And yet your claim that Reductio is invalid is just wrong. — Banno

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.

The problem is that this A, A→¬B∧B ⊢ ¬A was given as reductio ad absurdum. But this entails anything. — Lionino

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.

What if we reject (1) instead? Then A is made true, but it does not imply (B∧¬B). — Leontiskos

This is the path that @Banno and @TonesInDeepFreeze have chosen:

• (a→(b∧¬b)) → ¬a

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:

• Mt. Caradhras = reductio ad absurdum
• Mines of Moria = A modus tollens with only one premise

Note: For those considering the mines, two posts may be especially useful: first, second.

The grain of truth in Leontiskos' position is that reductio arguments need to be used with care. — 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.

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.~a → ~(b ∧ ~b) - contraposition (1)
4. ~a - modus ponens (2,3) — Count Timothy von Icarus

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:

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." — Leontiskos

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. — Banno

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?

- I understand perfectly well why many people call you a troll, but it isn't exactly the right word. The search for the right word goes on...

- I think it is silly, too. That's why I coined the term in a silly way. But you are the one who requires that sort of thing for the ¬A you wish to draw. My point in all this is that we should not allow contradictions into sentences.

If ¬(B∧¬B) is true, as it must be, then this is not a valid use of modus tollens. — Banno

See the original post where I already explained this interpretation:

Note that we could also do other things, such as treat the second premise as truth incarnate, but this is harder to see:

A→(B∧¬B)
¬(B∧¬B), but now conceived as "true"
∴ ¬A does not follow

...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). — Leontiskos

- Haha - it will take awhile to wrap my head around that, but "modus tollendo ponens" looks fun. :smile:

Why is the second line a quote? — Banno

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.

But A, A→¬B∧B ⊢ A is also valid — Lionino

Thank you. As I put it:

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). — Leontiskos

So the question is: how do we choose between either? Isn't it by modus tollens? — 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.

---

And what I am after is a straight forward explanation of what "FALSE" is. — 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:

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? — 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:

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. — 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:

Thus when Banno says that a contradiction (b∧¬b) is false, does he mean that it is false or that it is FALSE?... [argument continues on] — Leontiskos

First argument

• A→(B∧¬B)
• "(B∧¬B) is false"*
• ∴ ¬A

Second argument:

• A→(B∧¬B)
• ¬(B∧¬B)
• ∴ ¬A

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)

Rubbish. — Banno

The proof still exists from your heavily-edited post. Why are you editing posts long after they have been responded to? See:

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 — Banno

Leo, what is "FALSE"? — 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:

It doesn't. Explained last time you made this claim... — Banno

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).

what is "FALSE"? — Banno

I tried to look at this here:

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

Right.

The matter with modus tollens is that q could be otherwise, while in reductio it is not the case by definition. Then again, I don't think it is meaningful or interesting. — Lionino

It's the thing I've been going on about the whole time.

I don't see a meaningful difference. — 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: <p→q; ¬q; ∴¬p>.

What I claim is that what is at stake is not a bona fide modus tollens:

a→(b∧¬b)) is True, (b∧¬b)) is False, therefore a is False (from «1»). — 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:

• A→(B∧¬B)
• ¬(B∧¬B)
• ∴ ¬A

Are they both valid? As I said:

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. — Leontiskos

So I ask again: How is it that something and its negation can both [function as the second premise of a modus tollens]? — Leontiskos

Edit:

Earlier quote in context:

We can apply Aristotelian syllogistic to diagnose the way that the modus tollens is being applied in the enthymeme:

((A→(B∧¬B))
∴ ¬A

Viz.:

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

In this case the middle term is not univocal. It is analogical (i.e. it posses analogical equivocity). Therefore a metabasis is occurring. — 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?

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

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?

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.

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

Is this just obvious, ?

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.

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.

(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.

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.

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)"

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.)

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.

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.

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

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.

It's not a translation of the sentences discussed. — TonesInDeepFreeze

You said:

"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. — TonesInDeepFreeze

In saying this are you saying, among other things, that these two claims are not equivalent?

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 thinks his truth table has shown that (2) translates a reductio.)

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

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:

...and such is in line with ↪Banno'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). — 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?

Has everyone agreed by this point that 's truth table does not fully capture what a reductio is? (See bottom of post for truth table)

((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:

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.*

* A reductio requires special background conditions. In this case it would require the background condition that (1) is more plausible than (2). — 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:

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". — Lionino

And it is also related to Tone's point

"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. — 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, " <If A implies B & ~B, then A implies a contradiction> is true, and it is a translation of the sentences, not a statement about them."

Or if you think it is only truth-functional if it fits in a truth-table: — Banno

'non-particular' is your word. It's up to you to say what you mean by it. — TonesInDeepFreeze

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?

(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. — TonesInDeepFreeze

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:

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).

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.

Sorry - falling behind in this thread.

↪Leontiskos I solved my main problem just right above. — Lionino

I don't know if you saw my edit, which may now be redundant:

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

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 — 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.

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

Very good. This is another instance of the wrinkle that is created when the contradiction is allowed.

...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. — 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.

(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)) — Lionino

Good. This back to 's point about ¬¬a.

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:

(3) A ... toward a contradiction — 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.

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? — Lionino

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:

I think I finally solved my own problem. When translating it to natural language, I was misplacing the associativity of the → operator.
So ¬(A → (B∧ ¬B)) is the same as (¬A) → (B∧ ¬B) — 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.

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. — sime

Thanks. This is what I was trying to remember but could not find online (i.e. the complexities surrounding proofs of ¬¬a).

The negation of a contradiction is always true, and being true it is implied by anything, true or false. — Count Timothy von Icarus

Yes, good. :up: Kreeft's point comes back.

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. — 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.

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. — Count Timothy von Icarus

Yep.

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. — Count Timothy von Icarus

I think metabasis is useful, but I don't close my eyes to the fact that it is metabasis:

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). — 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*.

Where — Lionino

Here:

1. A→(B∧¬B) assumption
2. A assumption
3. B∧¬B 1,2, conditional proof
4. ~A 2, 3 reductio — 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.