limiting yourself to truth-functional logician — Leontiskos

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.

You lie again. — TonesInDeepFreeze

Nonsense, and you entirely failed to answer the question:

...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? — 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.

In this are you saying that these two claims are not equivalent?

"If A implies B & ~B, then A implies a contradiction"
(a→(b∧¬b))→¬a — Leontiskos

One is a statement in the meta-language and the other in the object language. They are different levels of statement.

One is a statement in the meta-language and the other in the object language. They are different levels of statement. — TonesInDeepFreeze

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:

1. "If A implies B & ~B, then A implies a contradiction"
2. (a→(b∧¬b))→¬a

(My claim here is that (1) represents a reductio whereas (2) does not, even though ↪Banno thinks his truth table has shown that (2) translates a reductio.) — Leontiskos

Unless you think you can use the word "particular" without having any idea what it would mean for something to be non-particular? — 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.

I think your charges of "misrepresentation" are all bosh — Leontiskos

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.

You lied — TonesInDeepFreeze

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:

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. — 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.

I don't know if I missed this or if it was an edit, but:

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. — 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:

A contradiction is an assertion of Propositional Logic that is false in all situations; that is, it is false for all possible values of its variables. — 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.)

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. — 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.)

From ~(A -> (B & ~B)) we infer that A implies no contradictions.

From (~A -> (B & ~B)) we infer that A implies no contradictions. — TonesInDeepFreeze

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:

From there things start to make more sense. — 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. — Leontiskos

Well I forgot what was the stake of the discussion you all are having, but

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. — 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.

Checking the natural language equivalent of logical terms.

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.

The truth-functionalist is likely to object to me, “But your claims are not verifiable within classical logic!” Yes, that is much the point. When we talk about metabasis eis allo genos, or contradiction per se, or reductio ad absurdum, we are always engaged in some variety of metalogical discourse.

When I said things like:

Thus when Banno says that a contradiction (b∧¬b) is false, does he mean that it is false or that it is FALSE?
— 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). — 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:

I understand that you'd think that B∧¬B should be able to be replaced by any proposition P, but that is not the case.

Example:
(A∧(B∧¬B))↔(B∧¬B) is valid
But (A∧C)↔C is invalid. — 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:

• P→Q
• ∴¬P

The revised textbook would need to add an asterisk: "This is a fallacy*."

" *Except in that case where Q = (b∧¬b)"

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. — Lionino

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.

And in each of the invalid cases if "B" could be made necessarily false they would presumably hold. — Leontiskos

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.

I'd like to explore this idea next:

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. — Lionino

(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. — 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.

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. — Leontiskos

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.

I'd like to explore this idea next:

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.
— Lionino — Lionino

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.

Retracted. It is dumb.

I'd like to explore this idea next — Lionino

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:

-Any consequent which is false proves the antecedent
-(B∧¬B) is a consequent which is false
∴ (B∧¬B) proves the antecedent — Leontiskos

-

Example:
(A∧(B∧¬B))↔(B∧¬B) is valid
But (A∧C)↔C is invalid. — Lionino

Perhaps it is right to say that the contradiction introduces exceptions to invalidity but not to validity.

some rules of classical logic to come into conflict — Leontiskos

Such as?

-Any consequent which is false proves the antecedent
-(B∧¬B) is a consequent which is false
∴ (B∧¬B) proves the antecedent — 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.

A→FALSE — 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".

Another way to read the first argument, and the one I prefer*, is as follows:

A→ABSURD
∴ "A cannot be affirmed" — 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.

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. — TonesInDeepFreeze

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.

That was my interpretation of Banno, not Banno himself. — Leontiskos

Presenting a statement that someone has not made is not presenting a translation.

I think your charges of "misrepresentation" are all bosh — Leontiskos

I agree with Tones that you habitually misrepresent positions that are counter to your own, here and elsewhere.
_________________

Has everyone agreed by this point that ↪Banno's truth table does not fully capture what a reductio is? — 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 ~ρ given a proof of (μ ^ ~μ) with ρ as assumption, a form displayed in the truth table.

