Prove by Natural deduction that $\lnot\exists xP(x)\rightarrow\forall x\lnot P(x)$

Question

I have the following problem:

Prove by Natural deduction in First Order Logic that

$\lnot\exists xP(x)\rightarrow\forall x \lnot P(x)$

I tried to prove it using the Contradiction Theorem but I got stuck.

Probably I am missing something.

Answer 1

Under the premise $\neg\exists x P(x)$, assume $P(x)$ and follow from this contradiction.

Answer 2

Actually, you don't need to use Contradiction Theorem!

The following is a derivation of the formula $\lnot\exists x \, P(x) \to \forall x \, \lnot P(x)$ in first-order natural deduction without using the rules RAA (reductio ad absurdum) and EFQ (ex falso quodlibet):

This means that the formula $\lnot \exists x \, P(x) \to \forall x \, \lnot P(x)$ is provable not only in first-order classical logic, but also in first-order intuitionistic (and minimal) logic.