# Universe of discourse in quantifiers

I am reading "How to prove it" by Velleman and am currently in the chapter about quantifiers. I don't understand why $$\exists x \in A (P(x))$$ is equivalent to $$\exists x~(x\in A \wedge P(x)),$$ whereas $$\forall x \in A (P(x))$$ is equivalent to $$\forall x~ (x\in A \rightarrow P(x)).$$

I don't understand why for the existential quantifier the two statements are "anded" whereas for the universal quantifier the conditional is used.

• Negate and compare. Commented Apr 23, 2023 at 10:19
• @AnneBauval Ok when I negate I get the following $\neg \forall x (x\in A \rightarrow P(x))$ is equivaltant to $\exists x (x\in A \land \neg P(x))$ How does this help? I can see that I have an and inside now. Commented Apr 23, 2023 at 10:30
• @AnneBauval Ok I get it now. because now I can pull out the the first to get $\exists x \in A(\neg P(x))$ which is equivalent to $\neg \forall x \in A (P(x))$ and then I negate on the initial statement and this statement to see that they are same. Am I right? This is really beautiful! Thanks! Commented Apr 23, 2023 at 10:34
• Precisely! And all this is consistent with $\neg[\forall x\in A(P(x))]$ being equivalent to $\exists x\in A\neg P(x).$ But I see my reply to your first comment comes too late. Perfect! Commented Apr 23, 2023 at 10:37
• Fun fact: If you accept that there exists no universal set (see Russell's Paradox) then $\exists x:[ x\in A \implies P]$ is true for any set $A$ and proposition $P$. Care should be taken when applying set theoretic existential quantifiers like this to conditionals. Commented Apr 23, 2023 at 14:10

$$\exists x\in A~\big(P(x)\big)$$ versus $$\exists x~\big(x\in A\land P(x)\big)$$
$$\forall x\in A~\big(P(x)\big)$$ versus $$\forall x~\big(x\in A\to P(x)\big)$$