/10.1023/B:MIND.0000005133.87521.5c
Likewise, 2+2=4 can't be the same thing as 2+2 is 4 in terms of computation for other reasons. There are an infinite number of programs with 4 as their output (given some input). But (6+2)/4 is not the same computation as 2+2. If programs just are their outputs then all sorts of completely different programs, some very quick to execute, other intractable, all end up being the same thing. But then why are some intractable, taking millions of years, and some take a millisecond if they are identical? Clearly they aren't identical in every way.
Hence why I said, "in some sense it might make sense to think of 2+2 is 4." However, this assumption seems flawed when one discussed computation in general and physical computation in particular.
But the "Scandal of Deduction," is about why we find the results of deduction and computation surprising and informative. We are physical beings. We do not compute in "no time at all."
That much I can agree with under a formalist/nominalist view of mathematics. If we take that mathematical objects are platonic objects, where, by definition, the two would be timelessly equivalent, does the Scandal of Deduction make mathematical platonism troublesome?
Another question, what do you think of the rebuttal:
I think the above comment can be exemplified in:
? There is a strand of DNA that makes a being a man (hypothetical simplified essentialist definition of 'man').
? By the laws of chemistry, every DNA degenerates after X years, DNA degeneration leads to death.
? Everyone with that strand of DNA dies after X years (follows from ?).
? Socrates has that strand of DNA, Socrates will die after X years.
If you think the above is an exemplification of the rebuttal, do you think it is a valid argument? The only way I see is by denying ? which is denying the laws of chemistry, which is back to the problem of induction (how do you know every DNA degenerates?).
Sum:
PS:
Quoting Count Timothy von Icarus
Is this about all computation then and what I wrote above irrelevant?
I think that is possible too, surely. Aren't perhaps "Elvis is a man does imply that Elvis is not both mortal and immortal" and "Elvis is a man does not imply that Elvis is both mortal and immortal" the same thing?
I would say that in natural language, the former would mean that "Elvis is a man" is a piece of information that tells us there is no such contradiction; while the latter that "Elvis is a man" is a piece of information that doesn't tell us there is a contradiction. Do the two mean different things?
I don't think so. However numbers exist, they don't seem exist in the way physical systems do, so this is not strictly an issue.
It is however relevant for how the assumptions of platonism affect how we think of physics. Folks like Gisin have argued, somewhat convincingly I think, that assumptions about math have been key to how physicists think of physics. In particular, these assumptions are used to support claims about eternalism or "the block universe" (as opposed to local becoming, the growing block universe, etc.). His argument is that intuitionist assumptions make a better grounding for physics.
I also think these issues could be used as an argument against platonism, something to the effect of: "the way number is actually instantiated differs radically from the assumptions of platonism." However the two aren't necessarily in conflict.
I don't think it gets around the problem of the premises having to include the conclusion. The conclusion still always follows with p=1 and must in a way be "contained" in what one starts with.
That quote also seems to assume some potentially problematic things about what constitutes a "natural law," but that's sort of aside the point.
It perhaps depends on how much you embrace pancomputationalism in physics and information theoretic explanations of the other special sciences. If we think number only appears in nature in terms of computation and process then that seems suggestive at the very least. But I don't think process metaphysics ultimately rules out more platonesque views like Hegel's doctrine of concept and essence, or the Patristics differentiation between logoi and Logos.
The entirety of this debate revolves around a vague OP:
If you are asking:
"Is 'A -> B && A -> !B' a contradiction (i.e., itself contradictory)?"
Then the answer is clearly no. If you are asking:
"Is 'A && A -> B && A -> !B' a contradiction (i.e., itself contradictory)?"
Then the answer is clearly yes.
Everyone is just interpreting the OP's question, because it is too vague, one way or the other.
https://en.wikipedia.org/wiki/Barbershop_paradox
The way I stated it was no more or less vague than it needed to be for it's relationship to that.
An NP-Complete problem is, among other things, one that has no known polynomial time algorithm/solution. The point being that your P(I) = O is an approximate solution, not a deterministic solution. If O is only an approximation of a solution then of course it deviates from the ideal solution, and from an isomorphic relation.
See:
Quoting Leontiskos
So:
Quoting Lionino
What is happening is apparently:
As noted earlier in the thread, a reductio is not representable in the object language, and therefore what you present is not a reductio in any formal sense.
The first thing to note is that the conclusion is ¬(P?Q). In the first place P and Q can both be false. But if we add an additional condition that they cannot both be false, ¬(¬P?¬Q), then to stipulate that one is true or false automatically determines the other value, and yet there is no principled reason to stipulate such a thing apart from mere stipulation. Or, if we stipulate that one is true, as you did, then we must accept that the other is false, even without the additional condition. Still, we have no reason to stipulate P instead of Q. There is no theory here or set of axioms, except in a purely stipulative and imaginary way. P and Q are exactly on a par as far as the formalization goes.
From the supposition we learn (P?¬Q), at which point P must be further asserted beyond supposition if we are to actually arrive at ¬Q:
(To suppose P and to assert P are here two different things)
..But it always goes back to the question of why we preferred P in (4) rather than Q.
Note too how this is different from a reductio:
Quoting Lionino
...insofar as your supposition produces no contradiction at all, and the thing you suppose is never rejected. This sort of underlines that you are merely supposing that something is true without any reason. This is what confused me in my initial reply. When you called your proof a reductio I assumed your supposition was being rejected.
I don't see it this way. I think the phrase, "A does not imply a contradiction," either means, "A implies something and that something is not a contradiction," or else, "Whatever is implied by A, it is not a contradiction." Both of these are examples of meta-language, and neither is represented by A?¬(B?¬B). The first means, "A implies B and B is not ¬A." The second means that whether or not A implies anything, it does not imply ¬A.
In any case we have to distinguish these two propositions:
Then why not reference that in the OP? Otherwise, it is simply too vague.
This is a poorly translated sentence into logic: the fact that allen never leaves the shop without brown does not imply that brown cannot be in the shop when allen is not there. On the contrary, it just means that when allen was present in the shop and then left (or brown was in the shop when allen was there), one can be sure that brown (or allen) left with him; which can't be expressed accurately without temporal logic. E.g., just because you know that "when my niece is with me in walmart, she will not leave the store without me", it does not follow that "when I am in Walmart, my niece is with me"...that's just bad logic. Even when considering the other axiom (that at least one is there) it does not follow that ¬A ? ¬B from the sentential form of the axiom.
If I grant the logical form which they are translating to (such that is the axiom is actually '¬A ? ¬B'), then it still the case, as I mentioned before, that there is no paradox at all: 'A ? ¬B' and '¬A ? ¬B' are not contradictory. Material implication is such that 'A ? B' is the same as '¬X ? Y'.
Am I missing something?
Because I wanted to talk about the question of their contradictoriness without the baggage of the riddle. I think most people got the correct answer, and understood the question, anyway.
I'm not sure what your getting at here. It's not an approximate solution, in the example the program spits out the shortest path between all the nodes, not an approximation. The computation is deterministic. It can only be completed in polynomial time by a non-Nondeterministic Turing Machine, but a deterministic Turing Machine can churn through checking every path, it just takes forever if the input is large.
And the NTM doesn't spit out approximations either, it's just able to take many branching paths through the computation, doing multiple actions based on a single instruction (I was always told to think of this as splitting the NTM into copies in a sort of branching process). People use estimates for these sorts of problems, but that's because they take too long to solve.
Quoting Count Timothy von Icarus
If we have a large number of nodes with infinite time then P(I)=O will produce the ideal solution, it will be unique, and in that case what you conclude no longer holds:
Quoting Count Timothy von Icarus
If we don't have infinite time then there is no sense in which P(I) is the same thing as O, but if we have infinite time then there is an important sense in which P(I) is the same thing as O.
But my point is that it isn't plausible to compare 2+2=4 to an NP-Complete problem and then claim that 2=2=4 suffers from the same limitations as the NP-Complete problem. 2+2 isn't NP-Complete. The basic objection to your argument would be, "Well I agree that P(I) is not the same thing as O, but it doesn't follow that 2+2 is not the same thing as 4."
Yes, given an "infinite amount of time," or an "eternal realm," it might make sense to think of these relations as eternal. Computation is inheritly stepwise though, so I am not sure this is particularly helpful to do. At any rate, no physical system computes in "no time at all." Human beings stop being able to see expressions as "other names" for their solutions at remarkably low levels of complexity.
Consider: 6X = 1929.121? (118 + 2/3) 1/918
The value of X is specified here. Is it obvious to you what it is? But surely you know what all the operators are and how to solve for X.
I don't think most people will even be able to glance at 175+39 and instantly tell what number it corresponds to. Hence our intuition that computation is informative
Anyhow, I think the Stroop Test (mentioned earlier) is a better illustration of the way in which cognition relies on communication.
I see now.
Quoting Count Timothy von Icarus
Quoting Count Timothy von Icarus
Quoting Count Timothy von Icarus
"2+2 has a relation with 4 that is informative."
Something that I can think of is that, for Peano, 4 is exactly S(S(0))+S(S(0)), which 2+2, which is S(S(S(S(0)))). I understand you will then reply that S(S(0))+S(S(0)) and S(S(S(S(0)))) are different things whose identity will be established by a computation.
Now, if I were to play the platonist, I would say that both S(S(0))+S(S(0)) and S(S(S(S(0)))) are symbols that point to the same thing. Just like "2" in Times New Roman and "2" in Arial are two symbols that, to an intelligent mind, reference the same universal. Though you will then reply that I must perform the computation to be able to reference the same universal. In that sense the two are not contradictory.
Were I to play the nominalist, S(S(0))+S(S(0)) and S(S(S(S(0)))) are both symbols that are constructed in relationship to each other (so in that sense your criticism that they are not quite the same is sensible) but that nevertheless are applied the same (in physical theories).
But even, and especially, for the nominalist, I think 2+2=4 can be informative while maintaning an identity. 2+2 and 4 are indeed not the same object, but as soon as you establish 2+2=4, you are giving a hint as to what kind of mathematics you are doing. That is, 2+2=4 doesn't give you all the information you need, as there are different arithmetics where 2+2=4, but 2+2=4 rules out many arithmetics; but that is why we have formulations of those different arithmetics and the person using them needs to know how to use them. Peano arithmetic is the one we use the most in everyday life doing taxes and splitting a bill. But we use module-60 arithmetic when adding up minutes of time. Within Peano arithmetic, 2+2=4 is the same, so it is indeed not informative within Peano arithmetic, but it is informative when Peano arithmetic is applied in real life: pushing two rocks against two rocks gives you four rocks. Peano arithmetic is applicable to this structure, thus the application gives us information.
The less obvious computations are [math] \int_{0}^{2}\frac{3x^2}{2}dx[/math] that gives us 4, but [math] \int_{0}^{2}\frac{3x^2}{2}dx[/math] is descriptive exactly of the speed of an accelerating car and knowing that a certain arithmetic also applies to the situation will give us the information of the total traveled distance.
Of course, this might be circular on the application level:
PA applies to a structure.
PA gives information about the structure.
The information we get is assumed in the first premise.
But I know everything you said about computation is still going to apply:
Quoting Count Timothy von Icarus
I don't know anything about either of those theories, so I can't say. But it doesn't feel like this conversation applies to mathematics per se.
Quoting Count Timothy von Icarus
Like the denial of LNC?
Note so that, hopefully, Leontiskos will desist from mischaracterizing my views: In all these posts: I am not claiming that classical logic is the only correct or useful logic. I am not claiming that classical logic accords with the many notions of logic and language in everyday discourse. And I do not claim that classical logic should not be critiqued. (Indeed, such critiques as from constructivism, predicativism, finitism, relevance logic, etc. are rich sources).) But when classical logic is being critiqued, it should not be mischaracterized, misrepresented or misconstrued, so explanations of how classical logic actually operates are productive. And giving explanations of classical logic does not imply advocacy for it or a presumption that it is the only formal logic that should be consulted. I admire the thought that goes into formal logic; I enjoy studying it; I appreciate its role in formalization of mathematics; I appreciate its use in such things as computing; I appreciate that it is the subject of much of philosophy of mathematics and that it generates rich questions in mathematics and philosophy. And, in my own limited way, I have studied other formal logics and have read and appreciated critiques of classical logic. I do not claim that anyone should even care about formal logic, but when people do talk about it, they should get it right, and it is eminently proper to remark when they don't.
Quoting Leontiskos
That is false. And it is said without basis. I know what I mean by the word. I use it in its everyday sense.
Quoting Leontiskos
What the concept of 'a particular contradiction' means? 'particular' is used in the everyday sense such as definition 1 at Merriam online.
Quoting Leontiskos
I don't use 'non-particular'. There are particular contradictions (such as "B & ~B") and there is a definition of 'is a contradiction'.
* A formula is a contradiction if and only if it is the conjunction of a formula and its negation.
* Sometimes 'contradiction' is also used in the sense that a formula is not necessarily a conjunction of a formula and its negation but is a formula that entails a conjunction of a formula and its negation. (For that sense, I usually use 'inconsistent'.)
