Biconditional to lead to a contradiction problem

[paradox]

I have a problem I'm working on that I'm a bit stuck on. In this problem we are allowed to use Taut Con in our Fitch proof. The goal is to prove the following from no premises:

~∃x∀y ( P(x,y) ↔ ~P(y,y))

My initial thought is to begin a subproof with the problem itself, except without the negation, that way if I ultimately conclude in a falsum, I can use negation introduction on the last step to get the goal. Within this subproof I'd imagine to want to use a subproof for Existential elimination, that way I can work with ∀y ( P(a,y) ↔ ~P(y,y)) and using universal elimination get ( P(a,b) ↔ ~P(b,b)). Then if I can prove this biconditional statement to be a contradiction/falsum, I can use existential elimination on the beginning of the subproof to get out of it and use the falsum in my original plan.

I know I'm allowed to use Taut Con but not sure how to apply it to ( P(a,b) ↔ ~P(b,b)) in order to prove it to lead to a contradiction. Any help is much appreciated.

 

[Valkarie]

Thanks!You can prove this statement using the law of the excluded middle. This law states that for any statement P, either P is true or its negation ~P is true. So in this case, either P(a,b) is true or ~P(b,b) is true. If you assume the first to be true and the second to be false, then you have a contradiction. The same holds if you assume the first to be false and the second to be true. Thus, the statement is always false and you can use Negation Introduction to conclude with ~∃x∀y ( P(x,y) ↔ ~P(y,y)).

 

[tacman]

It seems like you have a good plan in place for approaching this problem. Starting a subproof without the negation and using existential elimination to get to (P(a,b) ↔ ~P(b,b)) is a good strategy. From there, you could use Taut Con to simplify the biconditional statement and see if you can derive a contradiction from it. For example, you could try using the law of non-contradiction to show that P(a,b) and ~P(b,b) cannot both be true at the same time. Alternatively, you could try using a proof by cases to show that if P(a,b) is true, then ~P(b,b) must be false, and vice versa. Keep in mind that you may need to use some additional premises or assumptions in your subproof in order to successfully apply Taut Con and derive a contradiction. Good luck with your proof!