Quantifier explination: $\forall x(R\to Q(x))\equiv R \to \forall x (Q(x))$ I don't understand the proof for this problem and couldn't find the answer anywhere.
$$\forall x(R\to Q(x))\equiv R \to \forall x (Q(x))$$
 A: Think of it first from a concrete approach, in order to get a better feel for what you're looking to prove in the general case.
Example: (Domain of x is all people)
$R:$ It is raining. (Note that R is independent of any x)
$Q(x): x$ takes an umbrella or gets wet.
$\forall x(R \rightarrow Q(x))\tag 1$
$R\rightarrow \forall x(Q(x))\tag 2$
$(1)$ "says" Everyone, if it's raining, takes an umbrella or gets wet.
$(2)$ "says" If it is raining, everyone takes an umbrella or gets wet.
The important thing to note is that $R$, by itself, is a statement, and can, but need not be included in the bounds over which x is quantified.
Hopefully this explanation helps you to understand the general case. @Hagen von Eitzen has already given you the technical details of the proof you may be seeking.
In the future, when you encounter a question which overwhelms upon first site, break it down into smaller pieces. If you are being asked to prove anything about $x$, universally (in a given domain)...get a feel for what it looks like concretely,  in a particular interpretation of the domain of $x$ and predicates used, to try to get a sense of what you are working with.
A: Assume $\forall x(R\to Q(x))$. We want to prove $R\to \forall x Q(x)$.
So we assume $R$ and attempt to prove $\forall x Q(x)$.
So we let $x$ be arbitrary. For this $x$ we have, by specialization, that $R\to Q(x)$. As $R$, modus ponens gives us $Q(x)$. As $x$ was arbitrary, $\forall x Q(x)$. So we derived $\forall x Q(x)$ based on the hypthesis that $R$ holds. In other words, we have shown $R\to \forall x Q(x)$. This shows the $\implies$ direction of the desired equivalence.
Now assume instead that $R\to \forall x Q(x)$.
We want to prove $\forall x(R\to Q(x))$. 
So let $x$ be arbitrary. 
We want to prove $R\to Q(x)$.
So assume $R$. Then by modus ponens $\forall x Q(x)$.
In particular, $Q(x)$.
Thus we showed $Q(x)$ from $R$, i.e., we showed $R\to Q(x)$.
As $x$ was aribitrary, we showed $\forall x(R\to Q(x))$. This shows the other direction.