* With natural deduction, we sometimes refer to a pair of lines such that one is the negation of the other. But notice, that is merely commentary, as the formal deduction does not depend on the use of the term 'contradiction' but only on specifying a rule by reference to there being one line with a formula and another line with its negation.
Quoting Tautologies and Contradictions
That is the semantical side of the coin. 'contradiction' is usually defined syntactically (a formula and its negation) but it does follow that a sentence is a contradiction if and only if it is logically false.
Quoting Leontiskos
No, the conclusion of a reductio is of the form "~P" from the assumption "P", or for the non-intuitionistic version, "P" from the assumption "~P".
Quoting Leontiskos
Leontiskos can say whatever he likes, but he is terribly confused and uninformed about the subject.
I explained how reductio works in natural deduction, as just quoted by Leontiskos.
And formulation of the rules for natural deduction do not need to mention 'premise', 'assumption' or 'suppostion'. We may mention those words for convenience, but the specification of a natural deduction system does not require such words. Leontiskos would learn what a natural deduction system actually is - not his nutty imagination of what he thinks it is - by just reading a reliable text or article that covers the subject.
There is a difference between P being a member of G and P not being a member of G. That is not a "formalist" fiction.
I'll take time and effort to give even greater detail, though Leontiskos will likely ignore it or mangle it in the dysfunctional, electrically shorting food processor that is his brain when it spews garbage about logic.
One elegant way to formulate sentential natural deduction is with these rules that permit:
Enter P on a line and charge that line to itself.
If P, along with possibly other lines, shows Q, then infer P -> Q and charge it with all lines charged to Q except the line for P.
From P and P -> Q, infer Q and charge it with all lines charged to P and to P -> Q.
If P, along with possibly other lines, shows a formula Q and a formula ~Q, then infer ~P and charge it with all the lines used to show Q and to show ~Q except the line for P.
If ~P, along with possibly other lines, shows a formula Q and a formula ~Q, then infer P and charge it with all lines used to show Q and used to show ~Q, except the line for ~P. [not intuitionistic]
From P and Q, infer P & Q and charge it with all lines charged to P and to Q.
From P & Q, infer P or infer Q and charge it with all lines charged to P & Q.
From P or from Q, infer P v Q and charge it with all lines charged to P.
From P v Q, P -> R and Q -> R, infer R and charge it with all lines charged to P v Q and to P -> R and to Q -> R.
There is no mention of 'premise', 'assumption' or 'supposition'.
Those rules are equivalent with:
{P} |- P
If Gu{P} |- Q, then G |- P -> Q
If G |- P and H |- P -> Q, then GuH |- Q
If Gu{P} |- Q and Gu{P} |- ~Q, then G |- ~P
If Gu{~P} |- Q and Gu{~P} |- ~Q, then G |- P [not intuitionistic]
If G |- P and H |- Q, then GuH |- P & Q
If G |- P & Q, then G |- P and G |- Q
If G |- P or H |- Q, then GuH |- P v Q
If G |- P v Q, and H |- P -> R and J |- Q -> R, then GuHuJ |- R
There is no mention of 'premise', 'assumption' or 'supposition' nor, for that matter, 'contradiction'.
We don't infer that. Suppose A is "P & ~P", then A itself is a contradiction.
If ~(A -> (B & ~B)) is true then A is true. But if A is a contradiction, then ~(A -> (B & ~B)) is not true.
You are misstepping when you take ~(A -> (B & ~B)) to mean "A implies no contradiction" or that ~(A -> (B & ~B)) entails that there is no contradiction that A implies.
Otherwise, try to prove it. Assume:
~(A -> (B & ~B))
Then try to prove:
For every C, if C is a contradiction, then A does not imply C.
What are the assumptions you will use to try to prove it? Are you making a claim about what pertains in ordinary symbolic logic? Or are you making a claim about how statements of such forms are understood in certain everyday contexts? If you are making a claim about what pertains in ordinary symbolic logic, then your attempted proof should not include premises that are not derivable in context of ordinary symbolic logic.
Quoting Lionino
It is not well formed.
But we could have a notation P-/->Q, but what would it mean if not ~(P -> Q)?
One may instruct as to what does and does not obtain in classical logic without claiming that classical logic is the only credible logic. If the subject is classical logic, then we need to not make false claims about it, no matter what our views about it may be. So, if (made up example) someone said, "in classical logic, RAA is a rule for checking the well-formedness of expressions" then one doesn't have to commit to any particular view of classical logic in order just to point out that that is not what RAA is and not how RAA works.
Now, in this discussion, there has been talk about classical logic qua classical logic. And so it is eminently proper to point out when classical logic is misconstrued or misrepresented. Just as one doesn't have to be an intuitionist or advocate for paraconsistent logic, etc. to point out when intuitionististic logic or paraconsistent logic, etc. is misconstrued or misrepresented. How can that not be eminently reasonable? And it is eminently unreasonable, irrational and anti-intellectual to try to arrogate a presumed dialectic high ground by dissuading people from pointing out how classical logic actually does operate.
Quoting Leontiskos
That is so blatantly, ridiculously, shamefully wrong. It could be written only by someone who doesn't know jack about the subject. It is the opposite of the truth. Classical sentential logic is based on compositionality and not reducing all arguments to merely atomic formulas. And classical predicate logic extends with quantifiers.
Again, Leontiskos doesn't know jack about this subject.
When we say "It is not the case that P->Q implies ~P" it is understood that that means "It is not the case that for all formulas P and Q, P->Q implies ~P". 'P' and 'Q' are understood to be sentential variables and that in informal discussion, the quantification 'for all' is implicit. Of course we understand that P->Q implies ~P for certain P and Q while it is not the case that for all P and Q, P->Q implies ~P. And Leontiskos's claim about this - "some rules of classical logic to come into conflict" - is patently false.
Anyone who has studied even an introductory text in the subject would understand that. The study of the subject is not responsible for Leontiskos's ignorance about it.
There's more and more and more misconception and ignorance in Leontiskos's posts. One cannot even keep pace with them.
1 P {1}
2 Q {2}
3 ~(P & Q) {3}
4 P & Q {1,2}
5 ~P {1,2,3}
that is valid. {1,2,3} |= ~P
1 P {1}
2 Q {2}
3 ~(P & Q) {3}
4 P & Q {1,2}
5 ~Q {1,2,3}
that is valid. {1,2,3} |= ~Q
1 P {1}
2 Q {2}
3 ~(P & Q) {3}
4 P & Q {1,2}
5 ~~(P & Q) {1,2,3}
that is valid. {1,2,3} |= ~~(P & Q)
RAA permits only valid inferences. The fact that there are different valid arguments we may choose from does not vitiate that RAA permits only valid inferences.
Contraposition is a one formula version of the rule RAA.
With a natural deduction system with RAA, we can derive contraposition as an axiom.
With an axiom system with contraposition as an axiom, we can derive RAA as a rule.
I've chosen to correctly report that that is a truth table tautology and a theorem of ordinary symbolic logic. Will Leontiskos's tribunal find me guilty of the crime of being a "truth functionalist" for that?
That is a confusion of someone who doesn't know jack about the subject.
The formulas of the object language are written in the object language and referred to in the metalanguage. A proof mentions formulas from the object langugae, but the proof is a sequence of formulas (or tree of them, or a sequence of pairs of formulas and sets of line numbers, etc., depending on the system). The sequences are not written in the object language. For example, a sequence as written by:
1 P -> Q {1}
2 P {2}
3 Q {1,2}
mentions the formulas P -> Q, P, and Q, but also 1, 2, 3 and {} that are in the meta-language.
RAA is no different in that respect from such proof forms a modus ponens, conjunction introduction, etc.
I didn't say that. Leontiskos says he hardly reads my posts, but not so hardly to stop him from putting words in my mouth. Leontiskos is a strawmaner extraordinaire.
I didn't say "more than 1". Leontiskos pulled that from thin air.
Indeed, here is an RAA from even the empty set of sentences:
1. P & ~P {1}
2. P {1}
3. ~P {1}
4. ~(P & ~P} { }
Indeed, every logically true sentence is provable from the empty set of sentences, and the negation of every logically false sentence is provable from the empty set of sentences.
And I didn't say "the inconsistency of some assumption given a set of premises" (or if I did, then I typed incorrectly). What I said is that RAA shows that a sentence is inconsistent with the set of premises. There's a world of difference between those.
You're welcome for that. (Not too very bumptious of me.)
I haven't "chosen" it except that it is:
a theorem of sentential logic
a tautology
a symbolization, in one formula form, of certain common reasoning
Quoting Leontiskos
RAA is valid in the same way any other rule is valid:
A rule is valid if and only if application is truth preserving, which is to say that any interpretation in which the premises are true is an interpretation in which the conclusion is true.
By showing a derivation:
1. P -> (Q & ~Q) {1}
2. P {2}
3. Q & ~Q {1,2}
4. ~P {1}
RAA is a rule. If it is a primitive rule, then there's no call to prove it. If it is a derived rule from primitive rules, then we prove that any inference with the derived rule can be formulated with the primitive rules only.
A natural deduction system would have RAA as primitive. Ordinarily an axiom system would have RAA as derived.
Or, you might ask "how do we know that a rule is valid"? Well, we prove:
With any application of the rules, any interpretation in which the premises are true is an interpretation in which the conclusion is true.
That is the soundness theorem.
I answered it.
Choose in what sense?
Do you mean whether we should claim A or claim ~A?
I wouldn't claim either. The premises are inconsistent, a fortiori the arguments are not sound. I would claim a conclusion only from an argument that is not only valid but is sound, and especially not from an argument in which the needed premises are inconsistent.
Quoting Lionino
No. Here is an RAA:
1. (S & ~P) -> (B & ~B) {1}
2. S {2}
3. ~P {3}
4. B {1,2,3}
5. ~B {1,2,3}
6. ~~P {1,2}
Or if the version of the rule is to go through a conjunction B & ~B instead of B and ~B on separate lines:
1. (S & ~P) -> (B & ~B) {1}
2. S {2}
3. ~P {3}
4. B & ~B {1,2,3}
5. ~~P {1,2}
With the non-intuitionistic form we can have the sentence on the last line be P.
When someone says a falsehood negligently, especially with the intent to discredit another, that too is a lie. And when someone says a falsehood that they have committed to belief by willful intent, that too is a lie. And when someone has clearly been shown that what they've said is false but continue to say it anyway, that too is a lie and an egregious lie.
And notice that I did not say that Leontiskos lied the first time he made the false claims about me. Rather, I called to his attention that I have said the very opposite of what he claims I've said. But then he continued to make the false claims. At that point I securely said they are lies.
Quoting Leontiskos
But not ignoring them enough to not lie about them.
Quoting Leontiskos
My reply:
Quoting TonesInDeepFreeze
Quoting Leontiskos
Ah, the time honored tactic of pretending to be above a dispute by perpetuating it. And pointing out that Leontiskos is lying when he says I said the opposite of what I actually said is not silly. Rather it is mighty proper.
Quoting Leontiskos
Time, energy and interest allowing, I answer sincere and coherent questions as best I can.
Then why do I have 13 new replies from you today, 11 of which are in a single thread? You're a spammer and I don't have time for this stupid shit. Get someone else to teach you how a reductio works. Maybe they can also teach you how to interact without spamming. Consider it a rule of life that when you spam people they ignore you. Adios!
How predictable of Leontiskos! His hypocrisy is remarkable. Arguably he has posted more than anyone in this thread. A significantly greater number of posts and characters than I have. And he made several posts about my posts and about me during a time when I was not posting in this thread, just as recently made several during a time when he was not posting. He exercised his prerogative that way, and I exercised mine.
A poster is not wrong for catching up to several posts in a thread all at once. How in the world would it be unreasonable for a poster to exercise the prerogative to post and to reply to posts after, for whatever reason, not having posted in a thread during several days?
No one has to confine themselves to Leontiskos's own posting times to reply to him so as to refrain from catching up at some later time.
/
And as to teaching how reductio works, I posted an exact specification how it works, in two formulations. Leontiskos could read that post, or better yet, get an introductory textbook that uses natural deduction, because sure as shootin' he ain't got a clue now, as I showed.
Answer: There are two related inference rules:
(1) If P, along with possibly other lines, shows a formula Q and a formula ~Q, then infer ~P and charge it with all the lines used to show Q and to show ~Q except the line for P.
(2) If ~P, along with possibly other lines, shows a formula Q and a formula ~Q, then infer P and charge it with all lines used to show Q and used to show ~Q, except the line for ~P. [not intuitionistic]
Those rules are equivalent with:
(1) If Gu{P} |- Q and Gu{P} |- ~Q, then G |- ~P
(2) If Gu{~P} |- Q and Gu{~P} |- ~Q, then G |- P [not intuitionistic]
The post you quoted there was before you joined these threads. So there is no connection to you. "We" there simply means "I" not bumptious of me, the greatly humble person I am.
Quoting TonesInDeepFreeze
Ok, that is the derivation. The source I quoted at least is correct when abriding it. The RAA however is not how it was being presented in this thread by others before, which is what I was trying to confirm.
