I was given the following formula $$\sim \exists A\in \mathcal G(x\in A) $$ where $$ \mathcal G $$ is a family of sets.
The solution says after having applied the quantifer negation law it should look like this $$\forall A\in \mathcal G(x\notin A) $$
I don't understand the logic behind this. If $$\exists A\in \mathcal G(x\in A)$$ is equivalent to $$ \exists A(A\in \mathcal G \wedge x\in A) $$ Surely $$ \sim\exists A(A\in \mathcal G \wedge x\in A) $$ should become $$\forall A\sim(A\in \mathcal G \wedge x\in A)$$ And applying DeMorgan's law onto it it becomes $$\forall A(A\not\in \mathcal G \vee x\not\in A)$$ But this looks nothing like the solution! Is my answer equivalent to the solution? If not could anyone please tell me what did I do wrong?
$\forall A(A\notin\mathcal{G}\lor x\notin A)$ is exactly equivalent to $\forall A\in\mathcal{G}(x\notin A)$. Note that $A\notin\mathcal{G}\lor x\notin A$ is logically equivalent to $A\in\mathcal{G}\to x\notin A$, and both
$$\forall A(A\in\mathcal{G}\to x\notin A)$$
and
$$\forall A\in\mathcal{G}(x\notin A)$$
simply say (a) that no member of $\mathcal{G}$ contains $x$ and (b) nothing at all about sets that are not in $\mathcal{G}$.