$\forall x\in A(\in\mathcal P(E)),\,P(x)\Leftrightarrow\forall x\in E,\,(x\in A)\wedge P(x)$?

Question

The question is in the title.

When I am working in $\mathcal P(E)$, do statements like $\forall x\in A,\,P(x)$ translate to $\forall x\in E,\,(x\in A)\wedge P(x)$ or to $\forall x\in E,\,(x\in A)\implies P(x)$?

Answer 1

Well, there is difference between existential ($\exists x$) and "for all" ($\forall x$) quantification:

$\forall x\big (x\in A\Rightarrow P(x)\big)$ is different than $\exists x\big(x\in A \wedge P(x)\big)$.