Are these two sentences logically equivalent? What i'm essentially asking is if the following statement is true:  

$ \forall x \exists y (R(x) \lor Q(y)) :\Leftrightarrow  \exists y \forall x (R(x) \lor Q(y)) $  

where $:\Leftrightarrow $ means that the two statements are logically equivalent.
(sometimes $\vDash$ and it's mirror are used for logical equivalence but can't find the mirrored symbol in mathjax

Edit: I suspect that it is not true but i'm trying to prove it and seems it should be.
Yes, the sentence is true — the two sides are equivalent, because both are equivalent to
$$\forall x \, R(x) \lor \exists y \, Q(y).
$$
If either predicate used both variables this wouldn't be so: in general, $\exists\forall \to \forall\exists$ but not conversely. But here, you can rearrange the quantifiers because: 


  If $v$ is not free in $p$ then
  $$\exists v\,(p \lor \varphi(v)) \equiv (p \lor \exists v\,\varphi(v))$$
  and
  $$\forall v\,(p \lor \varphi(v)) \equiv (p \lor \forall v\,\varphi(v)).$$
Nope; but one implication is true:
\begin{equation}
\exists y\forall x(R(x)\vee Q(y))\to \forall x\exists y(R(x)\vee Q(y)).
\end{equation}
It's more or less intuitive that this holds: when you have "$\forall x\exists y\dots$" this defines a function: given $x$, exists $x(y)$ such that $\dots$ (You need Choice to define this to be an actual function, so you can well-order the universe and take the least $x$ such that "$\dots$"). So if you have "$\exists y\forall x\dots$" you can define a constant function with output $y$. Therefore you have "$\forall x\exists y\dots$". With this you can see that the other implication is false beacause the function $x(y)$ can take different values.