Quoting TonesInDeepFreeze
We are all speaking non-intuitionistically here, which is standard at least in amateur circles.
Quoting TonesInDeepFreeze
The bulk of the debate here between Banno and Leontiskos (and me interjecting sometimes) is why say G |- ~P instead ~P |- G.
The debate gets to the point on this post https://thephilosophyforum.com/discussion/comment/918628 by me
then to this https://thephilosophyforum.com/discussion/comment/918657 by Leontiskos
then this https://thephilosophyforum.com/discussion/comment/918724 by me
then he replied with this https://thephilosophyforum.com/discussion/comment/919282 and this post right under https://thephilosophyforum.com/discussion/comment/919292
I did not further reply
I ask that you check those before replying on this specific point, as I don't want to go in circles or pointlessly extend the thread.
@Count Timothy von Icarus
I still would like some final thoughts on this post specifically.
Thanks for that. I don't mean to belabor this thread, but speaking naturally, I would say this.
A contradiction for logic is like a fork in the road. It forces us to choose one way or another, for we cannot choose both. A reductio ad absurdum is the combination of this fork along with a concrete choice in favor of one side of the fork. "Here is a fork/contradiction in the road, and we will choose this side." The <choice> or <uncharacteristic act of will> is what <separates a direct proof from an indirect proof>.
In a formal sense a reductio in no way determines that we must take one side of the fork rather than the other. Yet the reason a reductio is not usually controversial is because there usually is a set of axioms that both interlocutors are committed to, and a quality reductio will place those axioms in one side of the scale, thus persuading such interlocutors to choose the side of the fork that favors the axioms. I maintainas I said at the outsetthat a reductio involves a <background of plausibility> in order to adjudicate between the two sides of the fork. If there is no background rationale for adjudicating in favor of one side of the fork rather than the other, then the reductio will have no rational force <as a choice between (??~?) ? (??~?) >.
So what do you think now?
You asked:
Quoting Lionino
So I answered correctly that it is not. And I showed an actual RAA.
Yours is a valid inference, but it is not formulated as RAA.
Quoting Lionino
I mention it because the proof itself ends with ~~P. To get P requires additional steps that are not intuitionistic. And I mention, as added information, that they are not intuitionisitc because that is interesting and good to know. I wouldn't always belabor the point, but in this context it is stark that there are two forms: classical and intuitionistic, and in the natural deduction system I presented, we need both to achieve all the classical inferences.
And sometimes people do speak on behalf of intuitionism in this forum, as well as there are likely readers of threads that don't post in them.
Quoting Lionino
Whatever the bulk the debate is about, I presented an exact formulation of the rule a a reference for whomever might want to know exactly what the rule is. And as part of my explanations why certain objections to the rule are off-base.
We don't say: If Gu{P} |- Q and Gu{P} |- ~Q, then ~P |- G.
We don't say that because |- is meant to be equivalent with |=.
And it is not the case that for all G and P, we have:
If Gu{P} |= Q and Gu{P} |= ~Q, then ~P |= G
Formulating a rule: "If Gu{P} |- Q and Gu{P} |- ~Q, then ~P |- G" would be stupid. And, believe it or not, logicians try not to offer stupidities that would thwart the very intent of presenting a formulation.
Ugh.
From the post I quoted:
Quoting Lionino
That was from 13 days ago. I entered the thread 19 days ago.
So, "The post you quoted there was before you joined these threads" is false. And my remark "You're welcome for that" is apposite in the timeline:
From 14 days ago, though my comments about the point went further back, a detailed, informative post:
Quoting TonesInDeepFreeze
After that post, you finally understood the point (I had explained it previously) as you wrote:
Quoting Lionino
So, celebrating that information had been understood, I replied:
Quoting TonesInDeepFreeze
Later you wrote:
Quoting Lionino
Then I said, "You're welcome for that". And you are.
I'll call both of the two forms below 'RAA' (though different writers label them in different ways):
(1) If Gu{P} |- Q and Gu{P} |- ~Q, then G |- ~P
(2) If Gu{~P} |- Q and Gu{~P} |- ~Q, then G |- P [not intuitionistic]
In a natural deduction proof, any formula can be entered on a line and charged with that line number. It is not required that the formula be a member of a given axiom set. But some lines can be "discharged" so that even though the line was used, it is no longer counted on as being required for the inference. That is the essence of natural deduction. For example, an RAA:
(1) P -> (Q & ~ Q) {1}
(2) P {2}
(3) Q & ~Q {1,2}
(4) ~P {1}
Line (2) was discharged at line (4).
{1} in line (4) is the set of formulas that are charged to the conclusion ~P. So, proving ~P in this case depends only on line (1) and not on line (2). Line (2) was discharged.
And it is well understood that this also is correct:
(1) P -> (Q & ~ Q) {1}
(2) P {2}
(3) Q & ~Q {1,2}
(4) ~(P -> (Q & ~ Q)) {2}
Line (1) was discharged at line (4).
{2} in line (4) is the set of formulas that are charged to the conclusion ~P. So, proving ~P in this case depends only on line (2) and not on line (1). Line (1) was discharged.
If we want only inferences from a given set of axioms, then each member of the set of formulas that are charged to the conclusion of an argument must be a member of that set of axioms. But the discharged formulas do not have to be members of that set of axioms.
Of course, having (1) as an axiom and (2) as not is different from having (2) as an axiom and (1) as not. Of course, it is optional which we choose to work with.
And of course, one is free to choose which axioms to work with, or to add to add premises of interest. But the discharged lines are not counted in the "bottom line". And yes, of course, the choices of axioms or added premises are subject to whatever criteria we wish to subject them to. That we have that option is not a fault in RAA. It's not a matter of a fault in RAA but rather it's the obvious fact that we are free to choose whichever axioms or premises we want to choose.
You got it wrong. I know what I meant with my posts. "We" there refers to me, I was not talking about anyone else. The specific post you quoted did not help sort out the issue, specifically the nitpick on "translation", which is why I had to make a whole new thread for that topic specifically. In fact in my thread you corrected yourself about something midway through the discussion:
Quoting TonesInDeepFreeze
But feel free to take credit for whatever you want in your mind, it doesn't bother me.
I made a mistake. I meant to say:
"The bulk of the debate here between Banno and Leontiskos (and me interjecting sometimes) is why say G |- ~P instead of P |- ~G."
That is something that Banno expresses here:
Quoting Banno
It has happened before in the history of science where we had to reject G when finding out that Gu{P} is contradictory, because P was so evidently true.
Ok. That is much clearer than your other posts. I suppose I agree with what is conveyed. However, the RAA has formally either rho or mu as a premise, so no choice between the conjunct is needed within the RAA, the RAA is logical.
You can use the notation however you wish, but in my formulations, G is a set of formulas not a formula, and on the left side of the turnstile is a set of formulas while on the right side of the turnstile is a formula.
The rule:
(1) If Gu{P} |- Q and Gu{P} |- ~Q, then G |- ~P.
For concision, we can formulate that as:
(2) If Gu{P} |- Q & ~Q, then G |- ~P.
So, the point you are making could be stated ('\' stands for set difference):
(3) If Gu{P} |- Q & ~Q, then there could be an R in G such that
(G\{R}) u {P} |- ~R
and
(G\{R}) u {P} is consistent.
That is a formal way of saying what, as you mention, has been correctly said in this thread for a long time:
If ~P is derived by RAA, then it might be the case that there could be a premise R, different from P, such that we derive ~R instead.
And I gave specific examples of that in exact RAA formulation. My point in that regard has always been in agreement with anyone who correctly makes that observation.
But I don't take that to be a problem with RAA.
As I mentioned, RAA, along with the rest of the natural deduction rules, provide all and only the valid inferences (that is the completeness and soundness theorems).
[From previoulsy and adding to it:] Of course, the choice of axioms or added premises is subject to whatever criteria we wish to subject them to. That we have that option is not a fault in RAA. It's not a matter of a fault in RAA but rather we have the obvious fact that we are free to choose whichever axioms or premises we want to choose. It is good that natural deduction allows us to work from whatever axioms or premises we want to choose. The purpose of the deduction system is not to dictate our starting assumptions, but rather merely to permit all and only the valid inferences from whatever assumptions we choose.
Quoting Lionino
((P & R) -> (Q & ~ Q)) |- (P -> ~R) & (R -> ~P)
That is the idea with a pair of formulas P, R; while my formulation above is way of saying the same idea but generalized with any set of formulas G.
Indeed, his idea of looking at conjunction elimination, not just RAA, in this regard is insightful.
Quoting Lionino
I would think so. And that is what this basic logic provides.
Ugh^ugh (that's ugh to the power of ugh).
I lightly jibed, aiming at myself as much as anyone; and you turn it a thing. But since your revisionist attack on me is ill-premised I respond.
(1) I haven't claimed that you don't know what you mean. (2) I haven't claimed that you used the word 'we' in any particular way.
(3) You can look at the posts, and I as I mentioned some of them in my previous reply:
You asked me for a translation of "~(A -> (B & ~B))". So I obliged your request.
You claimed that "A does not imply a contradiction" is a translation.
I explained why that is not a translation of "~(A -> (B & ~B))", a few times. As you still didn't see the point, I gave even more details, to which you replied:
"Yes, granted. I used the word "translation" wrong in basically all of my posts."
(4) The matter of the translation is not a nitpick. You posted the incorrect translation at least a few times and presumably it was important enough to you to do that. So I was right on the money to note that it is not a translation and to explain why.
Quoting Lionino
Yes, to my credit, I corrected myself as soon as I realized I erred in a formulation. But that was not related to my correct explanations as to why "A does not imply a contradiction" is not a translation of "~(A -> (B & ~B))".
(5) In my previous post, I pointed out that you falsely claimed (why?) that my reply "You're welcome for that" was in response to a post from before I entered the threads. But my reply was from 13 days ago; I entered the thread 19 days ago. More importantly, my explanations as to why "A does not imply a contradiction" is not a translation of "~(A -> (B & ~B))" were also before that reply.
We know that the assumptions are inconsistent, not merely that one of them is false. But, of course, if a set is inconsistent, then for any interpretation, at least one of the members is false in that interpretation.
I couldn't tell since you often jump into a conversation that happened several pages before the last post of the respective thread.
Quoting TonesInDeepFreeze
:rofl: alright however you wanna fly it
I linked exactly to your post. No matter how many intervening posts, all you had to do was click on your name under my quote of you.
But you chose to (1) Make a false claim that I was commenting on a post that was made before I came into this thread, (2) Get the conversation completely wrong by falsely saying that I had not explained why your translation was incorrect, (3) Falsely claim that my point about the translation was a nitpick. (4) Try to get a cheap advantage out the fact that I corrected myself in a formulation, even though that is unrelated to the subject of the translation.
Quoting Lionino
It's not how I choose to see it. It actually is as I showed.
Quoting TonesInDeepFreeze
More like the point Leontiskos is making.
Quoting TonesInDeepFreeze
I am aware.
Yep. Good point.
This is for classical logic. I am puzzling, as a sideline, with how conjunction plays out in paraconsistent logics - Non-Adjunctive systems in which {A,B}?A?B). Ja?kowski's discursive logic, by way of example.
I wonder if, and how, RAA would function in such a case. Is it that two folk in discourse could agree that A?B imply C, yet not agree that ~C implies ~(A?B)...
Anyway, that's a whole different minefield.
I'm referring to your correct point (shared by you, Leontiskos, Banno, and me) that instead of refuting P, we could refute one of the other premises. You adapted that point to my notation. I just pointed out that your notation doesn't suit the way I was using the notation and so I rephrased your point so that it does accord with my notation.
That's good. I disagreed with it many pages ago, as I saw immediately that it's wrong. (Not too very bumptious of me to say. winky face emoji whatever.)
I don't think the word 'translation' has an exclusive definition in logic, and if it does, it doesn't matter, I wasn't using logical jargon, I meant 'translation' to mean exactly what it does everyday, bringing from one language to another, in that case from logic to English. Translation doesn't always mean translating literally term-by-term, which is what you did in your translations. The only reason I said I used translation wrong every time was to accept your definition and move on as to not waste time on semantics, it was tangential to my goal, which led to my thread, where in the end the only good translation provided was by a user in the 1st page.
Don't matter. I was talking about myself. If I recall correctly, I was thinking of specifically my post with the several examples compared line by line which was what finally established that that translation was not possible.
Amazing!
Whether tangential or not, it raised an important and interesting point. The reason yours was not a correct translation is instructive. And it wasn't a matter of definitions, let alone a definition identified with me. Meanwhile, my translation was utterly obvious, therefore dull, but the point of such a translation is not to entertain but to be as pinpoint faithful to the formula as possible. If it had been a creative writing exercise then I would have tried to come up with something a lot more colorful.
Quoting Lionino
Actually, this:
Quoting Lionino
So far, so good. There you're saying what I had spent a few posts explaining.
Quoting Lionino
That last clause is wrong, obviously. (Maybe you corrected it subsequently.)
Quoting Lionino
Right.
