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 $Q(x)$

  • 1
    $\begingroup$ We have that $∃x(P ∧ Q(x)) ↔ (P ∧ ∃xQ(x))$ when $x$ is not free in $P$. $\endgroup$ Mar 3, 2015 at 10:41
  • $\begingroup$ Here, I mean x is free in both P(x) and Q(x) $\endgroup$
    – Trung Ta
    Mar 3, 2015 at 13:45

2 Answers 2


There are in general no "good" restrictions on $P(x)$ and $Q(x)$ for the equivalence to be true.

The only cases where the equivalence hold is whenever:

  • 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 $Q(x)$.


Speaking from a 'common sense' point of view, since the formal answer given by wece is quite good, what would make one think that the reverse implication would ever hold in a general sense? Saying there exists something that makes one formal statement true and something that makes another formal statement true gives almost no information whatsoever on whether those two instances overlap (unless your universe of things consists of one element!)


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