My proof did not assume mt, it did assume contraposition and MP.

If you believe that's the same as assuming mt, then that means the proof of mt that uses those two assumptions is circular.

yes, using MP, not mt.

If it's circular, fine, give me one that isn't circular. I assumed, perhaps wrongly, that when I asked you for a proof of mt that you like, that you wouldn't like an explicitly circular one.

Give me a proof of mt you like that isn't circular.

so you believe the MP+contraposition argument is circular? It's just using mt to prove mt?

You want a proof of some argument Y that doesn't assume modus tollens.

You presumably have a proof of Modus tollens that you like, that doesn't itself assume modus tollens.

Whatever format of argument that proof takes - that argument that doesn't assume modus tollens, but proves Modus tollens - I can use that exact same format of argument to prove Y similarly without assuming Modus tollens.

I've already done it above, but I can do it again

if you give it to me I will show you how it makes perfect sense

you want a proof of ρ→(φ^~φ) , therefore ~ρ that doesn't assume modus tollens, if you give me your preferred proof of Modus tollens I can give that to you.

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

You don't know why I'm asking what?

MT can be derived from MP and contraposition — Lionino

Ok, I'm going to assume you mean this proof (the one wikipedia lists as "Via contraposition"):

1
P→ Q (Given)
2
¬ Q (Given)
3
¬Q →¬P (Contraposition (1))
4
¬ P (Modus ponens (2,3))

This is the proof of modus tollens that you like - it proves modus tollens without assuming it, correct?


So it's pretty straight forward to use the same format, I'll take my previous argument which assumes Modus Tollens:

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

And reformat it to be in the style above, the proof that you like of modus tollens that doesn't assume modus tollens:

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

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?

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

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.

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.

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?

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)

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.

Hey you got it!

didn't, you are referring to this ↪Lionino, which I already said was a copypaste mistake, it has been edited. I don't see what the issue is. — Lionino

The issue is you said you never wrote it, but you did write it. I understand it's a mistake. Therefore it's not correct to say you never wrote it, it's correct to say you wrote it by mistake.

"Any operator" is not any mathematical operator you want. — Lionino

I don't know the rules of that game, my bad

It is like 2(x*y)=2x*y, but 2(x*y)≠x*2y — Lionino

This doesn't make sense if * is "any operator" either. Replace * with + and 2(x*y)=2x*y is not true

I don't see where I did that. — Lionino

You wrote
¬(A → B) is the same thing as ¬A→B

It is like 2(x*y)=2x*y, but 2(x*y)≠x*2y — Lionino

Is * multiplication here? I don't think this is right either.

I never replied positively to the question — Lionino

Well you gave what certainly looked like an affirmation. If I ask you "is lemonade your favourite flavour", and you say "lemonade is the same as my favourite flavour", most people are gonna think that's pretty much a "yes" to the question.

understand that you'd think that B∧¬B should be able to be replaced by any proposition P — Lionino

Me? You understand that I think that?

But what just happened is that you did that, and I told you it's incorrect...

¬(A → B) is the same thing as ¬A→B — Lionino

That's what I was asking, thank you.

I don't believe that's correct.

I was misplacing the associativity of the → operator.
So ¬(A → (B∧ ¬B)) is the same as (¬A) → (B∧ ¬B) — Lionino

Do you believe for, for all statements (A -> B), you can do ¬(A -> B) and transform that into ¬A -> B?

material implication is an example of the principle of explosion — Leontiskos

I don't think I claimed that. But as you're eager to reject basic reason, I'm not going to be one to stop you.

Another way to think about it is, "The only way you can be CERTAIN that A doesn't apply a contradiction is if you know A is true."

We are very far beyond Wikipedia at this point. At this point one can no longer simply appeal to authorities and logic machines. — Leontiskos

No, you asked for the rule of inference from classical logic - it's right there, common knowledge in wikipedia. I don't see any good reason why my answer should be considered unacceptable, other than you just don't want to accept it. You asked for the rule, that's it.

What rule of inference in classical logic are we appealing to? — Leontiskos

Funnily enough, the rules of inference we're appealing to are in fact the very first ones listed on the Wikipedia page:

https://en.wikipedia.org/wiki/List_of_rules_of_inference
Reductio ad absurdum

What rule of inference in classical logic are we appealing to? — Leontiskos

rule of noncontradiction, no?

What about a system of logic whereby, if the antecedent of an implication is false, rather than that making the entire statement of implication true it makes the entire statement meaningless, or undefined, or in flux, or something else like that? Something which is neither true nor false?

I'm interested in a system of symbolic logic that doens't deviate that drastically from what we normally mean by those expressions - a system of logic where you can say "I don't think A implies (C and ~C)" without simultaneously saying "A is true".

Maybe that system of symbolic logic is just... English? Normal logical language? Idk.

I can kind of explain it.

It seems as though, the right thing to say about basic classic symbolic logic is that EVERY statement is either true or false. So, you claim A, that's either true or false, period.

If A is false, then A implies anything. You can check the truth-table on implication: A -> B is always true if A is false. https://math.stackexchange.com/questions/1306280/implication-truth-table

So, no matter what B is, if A is false, A implies B - even if B ic (C and ~C).

So, if you KNOW that A doesn't imply (C and ~C), but you also know that if A was false, A has to imply (C and ~C) by the fact that anything follows from falsehood, then you must know that A must be true.

This makes sense in the universe of classic symbolic logic, where everything has explicit truth values and implication means what it means there.

I still can't make sense of it. — Lionino

This is one of those funny places where symbolic logic seems to take a detour from what we mean in natural language.

I mean I used a hand-written "grid", just not one like that multi-layered one you posted. I had columns where i recorded information about each spot.

It was really annoying when I got to the point where I could no longer straight-forwardly deduce any answers, though, and had to copy my grid by hand to test out making certain assumptions.

In your conclusion you rejected assumption (2) instead of assumption (1). — Leontiskos

I think calling them both assumptions has led to your confusion. Premise 1 is more of a GIVEN than an assumption.

We start out the scenario with it GIVEN that a -> (b and ~b). With that given, we say "let's see what happens when I assume a is the case". What happens is a contradiction, so we take a step back and realise, if it's given that a -> (b and ~b), a must be false.

It's quoted in a post earlier in this thread. — TonesInDeepFreeze

There are many many posts in this thread. I don't have any means of efficiently searching for it, so that's why I'm asking you. If you would prefer not to link me up for whatever reason, I suppose I'll just have to accept that.

That is what the paper says. The paper is incorrect. — TonesInDeepFreeze

There's a paper that says the premises prove the conclusion of this argument?

Dogs have four legs, and Lassie has four legs, therefore Lassie is a dog

I want to see this paper, could you link me?

modern symbolic disagrees that the Lassie argument's premises do not prove the conclusion. — TonesInDeepFreeze

Does that mean modern symbolic logic thinks the premises do prove the conclusion?