I like that word. When I was a kid, my dear friend was a San Francisco Giants fan amidst all the other kids who were Dodgers fans. He chose that path as yet another of his exercises in non-conformity. When he got to be the captain of one of the pickup teams, he named them "The A-Mays-ers". That was a wry little bit in a dull and gray time of life. Mays died recently. So deservedly a beloved man.
Then last week I was getting tech support on the phone, and the tech agent would say "Amazing!" in a kind of lilting modern way when I executed each step in the walkthrough so that I got to the right element each time. As if each click to the next element was a cause for a moment of minor celebration. I found that charming.
AT&T? Or perhaps Proximus? They are trained to do exactly that with customers in the US.
Neither of those, but what you say is interesting. I guess it is a thing.
It is. I did a gig at tech support years ago, it was part of our training to basically baby the customer and quite literally treat it like a king, saying "No" to the caller was strictly forbidden.
But the quality of tech support and customer support in general has plummeted over the last several years. Even when agents aren't pressured to wrap up the calls quickly, some of them are atrocious - not listening to you, mindlessly giving you steps that couldn't be expected to address your particular problem, and giving misinformation and lying. And resisting to the last breath transferring you to a tier 2 tech. And of course, long wait times, calls cut off, and failed promises to continue pursuit of a resolution with a followup call. And then, after you've been through weeks of runaround in multiple calls, chats and emails, they have the gall to send you an email survey asking how they did, and keep pestering you with even more emails. On the other hand, there are some companies that provide sterling support; agents that will stay on the phone and give you all kinds of extra info and tips.
By the way, your comments about dictionaries led me to discover that, unlike years ago, OED is free online. It definitely is a much richer resource than Merriam. Thanks for that. I'll be using them side by side from now on (and with my old print edition of unabridged Merriam).
"A dialetheia is a sentence, A, such that both it and its negation, ¬A, are true. If falsity is assumed to be the truth of negation, a dialetheia is a sentence which is both true and false. ....Dialetheism is the view that there are dialetheias. If we define a contradiction as a couple of sentences of which one is the negation of the other, or as a conjunction of such sentences, then dialetheism amounts to the claim that there are true contradictions."
The Aristotelians here don't cotton to it.
Rho is assumed and Mu is supposed, and if someone doesn't know the difference between an assumption/premise and a supposition then they won't understand a reductio. As I have argued at many points throughout this thread, the difference between an assumption and a supposition is a meta-linguistic difference. The object language cannot account for this difference.
The modus tollens and the reductio are two different things:
You can say that "the RAA is logical," but the fact remains that B3 is not as secure as A3. There are many ways to explain why, but I will stick with my original answer that an assumption and a supposition are only different at the level of meta-logic. A3 follows without any recourse to the meta-language or to meta-logic. B3 does not.
Quoting Leontiskos
...Refactoring this idea for a proper reductio:
(In order for the RAA to function as a dialogical proof one's interlocutor must agree that C1 is more plausible than C2.)
That's obviously not the reductio.
Given Apokrisis' demonstrations that you don't engage in good faith (as most of us already knew), I've simply put you on ignore, where you belong.
Quoting Leontiskos
It's not even a valid argument.
I do not think the bad faith is mine alone.
Here is RAA in exact formulation:
Quoting TonesInDeepFreeze
What is "secure" supposed to mean? At least RAA is as secure as modus tollens in the sense that they are sound.
For all sets of formulas G and formulas P, if G |- P then G |= P. That is:
For all sets of formulas G and formulas P, if G |- P then for all models M, if M is a model of G then M is a model of P. That is:
For all sets of formulas G and formulas P, if P is derivable from G, then any interpretation in which all the members of G is true is an interpretation in which P is true. That is:
Classical natural deduction (which includes RAA as a primitive rule, modus tollens as a derived rule and contrapostion as a derived theorem schema) does not permit inference of a false conclusion from true premises.
No matter whether we start with RAA as primitive and derive modus tollens and contraposition, or start with modus tollens as primitive and derive RAA, or start with contraposition and derive RAA and modus tollens, we arrive at the exact same set of allowable inferences.
Without the word 'assumption':
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1}
G is {A -> (B & ~B)}
P is A
Q is B
G u {P} |- Q & ~Q, so G |- ~P.
Quoting Leontiskos
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~(A -> (B & ~B)) {2}
G is {A}
P is A -> (B & ~B)
Q is B
G u {P} |- Q & ~Q, so G |- ~P.
And just a few posts ago, I discussed the no one is disallowed from choosing what to refute with RAA, as long as the refutation is valid. That upholds the very purpose of the logic system: It shows proofs from sets of formulas but does not dictate from which set of formulas (whether called 'axioms' or 'premises' or not called anything other than a 'set of formulas') we may derive. That was discussed with Lionino, who does understand the point, in the specific sense of axioms. People are free to choose their axioms, and to refute formulas inconsistent with those axioms. If G is a set of axioms and P (not a member of G) is inconsistent with P, then G refutes P. And if H is different set of axioms and Q (not a member of H but a member of G) is inconsistent with H, then H refutes Q. That we have this choice is a good thing, not a flaw in RAA nor in any other natural deduction rule, as it permits working from different axioms.
Also, A is not made true in that derivation. A is true or not depending on a given model. There is no requirement that any line be true, not even the conclusion, in a given model. For that matter ~A could even be a logically true formula itself. A sound logic system only requires that any interpretation in which all the members of G are true is an interpretation in which P is true.
The problem is that this proof of yours is invalid:
Quoting TonesInDeepFreeze
Once I pointed it out you edited your post to try to inject some background conditions, but we both know that this proof is invalid. There is no rule of inference that allows us to draw (4) from (1) and (2).
To be clear, I was not faulting your formulation, but rather only showing, contrary to the other poster, that it does not require an appeal to 'premise', 'assumption', 'supposition' or even 'contradiction'.
I edited to provide more explanation. My edit did not alter the substance of what was there before the edit. And it was it was not "pointed out" at that time that the proof is invalid. My edit could not have been about a claim of invalidity, since there was no such claim of invalidity today.
It is valid.
Every interpretation in which "A -> (B & ~B)" is true is an interpretation in which "~A" is true. If the poster can't see that, then he can make a truth table to see it.
Quoting Leontiskos [fixed quote in edit]
Still the poster doesn't know what RAA is.
The argument doesn't draw (4) from (1) and (2). The argument draws (4) from (1), as (2) is discharged.
The argument is an instance of RAA. And RAA is sound.
But it's not.
Quoting TonesInDeepFreeze
All you are saying is, "??¬?," but this does not make the proof valid. What rule of inference do you think you used to draw (4)? (4) adjudicates the and-elimination.
Quoting TonesInDeepFreeze
Heh. Why is (2) "discharged" and not (1)?
I moved Universities between logic courses, and was first taught natural deduction and then axiomatic systems, without the difference being made clear. So I tend to mix the two styles, with somewhat idiosyncratic results.
Quoting TonesInDeepFreeze
Yep. The example Leo gave is not an example of RAA. And this:
Quoting Leontiskos
What to conclude except that Leo does not understand validity. Quoting Leontiskos
That he asks this is quite odd, since our purpose here was A -> (B & ~B) ? ~A.
Tones was quoting me, and he should have used the quote feature. If he had you would not have inadvertently agreed with me. Because as this thread shows, you would say any number of stupid things rather than do that.
Note that (4) is originally your conclusion, and we now both agree that it is invalid. My example was a quote from you, where you claimed to give a reductio.
That was obvious. A mere typo.
Quoting Leontiskos
I didn't.
Quoting Leontiskos
This is inane. (4) cannot be invalid on its own. The argument is valid in classical prop logic.
A proposition can be invalid qua conclusion, and that's precisely what I said. :roll:
Quoting Banno
Again and again the simple questions go unanswered:
Quoting Leontiskos
Quoting Banno
You quoted my words and then you said, "Yep..." You just didn't know they were my words. :roll:
Quoting TonesInDeepFreeze
And so, A -> (B & ~B) ? ~A. This is a valid argument.
You made the claim that this was RAA:
Quoting Leontiskos
Which, as Tones pointed out, leaves out 3:
Quoting TonesInDeepFreeze
It was this with which I was agreeing.
Quoting TonesInDeepFreeze
Quoting TonesInDeepFreeze
Tones thinks that ¬(1) and ¬(2) both follow from (1, 2, 3). It goes without saying that there is no rule of inference that forces one rather than the other. I would simply say that both of these proofs are invalid. There is no rule of inference to justify (4) on either count. This all goes to the misunderstandings of reductio ad absurdum in this thread, and in particular to Tones' recent claim that there is no need to advert to a difference between an assumption/premise and a supposition.
The conversation I am having with Tones revolves around <your argument>, which is an instance of the form of reductio that I gave.
Quoting Banno
"3" is present in the word "contradiction." :roll: You are and were nitpicking.
And again, as far as I can tell Tones was quoting me without using the quote feature, as he responded to a post where I said:
Quoting Leontiskos
And you would be wrong.
No, it isn't. You missed out a line.
The poster continues to indicate that he does not know what validity is in this context and that he is unwilling to read the posts to which responds. He skips that I just stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here.
Quoting Leontiskos
The poster seems to not know what validity is and that he is unwilling to read the post to which he responded. He skips that I stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here.
Quoting Leontiskos
The poster seems to not know what RAA is and that he is unwilling to read the posts to which he responds. He asks what rule is used, when the rule used is RAA, exactly as the rule is formulated.
Again, as I said, for concision we may state RAA without conjunction elimination:
If Gu{P} |- Q and if Gu{P} |- ~Q, then G |- ~P
is equivalent with
If Gu{P} |- Q & ~Q , then G |- ~P
If Gu{~P} |- Q and if Gu{~P} |- ~Q, then G |- P
is equivalent with
If Gu{~P} |- Q & ~Q, then G |- P
So, in this case:
(version 1)
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1}
is equivalent with
(version 2)
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. B {1, 2}
5. ~B {1, 2}
4. ~A {1}
Quoting Leontiskos
...much less 's half-baked reductio:
The poster continues to substitute rhetoric for argument, utterly failing to engage in rational argumentation or inferential reasoning. Why such a course is taken, one does not yet know. Diagnosis continues.
Quoting TonesInDeepFreeze
The poster seems to suffer from psychological delusions and grandiosity. When faced with simple questions he retreats into himself, opting for 3rd-person rhetorical strategies and failing to engage in inferential reasoning.
Quoting TonesInDeepFreeze
The poster continues to evidence a significant difficulty in using fairly basic forum features, such as quotes.
Quoting TonesInDeepFreeze
The poster continues to assert his baseless arguments without answering the question and providing the rule of inference he purports to use in order to arrive at conclusion (4). Ongoing observation recommended. He seems to have no understanding of the difference between his truth-functional formalisms and reality, or even how to properly utilize his formalisms.
For about the half-dozenth time:
I am not a "truth functionalist". I study and enjoy classical logic, and appreciate its uses. But I am interested in other logics. I do not say that classical logic is the only logic that can be studied, enjoyed and used.
But when classical logic is being discussed, especially critiqued, it is crucial to say what actually is the case with classical logic. And in bringing clarity to what classical logic actually is, one needs to explain. Providing such explanations does not make one a "truth functionalist".
https://thephilosophyforum.com/discussion/comment/922432
And "discharged" in scare quotes is silly and juvenile.
So many of your claims have already been debunked in this thread. The truth-table approach to reductio was dispatched almost ten pages ago!
Quoting Leontiskos
Quoting TonesInDeepFreeze
If you want to bring clarity you should explain what inference you used to draw (4). As it happens, truth tables don't adjudicate contradictions. I don't get to say:
(The fact that you think this sort of thing can be adjudicated by a truth table is proof that non-truth-functionality is in your blind spot.)
Quoting TonesInDeepFreeze
https://thephilosophyforum.com/discussion/comment/922468
Your example didn't mix styles. It was fine. All I did was show how to formulate the proof without 'premise', 'assumption', 'supposition' or 'contradiction'.
If you gave a definition of 'valid' in your sense, then we could evaluate your claims about it.
Meanwhile, in ordinary formal logic the common definition is the one I've stated.
Of course, we may discuss relative to different definitions. I don't at all insist that the ordinary definition is the only one that we may use. But when the discussion is about classical logic, especially a critique of classical logic, then we need to at least see what happens in classical logic with its definitions. RAA is valid in context of classical logic. But, again, if you would provide your definition, then, of course, we may find that RAA is not valid in that different context.
Also, with RAA, as with any rule, validity pertains to the relation between a set of formulas and a formula. But there is another sense 'valid' in ordinary logic too, which is that a sentence is valid if and only if it is true in all interpretations.
It has been answered again and again and again. The answer is:
RAA
Another inane misattribution. Nowhere have I said that, and certainly not in the post linked.
Look at the proofs exactly. They show that ~(A -> (B & ~B)) follows from (2), and ~A follows from (1).
Of course, (1) and (2) together are inconsistent, so both ~(A -> (B & ~B)) and ~A follow from (1) with (2).
Meanwhile, it's not needed to mention (3) since it comes merely by inference from (1) and (2).
Quoting Leontiskos
Quoting Leontiskos
Yes, in the exact sense that there is no inference rule that dictates what set G must be.
