I don't think there is any mystery around (A→(B∧¬B)) |= ¬A, if something implies a contradiction we may say it is false. My curiosity was more around ¬(A→(B∧¬B)). We know that ¬(A→(B∧¬B))↔A is valid, (A→(B∧¬B)) entails ¬A, and ¬(A→(B∧¬B)) entails A. Tones gave a translation of the latter as:
"It is not the case that if A then B & ~B
implies
A"
I still can't make sense of it.
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?
The negation of a contradiction is always true, and being true it is implied by anything, true or false. — Count Timothy von Icarus
Elvis is a man - A
Elvis is a man implies that Elvis is both mortal and not-mortal. - A → (B and ~B)
Therefore Elvis is not a man. — Count Timothy von Icarus
Where — Lionino
1. A→(B∧¬B) assumption
2. A assumption
3. B∧¬B 1,2, conditional proof
4. ~A 2, 3 reductio — Banno
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
The negation of a contradiction is always true, and being true it is implied by anything, true or false. — Count Timothy von Icarus
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
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
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
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
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
to:Elvis is a man – A
Elvis is a man does not imply that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is a man. – A
A, ¬(A → (B∧ ¬B)) entails A.
[...]
Elvis is not a man – ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is not a man – ¬A
¬A,¬(A→(B∧¬B)) entails ¬A.
[...]
Elvis is not a man – ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is a man – A
¬A, ¬(A → (B∧ ¬B)) entails A. — Lionino
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
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
I was misplacing the associativity of the → operator.
So ¬(A → (B∧ ¬B)) is the same as (¬A) → (B∧ ¬B) — Lionino
(i.e. "Suppose a; a implies a contradiction; reject a") — Leontiskos
(A -> B), you can do ¬(A -> B) — flannel jesus
¬(A → B) is the same thing as ¬A→B — Lionino
I don't believe that's correct. — flannel jesus
understand that you'd think that B∧¬B should be able to be replaced by any proposition P — Lionino
for all statements (A -> B), you can do ¬(A -> B) and transform that into ¬A -> B? — flannel jesus
I never replied positively to the question — Lionino
It is like 2(x*y)=2x*y, but 2(x*y)≠x*2y — Lionino
The very proposition of "there both a) is a self and b) is no self" has (a) and (b) addressing the exact same thing - irrespective of how the term "self" might be defined or understood as a concept, the exact same identity is addressed — javra
The point is that if there is no determinate entity that 'the self' refers to, if there is only the concept, and if there is no actual entity, then saying that we are speaking about the same thing is incoherent. On the other hand, if you stipulate that the self is, for example, the body, then what would A be in the proposition (A implies B) where B is 'there is a self' ? Let's say that A is 'the perception of the body': this would be 'the perception of the body implies that there is a self". 'The perception of the body implies that there is no self' would then be a contradiction to that. — Janus
It is like 2(x*y)=2x*y, but 2(x*y)≠x*2y — Lionino
You wrote
¬(A → B) is the same thing as ¬A→B — flannel jesus
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
"Any operator" is not any mathematical operator you want. — Lionino
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