Something that I read recently, very interesting, and I can't remember where, on the topic of logic, is that syllogisms can be said to be question begging (this is a point that has been made by philosophers in the past).
"All men are mortal; Socrates is a man; therefore, Socrates is mortal" is of no value, since we could not know that the premise, "All men are mortal" is true unless we already knew that Socrates is mortal. So we learn nothing from the syllogism.
https://philosophy.stackexchange.com/questions/104220/do-you-gain-further-truth-from-syllogisms

I didn't really talk about proving MT at any point, so I don't know why you are asking that.

RAA as it is portrayed in some Google image search results is more of a rhetorical move than a logical one, as formally it leads to explosion.

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.

I was only relating what Metaphysician Undercover said. — fishfry

I am aware.

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.

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.

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.

Because, if it was the case that he would have won either way, the shooting would have been entirely irrelevant.

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

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

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 .

Yes, I do, I literally said that almost word for word in my post. The same thing happened with Bolsonaro in 2018 and eventually people forgot about it.

How absurd does the world need to be for us to become existentialist overcomers? How meaningless does it need to be for us to become self-generating overmen? — Count Timothy von Icarus

Let's say the world is just the right amount. How does the Ethiopian child trapped in quicksand being eaten alive by vultures get to self-transcend? After the suffering, it just dies right away. Does it get to self-transcend in the afterlife?

Now 8 days ago.

Physical objects, on the other hand, can be truly unique in this physical universe (but almost never in the physical multiverse). — Tarskian

This place is not real.

(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.

.

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.

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.

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.

Just a tidbit, Claude 3.5 told me:

Yes, modus tollens relies on contraposition. Let me explain the connection:

Modus tollens is a valid logical argument form that follows this structure:
If P, then Q.
Not Q.
Therefore, not P.
Contraposition is a logical equivalence that states:
(P → Q) ≡ (¬Q → ¬P)
In other words, "If P, then Q" is logically equivalent to "If not Q, then not P."
Modus tollens uses contraposition implicitly:

It starts with the premise "If P, then Q."
When we observe "Not Q," we use contraposition to infer "If not Q, then not P."
Then we apply modus ponens to "If not Q, then not P" and "Not Q" to conclude "Not P."



So, while modus tollens is often presented as a distinct rule of inference, it can be seen as a combination of contraposition and modus ponens.

If Claude 3.5 is right (I feel it is), you are basically doing modus tollens there.

Well, I am taking it I have solved the thread. I will also take credit for extending it for 10+ pages by asking this:

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

Oh alright

This is the RAA, innit? :smile:

(S∧¬P)→(B∧¬B)
S
∴ P — Lionino

If a → c it does. Contraposition, flip em and switch em (reverse the order and negate both). — Count Timothy von Icarus

a → c, ¬a → ¬c does not entail ¬a

Modus Tollens: ρ→φ, ~φ ⊢ ~ρ
RAA: ρ→(φ^~φ) ⊢ ~ρ — Banno

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)?

To my mind the explosion only occurs if you don't reject either of the two premises. — Leontiskos

When arguments are given as A, B ⊢ C, that is "A is true, B is true, therefore C is true".

Alright, I got something from the Spanish version of a site that shall not be named:
RAA is ((S∧¬P)→(B∧¬B)) → (S→P) — this is valid and not explosive.
((S∧¬P)→(B∧¬B)), S entails P (valid).
((S∧¬P)→(B∧¬B)), S is not explosive.
Without S as a true premise, the argument doesn't follow.
S is a statement (or collection of them) that are know to be true, P is the statement to be proven.

(S∧¬P)→(B∧¬B)
S
∴ P

The criticism that applies here:

When we do a reductio
A, A→¬B∧B ⊢ ¬A is valid

But A, A→¬B∧B ⊢ A is also valid

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

does not apply to:
(S∧¬P)→(B∧¬B)
S
∴ P
because it is not explosive.

But that's because the two assumptions, A and A→¬B∧B, are inconsistent. — Banno

Well, yes, explosion.

The problem is that this A, A→¬B∧B ⊢ ¬A was given as reductio ad absurdum. But this entails anything. However, you gave the RAA as φ→(ψ^~ψ)⊢~φ, in which case there is no explosion. But then the question is: how do we prove φ→(ψ^~ψ)⊢~φ? If this is considered the proof of φ→(ψ^~ψ)⊢~φ, RAA depends on both MP and MT. The issue is that the proof is also explosive:

Tones said "Modus tollens and RAA are in a sense versions of each other and they both do the job.". To me, RAA depends on modus tollens.

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

I don't think that follows.
a → (b ∧ ¬b), ¬a → ¬(b ∧ ¬b) entails ¬a because of the contradiction.
But if we change (b ∧ ~b) to c: a → c, ¬a → ¬c does not entail ¬a .
(b ∧ ~b) is False, so ~(b ∧ ~b) is True, ~a implies ~(b ∧ ~b) but not the converse (3), so we can't derive ¬a from that.

Reductio: — Lionino

P→(Q∨R), ¬¬P, ¬Q, ¬R entails ¬P
But it also entails anything one wants, including Q, P, ¬J.

The person in charge for formalism in the SEP is Alan Weir. I am sure he would appreciate an email about that. Keep us posted.

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

And since we said ρ earlier, we choose ρ→~μ?

Even if that solves the problem between the conjunct, A, A→¬B∧B can entail quite literally anything. So I wonder why we choose ¬A as an entailment instead of quite literally anything, if not by modus tollens?
The image I posted doesn't help me much.

There are formalists who are in for it exactly for the anti-platonist element.

Ok, so not even an annexed file bumps it. Is the thread shadow banned?

Even though the last post was 14 hours ago, the main page says the last post was 3 days ago. What the hell?
Anyway, spiral:

Check.

Can you see the answer? — Banno

Can I? Yes. Do I? No.

Reductio:

https://www.nbcnews.com/politics/joe-biden/president-joe-biden-tests-positive-covid-19-rcna162435
BUDDY IS A GONER

ρ,μ ⊢φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ) — Banno

When we do a reductio
A, A→¬B∧B ⊢ ¬A is valid

But A, A→¬B∧B ⊢ A is also valid

So the question is: how do we choose between either? Isn't it by modus tollens?
p→q is True, q is False ⊢ p is False
a→b∧¬b is True, b∧¬b is False, ⊢ a is False

The "intuition" to perceive "mathematical objects" can be, and has to be, imagination of instatiations of those mathematical facts, even if it is just symbol manipulation. Even if math is just symbol manipulation, I can apply the relationship between those symbols to calculate the size of a building from its shadow. That is the intuition. When I think of a triangle I don't necessarily have the same picture as someone else. Anyone who thinks they are perceiving platonic objects directly is not paying attention.