Quoting Leontiskos
Wrong. By the definition of 'valid' in context of classical logic, they are valid. If you have a different definition 'valid', then of course, anything goes.
Quoting Leontiskos
Wrong. Both proofs provide valid inferences. The first proof validly infers ~A from (1), and the second proof validly infers ~(A -> (B & ~B)) from (2)
Quoting Leontiskos
The misunderstanding in this thread is yours, as you don't even know what RAA is, despite that I've exactly formulated it for you and explained that exact formulation.
Quoting Leontiskos
It's not merely a claim. I showed you exactly, with exact examples, several times now.
Not so. What has been clearly demonstrated is that you do not have a grasp of propositional logic.
That you do not understand validity, nor truth functionality, nor how to perform a deduction.
But perhaps you have unwittingly presented an account of the awkwardness of Aristotelian logic.
When we are talking about classical logic, we need to be clear as to what classical logic is and what is the case about it in terms of its formulations and definitions. So, for example, when I say 'valid' in this context, I mean 'valid' in the sense of classical logic.
If one wishes to use terminology, such as 'valid', in some other sense, then that is fine, and we can discuss in that context too. But we need to be clear in any instance which context we're in.
Thus, since this discussion has focused on classical logic and a critique of classical logic, unless otherwise stated, my remarks pertain to what is the case with classical logic given its formulations and definitions. And, again, that is not a claim or attitude that only classical logic is admissible, but rather that when we are examining classical logic, we need to at least start by knowing what it is - what its formulations are definitions actually are.
Annotations do help to follow along in the proofs. But, depending on the formulation of the system, annotations may not be necessary. In the proofs I gave, annotations are not necessary. And that goes along with the fact that the statements of the rules do not require mentioning 'premise', 'assumption', 'supposition' or 'contradiction'. Unfortunately, the other poster understands none of this and knows jack about this subject.
I've given exact information, and clear explanations, demonstrations, reasoning. There is no lacuna in rationality there. The fact that I also mention that the poster is ignorant, confused and specious doesn't vitiate the on topic content I provide.
The proof quoted is exactly correct. I've answered every central question. It's not my fault that you are unwilling to read the answers or are incapable of understanding them though they are clear and exact.
And for the 1000000th time, the rule is RAA.
How many times do people do have to tell you?:
The rule is RAA. The rule is RAA. The rule is RAA.
Do you not get it?
The rule is RAA.
Quoting Leontiskos
(1) I am hardly grandiose as I've said several times in this forum that I have only an intermediary knowledge of this subject and that I do make (corrected) mistakes.
(2) I have head on addressed the main contentions in this thread. Your replies often blatantly skip the key information, explanations and corrections given him.
(3) My reasoning has not been shown to be incorrect. On the other hand, your arguments are so often a jumble of ignorance, confusions and illogic.
(4) I've referred to you in the third person because you deserve to be referred to that way.
Quoting Leontiskos
Oh please! People often get snagged copying/pasting quote brackets. I almost always correct in edit though. That you try to make any kind of deal out that shows that you're quite the fundament.
For the half-dozenth plus one time:
Quoting TonesInDeepFreeze
I explicitly said that they are examples of RAA. And the examples when given were earlier explicitly said as examples of RAA. And they were given this time in direct response to questions about RAA.
Quoting TonesInDeepFreeze
For the millionth time:
RAA. RAA. RAA.
Can you not read the words 'RAA' when they appear over and over?
First, you don't need '~Q' there.
And I didn't say "see truth table" in the proof.
There are two separate things: the deduction system, (such as natural deduction) which is syntactical, and truth evaluation (such as truth tables), which is semantical.
But we have the soundness and completeness theorem that states that a formula P is provable from a set of formulas G in the deduction system if and only if there is no row in the truth table such that all the members of G are true and P is false.
The proof would be this:
1. P -> Q {1}
2. P {2)
3. Q {1, 2}
The rule applied there is modus ponens:
From P and P -> Q, infer Q and charge it with all lines charged to P and to P -> Q.
The truth table would be:
P true, Q true ... P -> Q true
P true, Q false ... P -> Q false
P false, Q true ... P -> Q true
P false, Q false ... P -> true
There are two rows in which both P and P -> Q are true, and Q is true in both of those rows. There is no row in which both P and P -> Q are true but Q is false, so modus ponens is valid.
But if you want to include ~Q:
Quoting Leontiskos
There's no "avert eyes", "don't allow 3 a seat"
1. P -> Q {1}
2. P {2}
3. ~Q {3}
4. Q {1, 2}
or if we are required to use (3):
1. P -> Q {1}
2. P {2}
3. ~Q {3}
4. Q {1, 2}
5. Q & ~Q {1, 2, 3}
3. Q {1, 2, 3}
Truth table:
P true, Q true ... P -> Q true ... ~Q false
P true, Q false ... P -> Q false ... ~Q true
P false, Q true ... P -> Q true ... ~Q false
P false, Q false ... P -> Q true ... ~Q true
There are no rows in which P -> Q, P, and ~Q are all true and Q is false. The argument from {P -> Q, P, ~Q} to Q is valid. ~Q is in the truth table.
An odd thing to say, since a contradiction will have "F" all the way down it's main operator
The rule and truth tables agree. They agree and are independent. They are independent in the sense that are formulated separately without reference to each other. But there is the theorem that connects them with an equivalency:
Definition: An inference from G to P is valid if and only if there are no rows in the which all the members of G are true and P is false.
Theorem: An inference is allowed by the rules if and only if the inference is valid.
Moreover, the truth table method is algorithmic, thus sentential logic is decidable. So, actually the truth table method, if appropriately formulated, can itself be used as a deduction system.
Quoting Wiki.
Yeah.
Another source (from search of 'history of sentential logic') says, "The truth table system for Sentential Logic was invented in 1902 by the American logician Charles Peirce to display how the truth of some sentences will affect the truth of others. Truth tables were rediscovered independently by Ludwig Wittgenstein and Emil Post."
This is original.
Why do you reject the claim that Peirce came up with truth tables? Why do you omit Post?
Why do you claim that the notions of logically true and logically false were original from Wittgenstein?
He's hopeless.
Several times he was told that the rule used was RAA, and the proofs were stated in situ as being RAA, and yet he keeps demanding that it hasn't been said what rule was used.
That's just for starters.
What is original is that Witti points out how to use a truth table to determine tautology or contradiction.
One source says Peirce came up with truth tables in 1902. If that is correct, then why rule out that he didn't also see that we can use them to make inferences and infer that a sentence is a logical truth or logical falsehood or neither? He had to have been a pretty smart guy, so it is unlikely that he would look at truth tables and not notice that we can use them to make inferences and check whether a sentence is logically true or logically false or neither.
See for example Irvine Anellis. I don't see that Anellis carries the case that Pierce's approach was complete. From the little I've seen Pierce used them to set out the permutations of three variables and so on, but I can't see anywhere that he made the connection with tautology and contradiction. I might be mistaken. But on a quick look around Anellis seems to be alone in his claim.
So unless a stronger case can be made, I'll credit Wittgenstein. In any case, it was the Tractatus that brought truth tables to the attention of the greater philosophical community. You are of course welcome to take a different opinion.
Tautologies: "For many years, commentators have recognized that Peirce anticipated the truth-table method for deciding whether a wff is a tautology.
/
Aside from the question of invention, some salient and useful points about sentential logic:
(1) Sentential logic is, in a certain exact sense, isomorphic with the Tarski-Lindenbaum algebra.
(2) There are 16 binary truth functions. And, for all n, any n-ary truth function can be reduced to a binary truth function.
(3) All 16 binary functions can be derived from just one binary function (either Sheffer stroke or Nicod dagger).
(4) The truth table method can be formulated algorithmically. Sentential logic is decidable. And since sentential logic is decidable, it suffices to have just one inference rule: If G tautologically implies P then G proves P (e.g. 'A Mathematical Introduction To Logic' by Enderton, though he does it by saying that all tautologies are axioms).
(5) Soundness and completeness.
Yep.
I don't know exactly what the author meant by "anticipated"
Meanwhile, may I take it that the point is made about the number of letters? He wouldn't have to display a truth table with n number of letters for every natural number n for us to grasp that there is no finite bound on the number of letters in a truth table.
The author says, "But the discovery by Zellweger of Peirces manuscript of 1902 does permit us to unequivocally declare with certitude that the earliest, the first recorded, verifiable, cogent, attributable and complete truth-table device in modern logic attaches to Peirce, rather than to Wittgensteins 1912 jottings and Eliots notes on Russells 1914 Harvard lectures." Whether the author properly makes the case for that would deserve more scrutiny of the paper.
I'll maintain that the cardinal step, to using truth tables as a device for determining tautology and contradiction, was taken by Witti.
Quoting TonesInDeepFreeze
Yep.
Quoting Banno
That might be the case; but hardly clear that it is.
It seems amazing that it wasn't invented a lot earlier. Such a simple idea by now. It shows how much we take for granted in intellectual products.
"And Richard Zach reminds us that "Peirce, Wittgenstein, and Post are commonly credited with the truth-table method of determining propositional validity.""
It seems doubtful to me that anyone who sees that a truth table tests validity would not remark that some sentences are true in all rows and some sentences are false in all rows, whether such sentences are given the names 'logical truth' or 'logical falsehood' respectively. Indeed, to say that we test for validity is to say that we test whether the sentence is true in all rows (i.e. whether the sentence is a logical truth (aka 'tautology')).
I wonder if there is a "deductive system" using truth tables - say a proof of the completeness and consistency of prop logic using only truth tables... Someone must have done it. Might have a look around.
What do you mean by "showing truth"? The paper shows a Peirce truth table with truth values.
completeness and soundness.
consistency follows from soundness.
Another exercise is proving the correctness of the everyday methods for addition, subtraction, multiplication and division. I have a book that shows some of it.
Sure, and that is where it seems to stop. Wittgenstein does the same thing in 5.101. Again, the novelty in the Tractatus is set out here:
Here Wittgenstein is pointing out that we can take any "grouping"* of elementary propositions and find if it is contradictory or tautological (or neither) by constructing its truth table.
I don't see anything like this attributed to Peirce by Anellis. I'd be surprised if Peirce's logic would enable such a construct.
Curious.
A side line - I'm finding Google increasingly useless at searching for minutia of late. Any subtlety gets lost in irrelevancies.
* - a wff.
Same.
Quoting Banno
Yet, the Anellis paper says: "[T]he discovery by Zellweger of Peirces manuscript of 1902 does permit us to unequivocally declare with certitude that the earliest, the first recorded, verifiable, cogent, attributable and complete truth-table device in modern logic attaches to Peirce, rather than to Wittgensteins 1912 jottings and Eliots notes on Russells 1914 Harvard lectures."
Anellis might not have adequately made the case for that assertion, as perhaps we would need to see Peirces's papers in more detail. But at least it must be allowed that Anellis may be correct. And, at least we see that Peirce was using truth tables.
Then you moved to a different claim:
Quoting Banno
But a truth table determines validity. And Peirce was using truth tables to determine validity. And 'valid' and 'tautology' are synonymous for sentential formulas. So Peirce was determining tautologousness.
Then there's the question of "complete". But what is meant by 'complete' in this context? Does it mean observing that there is no finite bound on the number of letters in a truth table? Is it a given that Peirce didn't observe that and Wittgenstein did?
Quoting Banno
No, he also showed the 16 truth tables for the 16 binary Boolean functions. And he did more work with truth tables. And we don't know the extent of his work without reading all that he wrote.
Quoting TonesInDeepFreeze
Quoting TonesInDeepFreeze
Does Anellis show explicitly that Peirce used a truth table in this way? I don't see that. In the diagram on p.61 he lists some values for three terms. In the diagram on p.62, he lists the possible values for binary connectives.
...not explicitly for determining the validity of any wff. Now in the absence of further evidence, it is reasonable to supose that Wittgenstein was the first to do this. What is absent is something showing that it had occurred to Peirce that the validity of a given wff can be shown by setting out it's truth table. Wittgenstein does set that out.
Now yes, it might have been that Peirce did understand this, but that is surmise.
So if you like, I overstated the case with "Wittgenstein created truth tables", but maintain that their present use has more of Wittgenstein about it than Peirce. You are welcome to differ.
I suggest it is time to move on.
Quoting TonesInDeepFreeze
The reason these are not RAA is because there is no supposition taking place (and again, Tones' original attempt in this thread did not suffer from this problem). Banno and Tones will not understand RAA until they understand that the first step of the reductio portion of a proof (the "supposition" or "assumption") is different from a premise.
For example:
[ ? (P v ~P)
1. __Suppose: ~(P v ~P)
2. __? ~P {from 1}
3. __? P {from 1}
4. ? (P v ~P) {from 1; 2 contradicts 3}
Quoting Leontiskos
The question here is the validity of a conclusion. See:
Quoting Leontiskos
A truth table does not adjudicate between (1) and (2). It does not perform the and-elimination of the reductio for us. What Tones is doing is just arbitrarily ignoring inputs to the truth table:
Quoting Leontiskos
To say ?Q instead of ?~P is to selectively consider the truth table for (1, 2), rather than the truth table for (1, 3). To think that a truth table settles the matter is to ignore the contradiction, which in this case is present in (1, 2, 3).
