# $\exists x\in A \ (P(x))$ versus $\exists x \ (x\in A \land P(x))$ [duplicate]

I have a pretty simple question on mathematical logic.

Consider two logical statements $$\exists x\in A \ (P(x))$$ and $$\exists x \ (x\in A \land P(x))$$. How are they logically different from each other?

• Apr 17 at 9:05

These two logical statements are the same, by definition of $$\exists x \in A ...$$

When working with quantifiers and sets, it's important to remember these two definitions: $$\exists x \in A (P(x)) \iff \exists x (x\in A \land P(x))$$ $$\forall x \in A (P(x)) \iff \forall x (x\in A \implies P(x))$$

Notice how it's a $$\land$$ when $$\exists$$ is used but $$\implies$$ when $$\forall$$ is used.

• +1 Fun fact: $~\exists x (x \in A \implies P(x))$ is always true. Not a very useful construct, but fun. See "Drinkers Paradox." Apr 17 at 15:10
• Thank you for your reply! I have a question: So, we know what $\exists x (Q(x))$ means. Are you saying that the definition of $\exists x\in A (P(x))$ is $\exists x (A \land P(x))$?
– RFZ
Apr 19 at 3:06
• So if $\exists x \in A (P(x)) \equiv \exists x(x \in A \land P(x))$ as you said. Then the negation of $\exists x \in A (P(x))$ would be $\forall x (x \notin A \lor \neg P(x))$. But I thought that the negation of $\exists x \in A (P(x))$ is $\forall x \in A (\neg P(x))$. But you see that they are not identical. What am I missing?
– RFZ
Apr 19 at 3:22
• They are identical, because of the rule of material implication, which says $\neg X \vee Y$ is logically equivalent to $X\implies Y$: \begin{align*} & \forall x (x\notin A \vee \neg P(x)) \\ \iff & \forall x (x\in A\implies \neg P(x)) \\ \iff & \forall x\in A (\neg P(x))\end{align*} Apr 20 at 2:04
• Also, yes I am saying the definition of $\exists x\in A (P(x))$ is $\exists x (x\in A \land P(x))$ Apr 20 at 2:05