For the two quantified expressions, what is the difference?
(a.1) $(\forall x \in A)(P(x) \Rightarrow Q(x))$
(b.1) $(\forall x \in A, P(x)) \Rightarrow (\forall x \in A, Q(x))$
Wouldn't they be logically equivalent, in that since $x$ can only hold a singular state value? The only way I found to disprove the two quantified sentences was by negation:
(a.2) $\neg(\forall x \in A)(P(x) \Rightarrow Q(x)) \equiv (\exists x \in A)(P(x) \wedge \neg Q(x))$
(b.2) $\neg \forall x \in A, P(x)) \Rightarrow (\forall x \in A, Q(x)) \equiv (\forall x \in A, P(x)) \wedge (\exists x \in A, \neg Q(x))$
For the two expressions to be logically equivalent the negation of the negation would have to be logically equivalent to the original sentences:
(a.3) $ \neg (\exists x \in A)(P(x) \wedge \neg Q(x)) \equiv (\forall x \in A)(\neg P(x) \vee Q(x)) \equiv \text{(a.1)}$
(b.3) $\neg ((\forall x \in A, P(x)) \wedge (\exists x \in A, \neg Q(x))) \equiv (\exists x \in A, \neg P(x)) \vee (\forall x \in A, Q(x)) \equiv \text{(b.1)}$
The issue here though is that, both the sentences (a.3) and (b.3) have differing truth values, so they must not be logically equivalent.
Is this an adequate explanation or am I missing something?
We can reduce (based, on that $x\Rightarrow y$ is shorthand for $\neg x \lor y$) question to distribution $\forall$ with respect to $\lor$ and note, that it is not true $$\left[(\forall x)( R(x) \lor S(x) )\right] \Rightarrow [(\forall x)R(x) \lor (\forall x)S(x)] \quad (\text{False})$$
consider $R$="number is even" and $S$="number is odd" for numbers from $\mathbb{N}$. So you cannot state $(a.1) \Rightarrow (b.1)$.
However, there is case when equivalence can be hold: there is when one of from your $P,Q$ not depend on $x$. $$[(\forall x)( R(x) \lor S )] \Leftrightarrow [(\forall x)R(x) \lor S] $$