Quoting only yourself might indicate a failure to address one's interlocutors. But I could make no sense of that last post anyway. Pointing out your errors has become too sad to continue. Especially if you are going to repeat things already refuted.
The simple version, for your benefit:
Two premises and an inference:
What can be drawn from these claims? <According to Tones> one can draw two different, contradictory conclusions (and this pertains to the misunderstanding of RAA). Further, when I ask Tones why he drew one conclusion rather than the other, he tells me to look at the truth table, which is the sort of nonsensical statement that I had thought only you were capable of, for earlier in the thread you had to stick your foot in your mouth any number of times over this same issue.
Edit: Nice to see you have taken to quite radical edits of previous post.
Just to sum up, here's the truth table:
If they did contradict each other, the third column would all be F's.
So as things stand, 41% of folk got it wrong. Pretty sad.
Here's an actual contradiction, for comparison:
That rare combination of hubris and senility. Gotta love it.
I would suggest reading Lionino's first post on page 1, but that would require reading. I see that in the last page or two you managed to misread all sorts of things re: Peirce and Wittgenstein.
Is this statement true or false:
Quoting Leontiskos
According to what definition are both proofs valid?
Or if you like, when I asked what rule of inference allows you to draw (4), you simply said, "RAA." What do you suppose the inference rule "RAA" means? "RAA" is no answer at all, and appealing to the mere name begs the question at hand.
On my reading reductio ad absurdum ("reduction to absurdity") is the idea that a supposition can be rejected if it leads to an absurdity. What is necessary for a reductio is <an isolated supposition> which can then be "reduced" to an absurdity. Without a supposition there can be no reductio, for premises are equal one to another. If two premises contradict then our system is inconsistent. A reductio is not what happens when premises contradict.
It was a misclick from my part. :confused:
As far as I can tell, Anellis outlines the case but does not make it fully. I have not claimed that Peirce developed the notion of truth tables sufficiently to deserve being called 'the inventor of the truth table regarding its use for testing validity of formulas with any finite number of letters'. Only that it is arguably the case that he did per Anellis's conclusion that is based on grounds adduced by Anellis but that seem to me to require more evidence from Peirce's writings. So I wouldn't so strongly conclude that Peirce did not develop the notion of truth tables sufficiently to deserve being called ''the inventor of the truth table regarding its use for testing validity of formulas with any finite number of letters'.
Quoting Banno
He does say:
"Richard Zach [...] reminds us that Peirce, Wittgenstein,
and Post are commonly credited with the truth-table method of determining
propositional validity.
and
"Lane [...] [tells us] that, For many years, commentators have
recognized that Peirce anticipated the truth-table method for deciding whether
a wff is a tautology."
Again, I'd need to see the writings.
Quoting Banno
And reasonable to be more cautious pending finding out more about Peirce's writings.
Calculus was invented by two people simultaneously. The Zermelo-Russell paradox was found by three different people independently. I think that putting 0s and 1s on a grid is much less special than those two.
Our usage of them however does come from Wittgenstein.
The proof is RAA since it fulfills the definition of RAA. I've shown that several times already.
/
Quoting Leontiskos
My reply was and is:
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1}
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~(A -> (B & ~B)) {2}
Look at the proofs exactly. They show that ~(A -> (B & ~B)) follows from (2), and ~A follows from (1).
Of course, (1) and (2) together are inconsistent, so both ~(A -> (B & ~B)) and ~A follow from (1) with (2).
Meanwhile, it's not needed to mention (3) since it comes merely by inference from (1) and (2).
/
Quoting Leontiskos
My reply to that was and is:
Quoting TonesInDeepFreeze
And, regarding a different example (modus ponens) given by the poster, but applicable to the current examples:
There are two separate things: the deduction system, (such as natural deduction) which is syntactical, and truth evaluation (such as truth tables), which is semantical.
But we have the soundness and completeness theorem that states that a formula P is provable from a set of formulas G in the deduction system if and only if there is no row in the truth table such that all the members of G are true and P is false.
The proof would be this:
1. P -> Q {1}
2. P {2)
3. Q {1, 2}
The rule applied there is modus ponens:
From P and P -> Q, infer Q and charge it with all lines charged to P and to P -> Q.
The truth table would be:
P true, Q true ... P -> Q true
P true, Q false ... P -> Q false
P false, Q true ... P -> Q true
P false, Q false ... P -> true
There are two rows in which both P and P -> Q are true, and Q is true in both of those rows. There is no row in which both P and P -> Q are true but Q is false, so modus ponens is valid.
But if you want to include ~Q:
"P?Q
P
~Q
? Q {See truth table for 1, 2; avert eyes from 3 at all costs. I repeat: do not allow 3 a seat at the truth table!}"
Leontiskos
There's no "avert eyes", "don't allow 3 a seat"
1. P -> Q {1}
2. P {2}
3. ~Q {3}
4. Q {1, 2}
or if we are required to use (3):
1. P -> Q {1}
2. P {2}
3. ~Q {3}
4. Q {1, 2}
5. Q & ~Q {1, 2, 3}
3. Q {1, 2, 3}
Truth table:
P true, Q true ... P -> Q true ... ~Q false
P true, Q false ... P -> Q false ... ~Q true
P false, Q true ... P -> Q true ... ~Q false
P false, Q false ... P -> Q true ... ~Q true
There are no rows in which P -> Q, P, and ~Q are all true and Q is false. The argument from {P -> Q, P, ~Q} to Q is valid. ~Q is in the truth table.[/quote]
/
Quoting Leontiskos
Actually, I've given the poster a full explanation. I'll give it again, since he is unwilling to read what he even responds to:
One conclusion is from one set and the other conclusion from another set.
~A is syntactically inferred from the set {A -> (B & ~B)} per RAA.
~(A -> (B & ~B)) is syntactically inferred from the set {A} pre RAA.
Those are two different inferences, and both of them are valid.
To see that they are valid, we can look at the truth tables.
Every row in the truth table in which the member of {A -> (B & ~B)} is true is a row in which "~A" is true.
Every row in the truth table in which the member of {A} is true is a row in which "~(A -> (B & ~B))" is true.
/
Quoting Leontiskos
By the definition I posted in this thread probably at least three times. Again:
An inference from a set of formulas G to a formula P is valid
if and only if
every interpretation in which all the members of G are true is an interpretation in which P is true.
For sentential logic, that is equivalent with:
An inference from a set of formulas G to a formula P is valid
if and only if
Every row in the truth table in which all the formulas in G are true is row in which P is true.
Hey, Leontiskos, read!
So then looking at either example:
Quoting TonesInDeepFreeze
Or:
Quoting TonesInDeepFreeze
Either way, your claim is not fulfilled. In the first two arguments (4) does not follow from the truth of (1) and (2), and in the third argument (4) does not follow from the truth of (1), (2), and (3). As I've noted multiple times, your appeal to the "truth table" involves an arbitrary selection of certain premises and an arbitrary exclusion of others.
For example, in the third example the "truth table" would be one where the first three premises are true, and no such truth table exists. Given that it doesn't even exist, it surely is not going to help us in the way you claim that it will:
Quoting Leontiskos
I'm replying to a bot programmed to not understand anything about this subject, not even to understand the inference rules of sentential logic nor how to read a truth table, and to skip recognition of replies already given.
Again:
Every row in which "A -> (B & ~B)" is true is a row in which "~A" is true.
Every row in which "A" is true is a row in which "~(A -> (B & ~B))" is true.
Every row in which both "P -> Q" and "P" are true is a row in which "Q" is true.
And the third example is not vitiated by the fact that (1), (2), (3) together are inconsistent:
(a) The inference is from (1) and (2) to (4). The fact that (3) is inconsistent with (1) and (2) does not entail that (1) and (2) do not imply (4).
(b) I previous showed a truth table that does include a column for ~Q. And again, in that truth table:
Every row in which both "P -> Q" and "P" are true is a row in which "Q" is true.
Moreover, indeed, there are no rows in which "P -> Q", "P" and "~Q" are all true, so, vacuously, every row in which "P -> Q", "P" and "~Q" are all true is a row in which "Q" is true.
Moreover, "Every row in which all the Xs are true is a row in which Y is true" is equivalent with "The is no row in which all the Xs are true and Y is false". In this case, there is no row in which "P -> Q", "P" and "~Q" are true and "Q" is false.
Leontiskos, if you're going to critique basic beginning logic, then know what you're critiquing!
Quoting Leontiskos
Don't gotta love the crude, disgusting ageism there, no matter what the other poster's age is.
From the moment Tones entered the thread there have been complaints about the way he comports himself. His infantile <third-person> nonsense was the most recent chapter in this book.
Indeed! Look at the actual posts to see who "comported themself" how.
Were there any that were not from you?
Seems that you cannot coherently address what he says, so you attack the man.
Actually, there've been other first insulters in this thread.
Go back and see. Test your a priori thesis for once.
I mostly ignore users who run into a thread shitting on everyone in sight who is not a mod, and that's what I largely did when Tones entered.
Check the actual record of the posts.
Quoting Leontiskos
You ignored him for twenty-odd pages?
"When Tones entered." But his primary complaint has been that I ignore his posts (and not a few times have I logged in to find more than a dozen new posts from Tones alone, which are naturally ignored).
I think you're just unwilling to consider a closer look at the logic machine. One can paper over the differences between RAA and other inferences and get along fine, but there are also interesting differences to be catalogued. And yes, RAA does present a very interesting juncture between ancient and modern logic. MP and MT are commensurable with ancient (and colloquial) logic in a way that RAA is not. RAA directly leverages the LEM in an entirely unique way. But none of my early posts were written with logicians like you or Tones in mind, so it does not surprise me that they did not resonate with you.
If one looks at the manner that a meta-logician justifies RAA it will quickly be seen that the justification is altogether different from the other rules of inference. Looking back at my phil logic text, the author's proof takes the form of mathematical induction applied to the various levels of RAA (i.e. the number of suppositions that an RAA utilizes).
I haven't found RAA to be the most interesting part of this thread, but it should be emphasized that the OP is not ideal material for RAA. RAA ideally requires a set of axioms, the scope of which is then extended over a new proposition. The OP is really not any such thing, and I maintain that the intuitive inference to ~A has more to do with MT than RAA.
(1) The worst thing about you is that you lie about me. And that you tried to wiggle out of that with a specious point about lying, to which I've responded to twice but which you ignore.
(2) The second worse thing about you is that you reply without understanding and, often, not even recognizing that your arguments based in ignorance and confusion had already been dispatched. How ludicrous that you kept asking "by what rule?" when over and over and over you had been told "by RAA".
You did it again in your latest post. You continue to complain that I posted many posts in a row when I've already answered: I was not posting in the the thread during a time when you were posting prolifically (including posts quoting me), so I caught up later, and at a mere fraction of the number of posts as yours. The fact that others weren't posting when I was catching up is not a fault of mine. There's no rule, and there should be no rule, that a poster can't later catch up in a thread. How hypocritical and juvenile of you to demand that you can go post a ton of stuff, including quoting another poster, but that the other poster bears fault for catching up later.
(3) The third worst thing about you is that you don't know Jackson Browne about this subject yet you roar your ignorant, over-opinionated confusions about it.
RAA is derivable from MT, and MT is derivable from RAA. [Edit: To be precise, RAA is derivable from MT and LNC, and MT is derivable from RAA.]
Moreover, the next post:
You don't know Jack Kennedy about this subject.
Pythagoras's proof that the diagonal of a square is not commensurate with a side is a quintessentially famous example of reductio. And, if I recall correctly, so is Euclid's proof of the irrationality of the square root of 2.
Internet Encyclopedia Of Philosophy (if it is correct):
"As indicated above, this sort of proof [Pythagoreas's proof] of a thesis by reductio argumentation that derives a contradiction from its negation is characterized as an indirect proof in mathematics. (On the historical background see T. L. Heath, A History of Greek Mathematics [Oxford, Clarendon Press, 1921].)
The use of such reductio argumentation was common in Greek mathematics and was also used by philosophers in antiquity and beyond. Aristotle employed it in the Prior Analytics to demonstrate the so-called imperfect syllogisms when it had already been used in dialectical contexts by Plato (see Republic I, 338C-343A; Parmenides 128d). "
/
Stanford Encyclopedia Of Philosophy (if it is correct):
"Both Zeno of Elea (born c. 490 BCE) and Socrates (470399) were famous for the ways in which they refuted an opponents view. Their methods display similarities with reductio ad absurdum, but neither of them seems to have theorized about their logical procedures. Zeno produced arguments (logoi) that manifest variations of the pattern this (i.e. the opponents view) only if that. But that is impossible. So this is impossible."
