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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 $\forall x\in E,\,(x\in A)\implies P(x)$?

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  • $\begingroup$ At the title's beginining, did you mean this instead: $\forall x\in A(\subset E),\,P(x)?$ $\endgroup$
    – ryang
    Dec 4 '21 at 15:23
  • $\begingroup$ @ryang thank you! it should have been $\in$. $\endgroup$
    – plum356
    Dec 4 '21 at 15:33
  • $\begingroup$ @ryang $\mathcal P(E)$ is the set of all subsets of $E$. $\endgroup$
    – plum356
    Dec 4 '21 at 16:35
  • $\begingroup$ Oh, power set. Why not just say subset of E instead of element of E's power set. $\endgroup$
    – ryang
    Dec 4 '21 at 16:44
  • $\begingroup$ @ryang the notes that i am using usually employ $\mathcal P(E),\,(\mathcal P(E))^2$ etc ... $\endgroup$
    – plum356
    Dec 4 '21 at 16:50
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Well, there is difference between existential and all-quantification:

$\forall x[x\in A\Rightarrow P(x)]$

and

$\exists x[x\in A \wedge P(x)]$.

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  • $\begingroup$ makes sense. thank you very much! $\endgroup$
    – plum356
    Dec 4 '21 at 16:34

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