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I have to prove the following statement : (A -> B) <-> ~A \/ B which is the decomposition of implication. I can use the ~~A -> A axiom as it is given in the exercise, but I'm stuck pretty early in the demonstration.

I start by splitting then introducing, then I've tried a bit of everything (right, left, apply the absurd axiom then introducing) but nothing looks convincing and I don't really know where to go from there...

Any suggestions ?

Philémon Berstel-Da Silva
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    What is exactly the axiom you are given and thing you have to prove? Do they both start with a quantification `forall A`? If the `A` is a variable that is the same in both statements you give, then your statement is not provable. – Meven Lennon-Bertrand Mar 31 '21 at 10:03

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From the axiom forall A, ~~A -> A that you were given, you can prove the law of the excluded middle, forall A, A \/ ~A. They are actually logically equivalent. If you do that, then it is easy for you to solve your assignment.

I don't give the full solution here because it is not good to give solutions to course exercises, and I promise you will learn from doing it.

But as a small hint, when trying to prove

Goal (forall A, ~ ~ A -> A) -> (forall A, A \/ ~ A).

the first steps are intros H A. apply H. intros C.

larsr
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