/
Wikipedia (though I don't always trust Wikipedia):
"This argument form traces back to Ancient Greek philosophy and has been used throughout history in both formal mathematical and philosophical reasoning, as well as in debate."
and
"Reductio ad absurdum was used throughout Greek philosophy. The earliest example of a reductio argument can be found in a satirical poem attributed to Xenophanes of Colophon (c. 570 c. 475 BCE).[8] Criticizing Homer's attribution of human faults to the gods, Xenophanes states that humans also believe that the gods' bodies have human form. But if horses and oxen could draw, they would draw the gods with horse and ox bodies.[9] The gods cannot have both forms, so this is a contradiction. Therefore, the attribution of other human characteristics to the gods, such as human faults, is also false.
Greek mathematicians proved fundamental propositions using reductio ad absurdum. Euclid of Alexandria (mid-4th mid-3rd centuries BCE) and Archimedes of Syracuse (c.?287 c.?212 BCE) are two very early examples.[10]
The earlier dialogues of Plato (424348 BCE), relating the discourses of Socrates, raised the use of reductio arguments to a formal dialectical method (elenchus), also called the Socratic method.[11] Typically, Socrates' opponent would make what would seem to be an innocuous assertion. In response, Socrates, via a step-by-step train of reasoning, bringing in other background assumptions, would make the person admit that the assertion resulted in an absurd or contradictory conclusion, forcing him to abandon his assertion and adopt a position of aporia.[6]
The technique was also a focus of the work of Aristotle (384322 BCE), particularly in his Prior Analytics where he referred to it as demonstration to the impossible (Greek: ? ??? ?? ???????? ?????????, lit. "demonstration to the impossible", 62b).[4]"
Quoting Leontiskos
Can you show this using Prop logic? If not, then why can't it be dismissed as an artefact of the limitations of Aristotelian logic?
The IEP article is surprisingly interesting.
That was addressed long ago in this thread.
If Gu{P} |- Q & ~Q, then G |- ~P
makes no use of LEM.
However
If Gu{~P} |- Q & ~Q, then G |- P
does require LEM.
If p ? ~p, then ? ~p
And then
@Leontiskos, is there anything in the article that corresponds to what you have been attempting to articulate?
It would be hard to dispatch Tones' army of strawmen. I think they are infinite.
Within the paradigm of classical propositional logic there is a certain parity between RAA and other rules of inference (although, as noted, there are also significant differences). But the way that a different paradigm conceives of reductio vis-a-vis direct inferences will not be the same as classical propositional logic. I almost guarantee that Aristotle will see a reductio as a metabasis eis allo genos (and part of the difficulty here is that an absurdity and a contradiction are not synonyms in the historical senses of reductio ad absurdum. Metaphysical and logical absurdities are both utilized historically under that name.).
Now RAA can certainly be used to derive other classical inferences, but a large part of our discussion in this thread revolved around the question of whether RAA can be derived from MT, and this is not at all apparent. This is the question that I was most interested in, because I think the inference to ~A is based in MT. On my view there is merely an analogy between RAA and MT, such that RAA is not an instance of MT. Again, this question was dealt with at some length in the middle of the thread, and the reason we ended by talking about RAA is because neither you nor Tones were comfortable arguing directly from MT.
Quoting Banno
What I mean is that when logic becomes purely formal, abstracted entirely from natural language, then RAA and the suppositions that attend it take on a more central role. It becomes primarily a way to elaborate and extend a system.
That is interesting and a bit mind-bending, but it goes to my point above that meta-logical justifications of RAA tend to be sui generis. IEP calls Whitehead and Russell's approach "idiosyncratic." I have no doubt that there are any number of creative attempts to justify reductio in classical propositional logic. It does not reduce as easily to the other rules of inference.
A more stark way to put the difference between a direct proof like MP and an indirect proof like RAA, is that in a dialogical context (which is my primary context) a MP cannot be rebuffed, but a reductio can. Laymen and logicians alike are on occasion apt to say, "An absurdity? A contradiction? So what? 'I contain multitudes'."
RAA is certainly a valid inference in classical logic.
Quoting Leontiskos
I'll invite you to set out an example. It might be helpful.
I'll invite you to derive RAA from MT as a way to engage with what I've already written.
Quoting flannel jesus
And the conclusion is ??(?^~?)?~p, one of the variants of RAA.
Others have been offered.
Read the articles.
Look up Pythagoras to start.
Quoting Leontiskos
It is apparent that RAA can be derived from MT and LNC. (Among non-dialetheists, LNC should be uncontroversial):
RAA:
If Gu{P} |- Q & ~Q, then G |- ~P
MT:
If G |- A -> B and G |- ~B, then G |- ~A
An instance of MT:
If G |- P -> (Q & ~Q) and G |- ~(Q & ~Q), then G |- ~P
LNC:
{} |- ~(Q & ~Q)
To derive RAA from MT and LNC:
Show:
If G |- P -> (Q & ~Q) and G |- ~(Q & ~Q), then G |- ~P
and
{} |- ~(Q & ~Q)
implies
If Gu{P} |- Q & ~Q, then G |- ~P
Suppose:
If G |- P -> (Q & ~Q) and G |- ~(Q & ~Q), then G |- ~P
and
{} |- ~(Q & ~Q)
Show: If Gu{P} |- Q & ~Q, then G |- ~P
Suppose: Gu{P} |- Q & ~Q
So G |- P -> (Q & ~Q)
{} |- ~(Q & ~Q), so, a foritori G |- ~(Q & ~Q)
So G |- ~P
The other direction drives MT from RAA:
Show:
If Gu{P} |- Q & ~Q, then G |- ~P
implies
If G |- P -> Q and G |- ~Q, then G |- ~P
Suppose: If Gu{P} |- Q & ~Q, then G |- ~P
Show: If G |- P -> Q and G |- ~Q, then G |- ~P
Suppose: G |- P -> Q and G |- ~Q
{P} |- P
So Gu{P} |- Q & ~Q
So G |- ~P
Quoting Leontiskos
It's more than hard. It's impossible. Because it's impossible to dispatch what doesn't exist
Quoting Leontiskos
Your guarantees or "almost" guarantees are as worthless as a day pass to a rickety ride park closed for serial safety violations fifty years ago.
If Gu{P} |- Q & ~Q, then G |- ~P.
It's merely a matter of showing that if G along with P proves a contradiction, then there are no interpretations in which the all the members of G are true and P is true.
Proving that is actually rather routine and dull, hardly creative.
What is supposed to be the point of that? Classical logic doesn't excuse contradiction.
RAA derived from MT and LNC. Done.
Quoting Leontiskos
No, in a natural deduction system it is not a mere "elaboration" nor "extension". It is crucial for proving negations; it is key to having a system that deals with negation. Without such rules, the system would not be complete, in the sense that there would be validities not provable.
It's done frequently.
For example, proving the deduction theorem (thus deriving the rule of '-> introduction') for a Hilbert style system. That's one of the key topics in an early chapter of just about any intro text in mathematical logic that uses a Hilbert style system. Cf., e.g., Enderton's 'A Mathematical Introduction To Logic'.
I think my insult was in fact in my thread, not on this one.
?es, I did. In my thread, I explored the question "On the flip side, can the English meaning of "A does not imply B" be converted to first order logic formulas?". My conclusion thus far is no.
I referred to the last clause in this quote as it is still posted:
Quoting Lionino
"A does not imply a contradiction" is not a true statement about "(A?¬(B and ¬B))".
First, this is not a derivation of RAA. It is a putative modus tollens that looks a little bit like an RAA. As I said, there are analogical similarities.
Second, this is precisely what we argued about in the middle of the thread, and no one was willing to accept this sort of argument as a straightforward modus tollens. Again, the whole reason you have been so laser-focused on RAA is because the MT is so unconvincing. Recall that in order to run this "modus tollens" one must conceive of the contradiction as false (or 'FALSE') in a manner that is sui generis for a non-simple logical formula. Earlier in the thread you characteristically begged off from entering into that meta-logical dispute, and as a consequence refused to try to prove ~A via modus tollens.
-
Setting this out more clearly:
A corollary of what I have been arguing in this thread is the idea that reductio ad absurdum is a kind of black sheep in the logical family, or that there is a measure of discontinuity between RAA and the other inferences of classical propositional logic, such that there is no straightforward derivation of RAA from these other rules of inference.*
Now someone like yourself who is unwilling to engage in meta-logical discourse will naturally have a hard time seeing my thesis, and for this reason my thesis was directed towards people like Count Timothy rather than you or Tones. You say that I have made a number of well-documented errors in this thread. This is assertion and hot air which can in no way be substantiated, but there is a way for you to show that my corollary is mistaken. The corollary is mistaken if there is a straightforward derivation of RAA from the other rules of inference in classical propositional logic. All you need to do is provide such a derivation.
* What I have said more recently is that the more purely formal a system is, the less this discontinuity of reductio ad absurdum is able to be recognized.
I did.
And his is okay too; all it needs is to be generalized as I did with G.
The more formal the system, the more precisely we can determine exactly what is and what is not permitted by it.
The formalization doesn't overstep the ordinary, informal argument form of reductio ad absurdum. Rather, the formalization requires that the form be adhered to strictly.
Meantime, look up Euclid and Pythagoras to see that the argument form does go back to the ancients. Moreover, take a look at the quotes from the articles about the argument form.
So, RAA must be derivable from non-intuitionistic MT without even invoking LNC. Though, of course, non-intuitonistic MT does entail LEM, while my proof does not.
Mates lists LNC as a theorem in his system with non-intuitionistic MT. So, with that theorem, we would continue with the proof I gave. A full proof would have to show the details in deriving LNC.
In a primarily inferential system like classical propositional logic some can be derived and some cannot (i.e. some rules of inference are technically superfluous). For example, if one has access to the disjunctive syllogism then they also have effective access to the classical modus tollens (over the material implication). So there will be unique and irreducible inference-axioms in any inferential system, but my claim is that RAA is uniquely unique.
The text that we used for philosophical logic was the second edition of Harry Gensler's Introduction to Logic, which is very wide ranging. The chapter that precedes, "Chapter 6: Basic Propositional Logic," is a chapter on induction. In the last two pages of that chapter he points out the irreducibility of modus ponens and brings in Aristotle (link). Note that the beginning of the first sentence is, "Some suggest that we approach justification in inductive logic the same. . ."
Now justifying the inference-axiom of modus ponens is not overly problematic (although it does require induction and/or intellection), but justifying the inference-axiom of RAA is rather more problematic. This difficulty has shown up often in this thread in the way that folks wish to continue treating RAA glibly even when more fundamental questions are being asked. This goes to my point that it is easier to reject a RAA than a MP. To accept the premises of a MP is to already have implicitly accepted the conclusion. Not so for an RAA, as Tones' claim demonstrates.
I know. I don't edit all posts that I regret. But I don't agree now with what I wrote on that post.
What you say here is blatantly erroneous.
There is nothing "putative" about the use of MP, and the resulting schema is an instance of RAA.
Quoting Leontiskos
Such derivations have been presented here by several folk, including the one from the IEP given above.
Quoting Leontiskos
And then repeated them, despite their being shown to be wrong.
Quoting Leontiskos
...for someone who has failed to engage correctly with formal logic this may indeed be so.
Quoting Leontiskos
In axiomatic systems RAA is derived.
Go to https://www.maths.tcd.ie/~odunlain/u11602/online_notes/pdf_propos.pdf and you will find an axiomatic system. Look at proof 10.5. It is a proof of (~A?A)?A in an axiomatic system. Sadly, again, you are simply wrong, RAA is not "irreducible". Whatever an "inference-axiom" is - did you mean "theorem"?
So what can be redeemed here? Perhaps this might help:
Quoting Banno
You've pulled a 180! Earlier you literally rejected my characterization of the argument as a modus tollens and said:
Quoting Banno
After you decried that you wanted nothing to do with the modus tollens, I replied:
Quoting Leontiskos
Again, this has all been addressed at some length in these earlier posts, including the problem for the modus tollens where (B?¬B) and ¬(B?¬B) are equally capable of furnishing the modus tollens with its second premise. At this point I am bored of revisiting old material as you continue to contradict yourself.
Quoting Banno
IEP proves my point, "Whitehead and Russell in Principia Mathematica characterize the principle of reductio ad absurdum as tantamount to the formula (~p ?p) ?p of propositional logic. But this view is idiosyncratic. Elsewhere the principle is almost universally viewed as a mode of argumentation rather than a specific thesis of propositional logic."
Quoting Banno
Which is of course not RAA, unless you want to follow Whitehead and Russell in their idiosyncratic view. You're just Googling at random to try to support a claim you made, a claim which you said was elementary.
See the response I gave:
Quoting Leontiskos
But that's a silly game anyone can play.
Eventually, though, one stops playing the onerous guessing game.
Here's the thing: nearly every one of your posts in this thread contains factual errors.
Assertions, assertions, and more assertions. And when asked to provide substantiation, you fly like a little bird, even when one goes ahead and does the setup work for you:
Quoting Leontiskos
Using this post here to list all the goofy thought experiments of analytic philosophy.
How it is like to be a bat (privacy of mental states)
Supertasks (Zeno's paradox)
P-zombies (mind-body dualism)
Mary's room (-)
Brain in a vat or virtual reality hypothesis (solipsism)
[/hide]
I don't think that's quite right, depending on what you meant by "generally".
Let's continue to ignore the OP's use of "implies" and consider what's going on when someone says this sort of thing.
