# When existential quantifiers distribute over conjunction

I know that, generally:

$$\exists x~~ (P(x) \land Q(x)) \implies \exists x~~P(x) \land \exists x~~Q(x)$$

But I wonder if is there any circumstances (by some restrictions of $P(x)$ and $Q(x)$) that the below holds?

$$\exists x~~ (P(x) \land Q(x)) \iff\exists x~~P(x) \land \exists x~~Q(x)$$

Please note that $x$ is free in both $P(x)$ and $Q(x)$

• We have that $∃x(P ∧ Q(x)) ↔ (P ∧ ∃xQ(x))$ when $x$ is not free in $P$. Mar 3, 2015 at 10:41
• Here, I mean x is free in both P(x) and Q(x) Mar 3, 2015 at 13:45

There are in general no "good" restrictions on $P(x)$ and $Q(x)$ for the equivalence to be true.
• either $\exists x, P(x)$ or $\exists x, Q(x)$ is false
• $\exists x, P(x)$ and $\exists x, Q(x)$ are true and for the same $x$, which is exactly what is expressed by $\exists x, (P(x)\wedge Q(x))$.
Notice that the comment of Mauro ALLEGRANZA belongs to my second case (whenever $P$ and $\exists x, Q(x)$ are true) since $x$ is not free in P and hence does not restrict the set of possible $x$ satisfying $Q(x)$.