Proof for $\forall x A(x) \Leftrightarrow \neg \exists x \neg A(x)$

Question

I try to proof, that $\forall x A(x) \Leftrightarrow \neg \exists x \neg A(x)$

I know how to prove, that $\forall x A(x) \Rightarrow \exists xA(x)$, but I don't understand how to get negation.

Answer 1

$$\forall x A(x) \equiv \forall x \lnot\lnot A(x) \equiv \lnot \exists x \lnot A(x)$$

Perhaps it makes more sense to you if we start from the right, and move to the left.

You'll need to remember that $$\lnot \forall x P(x) \equiv \exists x \lnot P(x)$$ and $$ \lnot \exists x P(x) \equiv \forall x \lnot P(x)$$

Answer 2

A proof by words.

$\forall x A(X) \implies \lnot(\exists x \lnot A(x))$

$\forall x A(x)$ means that for every $x$ in the domain, the proposition $A(x)$ is true. This means that no $x$ exists such that $A(x)$ is false. Which can be written $\lnot (\exists x \lnot A(X))$, because the statement $\exists x \lnot A(X)$ always evaluates as false.

$\lnot(\exists x \lnot A(x)) \implies \forall x A(X)$

There is no $x$ such that $A(x)$ is false. Therefore on the domain of $x$, all $x$ imply $A(x)$ is true.

Small addendum: I think $\lnot(\exists x \lnot A(x))\equiv (\lnot\exists) x \lnot A(x)$.