Prove: [∃xp(x)->∃xq(x)]->∃x[p(x)->q(x)] $[∃xp(x)\to∃xq(x)]\to∃x[p(x)\to q(x)]$
So I understand that if $∃x(p(x)\to q(x))$ is false then the whole statement would be false since it is an implication. In that case $∀x p(x)\wedge∀x \neg q(x)$. My teacher then said p(x) is always true and q(x) is always false. How did they know that? 
 A: It sounds like a semantical argument to show that the formula is valid.
Assume that is not; being a conditional, this amounts to assuming that the antecedent is true and the consequent is false, i.e. $\lnot ∃x[p(x)→q(x)]$ must be true.
But $\lnot ∃x[p(x)→q(x)]$ is equivalent to : $\forall x[p(x) \land \lnot q(x)]$.
$\forall x[p(x) \land \lnot q(x)]$ implies $\forall xp(x) \land \forall x \lnot q(x)$ and this is why your teacher says "$p(x)$ is always true and $q(x)$ is always false". 
This in turn implies $\exists xp(x) \land \lnot \exists xq(x)$ that is the negation of $\exists xp(x) \rightarrow \exists xq(x)$. 
In conclusion, the assumption that $∃x[p(x)→q(x)]$ is false contradict the assumption that $\exists xp(x) \rightarrow \exists xq(x)$ is true.
Thus, the formula is valid.
A: The basic idea is to expand and simplify the formula to get an idea what the condition is saying, $\exists x ~ P(x) \land \forall y ~ \lnot Q(y)$ and what the consequence is saying, $\exists z ~ \lnot P(z) \lor \exists w ~ Q(w)$.  And since those are tautologically compatible, twice combine a term from the condition and a term from the consequence form an expression like "$a \text{ or } \lnot a$".
$$\begin{array} {rcl}
%
\bigg( \exists x ~ P(x) \implies \exists y ~ Q(y) \bigg) \implies \bigg( \exists z ~ P(z) \implies Q(z) \bigg)
%
& &  a \implies b \equiv \lnot a \lor b  \\ \\
%
\lnot \bigg( \lnot \exists x ~ P(x) \lor \exists y ~ Q(y) \bigg) \lor \bigg( \exists z ~ \lnot P(z) \lor Q(z) \bigg)
%
& & \text{De Morgan's} \\ \\
%
\bigg( \exists x ~ P(x) \land \forall y ~ \lnot Q(y) \bigg) \lor \bigg( \exists z ~ \lnot P(z) \lor Q(z) \bigg)
%
& & \exists x ~ a(x) \lor b(x) \equiv \exists y ~ a(y) \lor \forall z ~ b(z) \\ \\
%
\bigg( \exists x ~ P(x) \land \forall y ~ \lnot Q(y) \bigg) \lor \exists z ~ \lnot P(z) \lor \exists w ~ Q(w) 
%
& & \land \text{ distributes over } \lor \\ \\
%
\bigg( \exists x ~ P(x) \lor \exists z ~ \lnot P(z) \lor \exists w ~ Q(w) \bigg) & & \exists x ~ a(x) \lor b(x) \equiv \exists y ~ a(y) \lor \forall z ~ b(z) \\
\land 
\bigg(\forall y ~ \lnot Q(y) \lor \exists z ~ \lnot P(z) \lor \exists w ~ Q(w) \bigg) & & \text{ and De Morgan's} \\ \\
%
\bigg( \exists x ~ P(x) \lor \lnot P(x) \lor \exists w ~ Q(w) \bigg) \\
\land 
  \bigg( \forall y ~ \lnot Q(y) \lor \lnot \forall w ~ Q(w) \lor \exists z ~ \lnot P(z)  \bigg) \\ \\
%
\bigg( \exists x ~ \text{True} \lor \exists w ~ Q(w) \bigg) \\ 
\land 
  \bigg( \text{True} \lor \exists z ~ \lnot P(z)  \bigg) \\ \\
% 
\bigg( \exists x ~ \text{True} \lor \exists w ~ Q(w) \bigg) 
%
\end{array}$$
Now you might want to apply $\exists x ~ \text{True} \equiv \text{True}$, but that is only the case in a nonempty universe.  Actually, that is basically the definition of a nonempty universe.  So, to continue, you have to make the non-empty universe assmption:
$$\bigg( \text{True} \lor \exists w ~ Q(w) \bigg) $$
$$\text{True}$$
If you follow this, then I suggest trying to establish the reverse condition, $\bigg(\exists x ~ P(x) \implies Q(x) \bigg) \implies \bigg( \exists y ~ P(y) \implies \exists z ~ Q(z) \bigg)$.