The easiest way to see this is to note that a reductio ad absurdum is not formally valid — Leontiskos

This 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 ρ,μ ⊢φ^~φ, we can write that ρ→~μ or we can write that μ→~ρ. (Tree proof 1; Tree proof 2)

Leo seems to think that choosing between ρ→~μ and μ→~ρ 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, ρ→~μ or μ→~ρ, is valid.

Indeed, the "problem" is not with reduction, but with and-elimination. And-elimination has this form
ρ^μ ⊢ρ, or ρ^μ ⊢μ. We can choose which inference to use, but both are quite valid.

We can write RAA as inferring an and-sentence, a conjunct:

ρ,μ ⊢φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ)

(ρ^μ) →φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ)
(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

ρ,ρ→μ ⊢μ

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? — Lionino

Such as your example indicates here: . Does classical logic not presuppose that such substitution is truth-preserving?

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. — Lionino

Yes, indeed.

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". — 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).

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. — Lionino

You are giving a reductio, so this all goes back to my points about reductios:

1. Dogs are fish.
2. Fish, among other things, is defined as not-mammals.
3. Dog is defined, among other things, as mammal.
4. "A mammal is not a mammal"
5. Contradiction; reject 1

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.

If this is right then (b∧¬b) introduces instances of formal equivalence that are not provable. — Leontiskos

Is this just obvious, ?

What is the supposed difference between "false" and "FALSE"?

Does classical logic not presuppose that such substitution is truth-preserving? — Leontiskos

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.

(B∧¬B) is ambiguous, and can be interpreted as p or as FALSE (i.e. always-false). — 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.

Why did you reject (1) and not (2) or (3)? The reductio is not formally valid in that tight sense. — 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.

Presenting a statement that someone has not made is not presenting a translation. — Banno

I literally said it was an interpretation, not a translation. I still see it as the better option.

Either inference, ρ→~μ or μ→~ρ, is valid. — Banno

This does not contradict what I have been saying. Many in this thread have been concluding ~μ, not (ρ→~μ) ∨ (μ→~ρ).

• A→(B∧¬B)
• ∴ ¬A

Rather than:

• A→(B∧¬B)
• ∴ (¬A ∨ ¬(A→(B∧¬B)))

This is on point.

and see that the choice is not in the reductio but in choosing between the conjuncts. — Banno

Tomato tomato. A reductio without choosing between them is not yet a reductio.

I literally said it was an interpretation, not a translation. — Leontiskos

Ok, Presenting a statement that someone has not made is not presenting an interpretation. :roll:

This does not contradict what I have been saying. — Leontiskos

Quite so. So what? It remains that RAA is a valid inference in classical propositional logic.

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. — Lionino

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.

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. — Lionino

My question then is whether we ever utilize (B∧¬B) without conceiving of it as a kind of P.

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. — 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?

My question then is whether we ever utilize (B∧¬B) without conceiving of it as a kind of P. — Leontiskos

If P can only be False, yes; otherwise, no.

So do we have a proof for ((a→(b∧¬b)) → ¬a)? — Leontiskos

Uh

Leo seems to think that choosing between ρ→~μ and μ→~ρ 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, ρ→~μ or μ→~ρ, is valid.

Indeed, the "problem" is not with reduction, but with and-elimination. And-elimination has this form
ρ^μ ⊢ρ, or ρ^μ ⊢μ. We can choose which inference to use, but both are quite valid.

We can write RAA as inferring an and-sentence, a conjunct:

ρ,μ ⊢φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ)

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. — Banno

I think Leontiskos is talking about choosing between the conjuncts, while Banno is correctly stating that reduction ad absurdum is formally valid.

I think the only way we can utilize logical inference is by using the modus tollens — 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.

Ok, Presenting a statement that someone has not made is not presenting an interpretation. — Banno

Here's the quote:

Thus when Banno says that a contradiction (b∧¬b) is false, does he mean that it is false or that it is FALSE? — 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!?

Quite so. So what? It remains that RAA is a valid inference in classical propositional logic. — Banno

Hmm? See:

A reductio without choosing between them is not yet a reductio. — Leontiskos