I was looking at some random math problem and needed to resolve
$$\bigg( \exists x ~:~ P(x) \bigg) \rightarrow \bigg(\exists y ~:~ Q(y) \bigg) \tag 1$$
by rewriting as an equivalent statement. I came to the conclusion:
$$\bigg( \exists x ~:~ P(x) \rightarrow \exists y ~:~ Q(y) \bigg) \equiv \bigg(\exists z ~:~ P(z) \rightarrow Q(z) \bigg) \tag 2$$
which is what I would like double checked.
I did so by dividing the universe into 4 disjoint (some possibly empty) sets: $$\begin{align} E_3 &= \{x ~:~ P(x) \land Q(x) \} \\ E_2 &= \{x ~:~ P(x) \land \lnot Q(x) \} \\ E_1 &= \{x ~:~ \lnot P(x) \land Q(x) \} \\ E_0 &= \{x ~:~ \lnot P(x) \land \lnot Q(x) \} \\ \end{align}$$
If $E_2$ is the only nonempty set, then (1) cannot be satisfied. Otherwise, any element of any other set provides witness to $x$ and $y$ in (1). So I get:
$$\begin{align} \bigg( \exists x ~:~ P(x) \rightarrow \exists y ~:~ Q(y) \bigg) &\equiv \bigg(E_3 \ne \emptyset \lor E_1 \ne \emptyset \lor E_0 \ne \emptyset\bigg) \\ &\equiv \bigg(\exists z ~:~ \lnot(P(z) \land \lnot Q(z))\bigg) \\ &\equiv \bigg(\exists z ~:~ P(z) \rightarrow Q(z)\bigg) \end{align}$$
Seems right to me. Can't think of any counter examples. Maybe there is a more direct proof that doesn't involve set comprehension? Any feedback is appreciated. Tag this not-homework.
In (2), we have $\Rightarrow$ but not $\Leftarrow$. Example to show that $\Leftarrow$ fails: $P(x)$ is "$x=1$" and $Q(y)$ is always false. Then the RHS of (2) is true (take $z=0$) but the LHS is false.