But (a→b)∧(a→¬b) being False simply means that A does not imply a contradiction, it should not mean A is True automatically. — Lionino

Here is the thing about this. Something not implying a contradiction does not mean that something is true. Of course. We have established that "A does not imply a contradiction" is not a good reading of ¬(a→b∧¬b). Yet "A implies a contradiction" is a good reading of a→b∧¬b. So if we say "A implies a contradiction" is false, it is the same as saying "A does not imply a contradiction", so saying (a→b∧¬b) is false is the same as saying ¬(a→b∧¬b) is true (see truth tables).
So, because "¬(a→b∧¬b) is true" is the same as "(a→b∧¬b) is false", ¬(a→b∧¬b) should indeed be read as "A does not imply a contradiction" when it is true.
I can't take this anymore.

The problem is that modus tollens can be proven syllogistically quite easily, but how do you prove that you may derive ~ρ from ρ→(φ^~φ)? — Lionino

Prove RAA without MT?

Interesting problem.

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.

Perhaps I caught my mistake:

"A implies a contradiction" is false, it is the same as saying "A does not imply a contradiction" — Lionino

The two might not be the same in natural language, perhaps there is the mistake, hence ¬(a→b∧¬b) still shall not be read as "A does not imply a contradiction" when it is true.

perhaps there is the mistake — Lionino

Thread solved again.

RAA proves (¬S v P). — Leontiskos

I don't understand.
(S∧¬P)→(B∧¬B)
S
∴ P is supposed to be the definition of RAA according to where I got it from.

.

Seems right, you can prove MT with contraposition and MP.

But you can do a disjunctive syllogism too.

And the disjunctive syllogism
1. a → (b ∧ ~b)
2. ~(b ∧ ~b)
3.~a V (b ∧ ~b) - material implication (1)
4. ~a - disjunctive syllogism (2,3)

Then again, this probably also amounts to the same thing.

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>.

(S∧¬P)→(B∧¬B)
¬P
∴ S — Leontiskos

We don't know whether P is true or not. We know S is true. S being true and P being false leads to a contradiction. Therefore we have ascertained that P is true. No assumption is needed or allowed.
Oh, by the way, S does not follow from (S∧¬P)→(B∧¬B) and ¬P, because that would be a contradiction. Perhaps you meant.
(S∧¬P)→(B∧¬B)
¬P
∴ ¬S
But the issue is that we already know S is true.
And: (S∧¬P) → (B∧¬B), ¬P does not entail ¬((S∧¬P) → (B∧¬B)), so you can't deny the first premise either.

Then again, this probably also amounts to the same thing. — Count Timothy von Icarus

Mayhaps.

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

That's not how contraposition works.

Edit. I see you corrected yourself already, nevermind.

The problem is that modus tollens can be proven syllogistically quite easily, but how do you prove that you may derive ~ρ from ρ→(φ^~φ)? — Lionino

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

Ah. Thanks. So, B and notB are negations of one another while "not B" in normal parlance means different than B but not necessarily contradictory to B.

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 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.)

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).

This is perhaps my favorite proof for the modus tollens thus far. The question is whether that second step justifies the modus tollens. — Leontiskos

This isn't a proof of Modus tollens. This is a use of Modus tollens.

You've been asking for 12 pages for a proof of Modus tollens?

When you say "we know S is true" you are stipulating — Leontiskos

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.

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

That is modus tollens. My question is why do we accept ~ρ from ρ→(φ^~φ) if not by modus tollens? I don't think we do. The RAA was given here as φ→(ψ^~ψ)⊢~φ, this statement seems to be justified by MT. The RAA I found elsewhere is (ρ∧¬μ)→(ψ^~ψ),ρ⊢μ, which is different .

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.

The issue is that we don't know S is true — Leontiskos

We do because S fully follows from the axioms of a theory.

if not by modus tollens? I don't think we do. — Lionino

Ok so we're playing a game (I don't mean that pejoratively, I like games) where we have to prove the conclusion without using modus tollens, is that right?

What are the rules of the game? Are we allowed to use the rule of non contradiction?

You said before that the proof for modus tollens is easy - does that proof obey the rules of this game? If so, which proof of Modus tollens do you like? There are multiple, I want to make sure I'm using the right one.

I will play this game, if you answer my questions then we can have a solution.

We do because S fully follows from the axioms of a theory. — Lionino

Which premise do you think provides us with such information?

(S∧¬P) does not favor S over ¬P in the case of a contradiction. If a contradiction follows from (S∧¬P) it is no more rational to reject ¬P than to reject S.

I think you are referring to background conditions that are not formally present, hence my point. If we really had a set of axioms and a theory instead of a single assumption, then you could say that it follows from a theory. In this case we do not have that.

Some users were debating the relationship between RAA and MT. If RAA is given as φ→(ψ^~ψ)⊢~φ, it is (to me) entirely dependant on MT. Not only that, but it is just MT with an omitted premise, as I think your post shows.
But if RAA is given as (ρ∧¬μ)→(ψ^~ψ),ρ⊢μ, it is a valid move on its own, distinct from MT.

That is all.

what proof of Modus tollens do you like? We can prove φ→(ψ^~ψ)⊢~φ without assuming Modus tollens is the case, but by instead directly using the proof of Modus tollens.

Which premise do you think provides us with such information? — Leontiskos

S.

(S∧¬P) does not favor S over ¬P in the case of a contradiction — Leontiskos

(S∧¬P), S does.
Edit: (S∧¬P)→contradict, S does.

what proof of Modus tollens do you like? — flannel jesus

Yours wasn't a proof of MT, it was MT itself. MT can be derived from MP and contraposition.

I didn't say mine was, are you reading the words I'm posting?

(S∧¬P), S does. — Lionino

And note that you supposed ¬P (which is the same as preferring S). Either way its a random pick for the second assumption. The point here is as I have said:

This is the formal conclusion, before the and-elimination step of the reductio takes hold:

A→(B∧¬B) {Assumption}
A {Assumption}
∴ (¬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. — Leontiskos

I'm asking you what proof you like - that's not a claim that mine is a proof of that. What proof of Modus tollens do you like?

Or disjunctive syllogism. I did verify that this is a way to prove MT.

But note that you supposed ¬P — Leontiskos

I didn't suppose ¬P.
I find out that (S∧¬P) leads to a contradiction within that theory. Not an assumption.
I know that S follows from the axioms of the theory. Not an assumption.
Conclusion: P.

Socrates is a man. Not an assumption.
Men are macroscopic. Not an assumption.
Socrates is macroscopic. Conclusion.