Does a conditional statement imply a universal quantifier? To me, it seems like the following two statements are equivalent
x ∈ Arbitrary set P(x)→ Q(x) = 
∀x ∈ Arbitrary set (P(x)→ Q(x))
For example, if x ∈ all people, P(x) stands for "x is a man", and Q(x) stands for "x isn't a female", then ∀x(P(x)→ Q(x)) can be thought of as saying "For any person x, if x is a man, then x isn't a female". But isn't that what P(x)→ Q(x) is already saying, just without the explicit "For any person x" and instead with the implications from x ∈ all people? I mean, if we are assuming P(x)→ Q(x) is a valid premise, then shouldn't x be allowed to range over all possible x, i.e, "any person"? 
Sorry if I'm not making too much sense and using incorrect notation; I've just started this stuff, probably obviously so lol.
 A: No, P(x) and $\forall x ~ P(x)$ are not always the same thing.  For example, $\pi > 7$ is not the same thing as $\forall \pi~\pi > 7$.
Free variables are implicitly quantified over their entire proof, but not necessarily any individual theorem.  Imagine the following proof of the statement "if $x < 5$, then $x < 7$" (assume we have already established $y < n \implies y < n + 1$ , and our universe is only natural numbers):
$$\begin{array} {r|l|l}
(1) & \quad x < 5 & \text{assumption} \\
(2) & \quad x < 6 & \text{From 1} \\
(3) & \quad x < 7 & \text{From 2} \\
(4) & x < 5 \implies x < 7 & \text{From 1 through 3} \\
\end{array}$$
Notice that from line (4), it would be appropriate to infer $\forall x ~ x < 5 \implies x < 7$.  However, from line (2), it would not be correct to infer $\forall x ~ x < 6$.  That is because there is an assumption made about the $x$ in line (2): the assumption is that $x < 5$.  In general, the meaning of a free variable is that it is universally quantified over the entire proof:
$$\forall x ~ \left(\begin{array} {r|l|l}
(1) & \quad x < 5 & \text{assumption} \\
(2) & \quad x < 6 & \text{From 1} \\
(3) & \quad x < 7 & \text{From 2} \\
(4) & x < 5 \implies x < 7 & \text{From 1 through 3} \\
\end{array}\right)$$
Due to the fact that $\forall x ~ (B \implies C(x))$ is equivalent to $B \implies \forall x ~ C(x)$ when $x$ is not a free variable in $B$, you can infer from line (4) $\forall x ~ x < 5 \implies x < 7$.