(1) If Smith wins the election, there will be a recession.
That claims some kind of link between one event and another. To contradict this claim, you deny that the link holds:
(2) If Smith wins the election, there may or may not be a recession.
That's the straightforward denial of (1).
But now consider
(3) If Smith wins the election, there will not be a recession.
That proposes another link, and I would suggest that in everyday reasoning the truth of (3) requires the falsity of (1), even though P?~Q does not entail ~(P?Q), which indeed does seem to be a problem for material implication.
After all, if P requires that ~Q, it can hardly require that Q.
I think people do recognize the difference even in everyday reasoning, and would accept that (2) is the simple contradiction of (1), and that (3), while also denying (1) a fortiori, is a much stronger claim.
It's clear with quantifiers too:
(4) All Englishmen are honorable.
(5) Not all Englishmen are honorable -- some are, some aren't.
(6) No Englishmen are honorable.
If no Englishmen are honorable, then it stands to reason that not all of them are, but that's a much stronger claim than simply denying that being English entails being honorable.
(Apologies if this was already covered.)
To bring this full circle, consider your post which started us off on this long trek:
Quoting Lionino
Or, "((A?B)?(A?¬B)) and (A?(B?¬B)) are the same formula."
Now it is surely true that < ((A?B)?(A?C)) and (A?(B?C)) are the same formula >.
But are these the same sort of equivalence?
I want to say that (B?¬B) and (B?C) are not meta-logically equivalent, and because of this the truth tables are misleading.
-
Or going back to this:
Quoting Lionino
Logically speaking, (A?C)?C is invalid for any (non-A) substitution of C. That's just what it means for a formula to be invalid. Yet when we substitute (B?¬B) for C it magically becomes valid. There is a kind of meta-contradiction here insofar as (A?C)?C is both invalid and not invalid. It is invalid according to the truth table, and it is not invalid given the fact that we can substitute some C which makes the formula valid.
This is what I was warning about in my first posts on metabasis eis allo genos:
Quoting Leontiskos
(Note that to say, "It works fine, just look at the truth table!," is exactly like saying, "The ethanol fuel works fine, just look at my gas gauge!")
Thanks for bringing it back to the beginning, which is what I was also trying to do in my last post.
Quoting Srap Tasmaner
This is what we were just talking about on page 8 of the related thread, "What can we say about logical formulas/propositions?"
Quoting Srap Tasmaner
Agreed, but I stand by my original claim that (1) contradicts (3).
Quoting Srap Tasmaner
Yes, this is the square of opposition. "Some S are P" and "No S are P" are contradictories, whereas "All S are P" and "No S are P" are "contraries." But "contraries," when used in this sense, are also contradictories.
But your point is well made. The two parties would be speaking "contraries" and not merely contradicting.
(For some reason I did not receive a notification of your post.)
Quoting Leontiskos
But when we place a contradiction in the consequent of a conditional it is no longer conditional (e.g. (A?(B?¬B)). So if it is a meta-principle of classical propositional logic that all conditionals are conditional, then allowing the contradiction has upended this meta-principle.
Thus when draws ~A, he has implicitly accepted that the conditional is no longer a conditional. The objection to his move is to say that anything which undermines the meta-principle that all conditionals are conditional should not be allowed into propositional logic. Ergo: we should not accept and affirm formulae which are known to contain contradictions. This objection is of course not open to the truth-functionalist who has no recourse to meta-objections (despite the fact that these internal contradictions destroy the formal nature of validity).
I made a similar point earlier:
Quoting Leontiskos
In other words, this is an invalid form:
(Note that these sorts of exceptions created by (b?¬b) are popping up in many places. For example, "(A?C)?C is invalid for any (non-A) substitution of C," and yet (b?¬b) creates an exception.)
One could of course be very analytical and simply say that we must choose between, say, the principle that all conditionals are conditional, and the permissibility of introducing standing contradictions into formulae. Still, I think it is fairly obvious that classical propositional logic is built for the former rather than the latter. There are special moves built-in to the systemin this case RAAthat provide a way to navigate such meta-difficulties, but unlike standard inferences RAA is "a mode of argumentation rather than a specific thesis of propositional logic" (IEP). RAA pertains to the boundary of the system, not the interior.
Quoting Leontiskos
In my opinion the creators of classical propositional logic never intended for the system to accommodate standing contradictions. ...Or in the language I used earlier, "internal contradictions," or contradictions contained within the "interior logical flow of argumentation."
Leontiskos uses my name for a link to his own post. That deserves a link to my reply to his post:
https://thephilosophyforum.com/discussion/comment/922505
Leontiskos is to serious philosophical engagement as salmonella is to a seriously good dinner.
What rule is used? What rule is used? What rule is used?
Questions, questions, questions.
RAA. RAA. RAA.
Answers, answers, answers.
Prove RAA from MT. Prove RAA from MT. Prove RAA from MT.
Challenges, challenges, challenges.
Proof of RAA from MT and LNC. Proof of RAA from MT and LNC. Proof of RAA from MT and LNC.
Challenge met, challenge met, challenge met.
What are 'B' and 'C'?
If 'B' and 'C' are atomic sentences, then
(B?¬B)
and
(B?C)
are not equivalent.
Buy if 'B' and 'C' meta-variables ranging over sentences, then we note that it is not the case that for all sentences B and C, we have that (B?¬B) and (B?C) are equivalent. But that doesn't entail that there is no sentence C such that (B?¬B) and (B?C) are equivalent. For example, we may instantiate both 'B' and 'C' to a sentence P (where 'P' is not a variable but is an atomic sentence), and (P & ~P) and (P & ~P) are not equivalent and indeed the same. Or, we could instantiate 'B' to P and 'C' to Q & ~Q (where 'Q' is not not a variable but rather an atomic sentence), and (P & ~P) is equivalent with (P & (Q & ~Q)).
Truth tables are not "misleading" in this regard.
(1) "If Smith wins the election, then there will be a recession."
(P -> Q in ordinary symbolic logic)
That is understood as asserting a relation (perhaps necessity, causality or other relevance) between the antecedent and consequent.
But it does agree with the material conditional in one sense: The second row of the truth table for P -> Q (Smith wins but there will not be a recession) is not the case. That is, if "If Smith wins the election, then there will be a recession" is true, then it is not the case that both "Smith wins the election" is true and "there will be a recession" is false.
(2) "It is not the case that if Smith wins the election then there will be a recession."
(~(P -> Q) in ordinary symbolic logic)
That is understood as denying that there is a relation (perhaps necessity, causality or some other relevance) between the antecedent and consequent.
But it seems not to suit the material conditional at all: No row of the truth table for ~(P -> Q) (equivalent with P & ~Q) pertains, since "It is not the case that there is a (necessary, causal, or other relevance) relation between Smith winning the election and there being a recession" would not be understood as being equivalent to "Smith will win the election and there won't be a recession".
(3) "If Smith wins the election, there will not be a recession."
(P -> ~Q in ordinary symbolic logic)
That is understood as asserting a relation (perhaps necessity, causality or other relevance) between the antecedent and consequent.
But it does agree with the material conditional in one sense: The first row of the truth table for P -> ~Q (Smith wins and there will be a recession) is not the case. That is, if "If Smith wins the election, then there will not be a recession" is true, then it is not the case that both "Smith wins the election" is true and "there will be a recession" is true.
Tabulating:
(1) implies not (2).
(1) implies not (3)
(2) implies not (1)
does (2) bear upon (3)?
(3) implies not (1)
does (3) bear upon (2)?
So:
(1) and (2) contradict each other.
(1) and (3) contradict each other.
(2) and (3) don't bear upon each other?
Quoting Srap Tasmaner
It's a problem for material implication if material implication were required to accord with much of everyday reasoning.
Quoting Srap Tasmaner
That's true. But I don't see its connection with the analysis.
Quoting Srap Tasmaner
Okay.
Quoting Srap Tasmaner
(3) denies (1). But what a fortiori are you referring to? In what sense is (3) stronger than (2)? Does 'stronger than' include that (3) implies (2)? Does it?
What does "conditionals are conditional" mean?
I don't know what the poster has in mind, but, of course, with A -> (B & ~B), the consequent is not contingent. That doesn't "upend" any principle of classical logic.
How refreshing to find a topological analysis of RAA and natural deduction. Advanced stuff indeed.
A sentence S is invalid if and only if there is an interpretation in which the sentence is false.
Quoting Leontiskos
Not "magically".
If 'A' and 'C' are meta-variables ranging over sentences, then there are instances of
(A & C) <-> C that are invalid and instances of it that are valid.
But if 'A' and 'C' are atomic sentences, then, of course
(A & C) <-> C
is invalid.
There's no more "magic" to that than:
x = y
is true for some values of x and y and not for other values of x and y.
/
Bottom line: We must not conflate two kinds of usage of letters:
(1) Letters as meta-variables ranging over sentences.
(2) Letters that are themselves atomic sentences.
/
Instance after instance after instance in which Leontiskos thinks he's pointed out some problem with symbolic logic, it turns out the real problem is that he's ignorant of the basics of the subject.
They are equivalent formulas, but not the same formula.
I think the matter of bringing logical propositions into English and vice-versa is still quite meaningful. After all, in this thread:
Quoting javi2541997
One of the goals of the puzzle was to state the riddle in logic. Gödel's ontological proof has renditions of logical statements into plain English. The logical argument has been verified as valid, so we know that the plain English argument is valid too.
But then, I now think that «not A without B» is the best way to translate A?B, not «A implies B» or «if A, B», whilst «A without B» for ¬(A?B); which somewhat agrees with the white bullet-points of this slideshow I found:
If I am right then it is very likely gibberish in logic as well. It is at least clear that no one knows what it is supposed to really mean.
Quoting Lionino
I'm still not sure, but I would say, along the lines of my two previous posts, that an understanding of (1) does not necessarily provide any understanding of (2), whether that understanding has to do with logic or English:
To think that (1) must provide an understanding of (2) is to think that (B?¬B) is always substitutable for P, which it is not. And again, I think this throws the original derivation of A?(B?¬B) into doubt in the first place.
NB: The riddle of the metabasis is the riddle of when we can take (B?¬B) as P and when we cannot. If we don't know the answer to this question then none of the reasoning in this thread which treats (B?¬B) as P is secure.
s be the first sentential constant:
f <-> (s & ~s)
That is not "gibberish".
Moreover, in some systems 'f' is primitive and '~' is defined by:
~P <-> (P -> f)
/
Again, if 'A' and 'C' are variables ranging over sentences, then
A -> (B & ~B) comes from A -> C by instantiation of 'B & ~B' for 'C'. There is nothing problematic with that.
A month ago I was talking about the implications of interpreting (B?¬B) as 'FALSE' and all I received were superficial objections from the very people who were interpreting it in this manner but had not yet recognized it. I'm guessing those posts will have a very long shelf life.
I was not a person who objected to defining falsum as 'B & ~B'.
There are some distinctions to be made however.
In the context of my post, falsum is a symbol, not a truth value. It is a sentential constant. The value 'false' is not a symbol.
But the truth value of falsum is false in all interpretations. Just as the truth value of 'B & ~B' is false in all interpretations.
This thread quickly turned into a discussion of (a?(b?¬b)), and a few of us raised the question of whether (b?¬b) is a "genuine proposition," to use Kimhi's language. First was Count Timothy:
Quoting Count Timothy von Icarus
Then I argued <here>, that (b?¬b) is not ontologically possible. Following Kimhi I would say that it is also not psychologically possible. The truth-functionalists trying to stick to their guns could only make sense of (b?¬b) as FALSE (or falsum, which is the same thing). As I pointed out multiple times, false is different from contradictory/absurd/impossible. And as I also argued multiple times, sentences containing (b?¬b) are not well-formed in the specific sense I laid out. In Kimhi's language (b?¬b) is not a genuine proposition. It will not do to say, "Ah, Kimhi doesn't understand Fregean logic." The point is that if Fregean logic thinks (b?¬b) is a genuine proposition, so much the worse for Fregean logic.
But I would say that even in propositional logic things like (b?¬b) are never part of the object language (and again, RAA ferrets them out). They only represent a sort of second-order psychological act, and they can only be asserted by those who do not understand that they are asserting a contradiction. But when someone contradicts themselves, the sense of what is asserted is crucially different from the sense of what is objected to, for the person asserting does not recognize the contradiction within their claim.
IMO, the issue is the reduction of logic to "formal logic," the form of argument, without any concern for the matter of an argument (what Scholastics called "material logic).
As Aristotle says:
And St. Thomas says something similar in the commentary on the Metaphysics
[Quote]
It is impossible for anyone to actually adopt or believe the view that one and the same thing both is and is not in a given respect, even though some have attributed this opinion to Heraclitus. For while it is true that Heraclitus said this, yet it was not possible for him to believe what he said.Nor is it necessary that everyone has in mind or really believes everything that they say.Nor is it necessary that everyone has in mind or really believes everything that they say.
[/quote]
But if "logic" means only the study of words, language games, and not discourse/argument, then contradiction plays a different role, for surely we can speak the assertion of a contradiction, even if we cannot believe it.