Determine whether or not $∀x[p(x) → q(x)]$ and $[∀xp(x)] → [∀xq(x)]$ are logically equivalent. 
Determine whether or not $∀x[p(x) → q(x)]$ and $[∀xp(x)] → [∀xq(x)]$ are logically equivalent.

I believe that they are not equivalent, but that is just an assumption. I am not sure how to go about proving this.
 A: To Answer this, let us take two statements:
$p(x)=$ "Person who likes Mathematics"
$q(x)=$ "Person who like math.stackoverflow"
Equation 1:$ ∀x[p(x)→q(x)]$
It says "Its always the case that if a person likes Mathematics, they will like math.stackoverflow". It may be sometimes "false" as there may be some person who likes Mathematics but not Conclusion.
Equation 2: $[∀xp(x)]→[∀xq(x)]$
It says that "If its always the case that all person likes Mathematics then its always the case that they will like math.stackoverflow.
As Hypothesis cannot necessarily be always true, Equation 2 will always be true.
A: Let $P$ be the set of all $x$ where $p(x)$ is true.
Let $Q$ be the set of all $x$ where $q(x)$ is true.
Let $\Omega$ be the set of everything.
$\forall x ~ (px \implies qx)$ is equivalent to $P \subseteq Q$.
$(\forall x ~ px ) \implies (\forall x ~ qx)$ is equivalent to $P = \Omega \implies Q = \Omega$.
Can you think of possible sets $P$ and $Q$ where $P \subseteq Q$ is true and $P = \Omega \implies Q = \Omega$ is false?
A: Let domain be set of all natural numbers.
Let $p(x)$: $x$ is prime and $q(x)$ : $x$ is odd.
Here $\forall x [p(x) \implies q(x) ]$ is false because 2 is prime but not odd.
 While $\forall x p(x) \implies \forall x q(x) $ is true vacously as hypothesis is not true.
