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. — Lionino

I edited that post a bit, perhaps after you read it. For example:

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. — Leontiskos

This isn't a proof of Modus tollens. — flannel jesus

I was not trying to say it was.

ρ→(φ^~φ) (premise)
~(φ^~φ) (law of non contradiction)
:. ~ρ (modus tollens) — flannel jesus

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 (φ^~φ) and ¬(φ^~φ).

<I believe the reductio has failed and that the only strict way to draw ¬A is by using the modus tollens.>

• A→(B∧¬B)
• ∴ ¬A {modus tollens}

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:

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. As I said earlier... — 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:**

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." — 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 don’t 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.)

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. — Lionino

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:

• A→(B∧¬B) {Assumption}
• A {Supposition}
• ∴ ¬A {reductio ad absurdum}

But we could equally say:

• A {Assumption}
• (A→(B∧¬B)) {Supposition}
• ∴ ¬(A→(B∧¬B)) {reductio ad absurdum}

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:

1. A→(B∧¬B) {Assumption}
2. A {Assumption}
3. ∴ (¬A ∨ ¬(A→(B∧¬B)))

...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.

But the issue is that we already know S is true. — 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.

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. — Lionino

What's at all wrong with this?:

(S∧¬P)→(B∧¬B)
¬P
∴ ¬S

We're going in circles. Time to go <meta>.

This is the RAA, innit? :smile: — Lionino

I was already convinced that RAA is insufficient. That as you say:

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

RAA will not prove ¬A.

(S∧¬P)→(B∧¬B)
S
∴ P — Lionino

The logic of the RAA proves (¬S v P), and the RAA choses one or the other.

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.

...

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... — Leontiskos

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.

- I'm quite serious. See my edit to that post, which may help you.

Straight RAA does not require the "and elimination". It's an additional step when there are multiple assumptions. — Banno

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:

A→ABSURD
∴ "A cannot be affirmed"

...

Introducing ABSURD in the way I did above destroys the LEM of classical logic. — Leontiskos

While you are there, what does "FALSE" mean? — Banno

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:

how do you prove that you may derive ~ρ from ρ→(φ^~φ)? — Lionino

I consider it an open question as to whether this question is answerable.

So, what is a "direct proof"? I gather you think using MT is direct, but RAA isn't? WHat's the distinction here? — Banno

Modus tollens requires no "and-elimination" step. Is that a good way to put it in your language?

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. — Count Timothy von Icarus

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.

And then we can do the same thing assuming ~B. That covers all our options assuming LEM. — Count Timothy von Icarus

I am wondering if allowing contradictions in this way fiddles with the LEM, but this is just conjecture.

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). — 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.

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