TITLE: Show that $\lnot\exists x\in A P(x) \equiv\forall x\in A\lnot P(x)$
QUESTION [1 upvotes]: I need to show that: $$\lnot\exists x\in A P(x) \equiv\forall x\in A\lnot P(x)$$
when I start from LHS, should I write(1): $$\lnot\exists x (x\in A \to P(x))$$
or(2): $$\lnot\exists x (x\in A \land P(x))$$
because after simplification in first case I get: $$\forall x(x\in A \land\lnot P(x))$$
whereas in the second case I get: $$\forall x(x\in A \to\lnot P(x))$$
I'm confused, because I know that in general: $$(p\to q) \equiv \lnot(p\land\lnot q)$$
therefore I don't know whether assuming (1) or (2) is right. 
Could you please clarify that for me?

REPLY [1 votes]: The logical forms of the statements $\exists \, x \!\in\! A \, P$ and $\forall \,x \!\in\! A \, P$ are respectively:
\begin{align}
 \exists x \, (x \!\in\! A \land P) \qquad \text{and} \qquad \forall x (x \!\in\! A \to P).
\end{align}
In your case, then, $\lnot \exists \, x \!\in\! A \, P(x)$ stands for $\lnot \exists x (x \in A \land P(x))$, which by De Morgan is equivalent to $\forall x \,\lnot (x \in A \land P(x))$, which  by De Morgan (and the fact that $p \to q \equiv \lnot p \lor q$) is equivalent to $\forall x \, (x \in A \to \lnot P(x))$, which is the logical form of $\forall \, x \!\in\! A \, \lnot P(x)$.
{"set_name": "stack_exchange", "score": 1, "question_id": 3690890}