It can be shown that if

\[ \begin{aligned} \forall x .(\exists y. p(x,y) \implies q(x)) \end{aligned} \]

holds, then

\[ \begin{aligned} \forall x .\forall y.(p(x,y) \implies q(x)) \end{aligned} \]

holds. Do you see why ?

Turns out it's due to the simple reason that

\[ \begin{aligned} \forall x .(\exists y. p(x,y) \implies q(x)) \end{aligned} \]

is not the same as

\[ \begin{aligned} \forall x .(\exists y. (p(x,y) \implies q(x))). \end{aligned} \]

In particular, the sentence

\[ \begin{aligned} \exists y. p(x,y) \implies q(x) \end{aligned} \]

does not mean there exists \(y\) that the implication holds. Rather, it means if there exists \(y\) so that \(p(x,y)\) holds, then \(q(x)\) holds. In other words, the quantifier "exists y" applies only to the antecedent and not to the consequence.