What are some examples of ∃x (P(x) <-> Q(x)) and (∃xP(x)) <-> (∃xQ(x)) not being logically equivalent to each other

Question

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.

Answer 1

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.

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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).