What are some examples of ∃x (P(x) <-> Q(x)) and (∃xP(x)) <-> (∃xQ(x)) not being logically equivalent to each other I just can't think of any examples for P(x) and Q(x) where this would work out. I can think of it the other way around, for examples if p(x) was x>5 and q(x) was x<2 , then they can never be together and still make sense. However, I can't think of an example where they wouldn't be able to be separated from each other.
 A: Suppose $Q(x)$ is just "$\neg P(x)$." Then
$$\exists x(P(x)\iff Q(x))$$ is false, no matter what $P$ is; however, $$(\exists x(P(x)))\iff (\exists x(Q(x)))$$ could still be true, and in fact will be true unless $P$ either holds everywhere or fails everywhere.
For a concrete example, work in $\mathbb{N}$ and take $P(x)$ to be "$x$ is even" and $Q(x)$ to be "$x$ is odd." Then:


*

*$\exists x(P(x)\iff Q(x))$ is false: there is no number which is both even and odd, and no number which is neither even nor odd.

*Meanwhile, $(\exists x(P(x)))\iff (\exists x(Q(x)))$ is true since both "implicands" are true: there is a number which is even, and there is a number which is odd.


*

*It might be clearer here to change the bound variable so that there's no possibility of confusion, that is, write "$(\exists x(P(x)))\iff (\exists y(Q(y)))$ instead. This doesn't change the meaning, but it makes it clearer that the two pieces are separate.




And that's not the only way for the two statements to be inequivalent.
For another example, suppose $P$ holds sometimes but not always, and $Q$ never holds. Then $\exists x(P(x)\iff Q(x))$ is true ($P$ fails somewhere and $Q$ fails everywhere, so there's some element which both $P$ and $Q$ fail to hold on simultaneously), but $(\exists x(P(x)))\iff (\exists x(Q(x)))$ is false (the first implicand is true but the second implicand is false).
