Can we move quantifiers $\forall,\exists$ to the front if there aren't variable collisions? Let's say I have a formula $\phi$ in First Order Logic, which contains some existential and universal quantifiers quantifiers $\exists, \forall$.
Supposed that in $\phi$ there are no collisions in variables names.
Can we take all the quantifiers of $\phi$, move it on the left (in the same order that appears in $\phi$)?
Example:
$$\phi_1 = \forall x P(x )\rightarrow \exists y T(y) \\
\phi_2 = \forall x\exists y P(x )\rightarrow  T(y) \\
\phi_1 \overset{?}{\equiv} \phi_2$$
Obviously, I'm interested in the general case, not in this one, which is shown just to express the question using an example
I looked this up on Google, and I found many material which talks about prenex normal form, which solves this problem using skolemization, however I can't find an answer to this direct question, and skolemization makes it valid for every instance of $\phi$ that I try it on (but obviously trying many times is not a valid proof )
 A: The answer to your general question is very much "no", because moving a quantifier past a connective can change the quantifier. For instance, $\neg \forall x \phi(x)$ is equivalent to $\exists x \neg\phi(x)$ and in the vast majority of cases is not equivalent to $\forall x \neg \phi(x)$.
A: In this very particular case, NO, $\phi_1$ is NOT equivalent to $\phi_2$, but the overall "intuition" is correct, except for one edge case that was overlooked.
Let me elaborate. It is indeed the case (assuming no name collisions, obviously) that:
$\forall x Px \land \forall y Qy \iff \forall x \forall y(Px \land Qy) $
$\forall x Px \land \exists y Qy \iff \forall x \exists y(Px \land Qy) $
$\exists x Px \land \forall y Qy \iff \exists x \forall y(Px \land Qy) $
$\exists x Px \land \exists y Qy \iff \exists x \exists y(Px \land Qy) $
$\forall x Px \lor \forall y Qy \iff \forall x \forall y(Px \lor Qy) $
$\forall x Px \lor \exists y Qy \iff \forall x \exists y(Px \lor Qy) $
$\exists x Px \lor \forall y Qy \iff \exists x \forall y(Px \lor Qy) $
$\exists x Px \lor \exists y Qy \iff \exists x \exists y(Px \lor Qy) $
This rule does NOT hold true when we are talking about the first predicate of an implication ($a\implies b$), rather than disjunction ($a \lor b$) or conjuction ($a \land b$). E.g. in your case, $\forall x Px \implies \exists y Qy$ is NOT equivalent to $\forall x Px \implies \exists y Qy$. Rather, by rewriting the implication as a disjunction, we can show that:
$\forall x Px \implies \exists y Qy$
$\iff \neg \forall x Px \lor \exists y Qy$
$\iff \exists x \neg Px \lor \exists y Qy$
$\iff \exists x \exists y (\neg Px \lor Qy)$
$\iff \exists x \exists y (Px \implies Qy)$
Ultimately, we were able to show that:
$\forall x Px \implies \exists y Qy \iff \exists x \exists y (Px \implies Qy)$
In a similiar fashion, we can show all the other cases as well:
$\exists x Px \implies \exists y Qy \iff \forall x \exists y (Px \implies Qy)$
$\forall x Px \implies \forall y Qy \iff \exists x \forall y (Px \implies Qy)$
$\exists x Px \implies \forall y Qy \iff \forall x \forall y (Px \implies Qy)$
So yes, your initial intuition does hold true, as long as there are no implications involved. If there is an implication ivolved, then you will have to "switch" the quantifier of the first predicate when moving the quantifiers around, while the quantifier of the second predicate remains the same, just like with disjunction and conjuction.
After reading your question, I am not too sure if you were only interested in knowing the rules for "moving" quantifiers, or if you wanted a (semi)formal proof for it as well. If you are also interested in having a proof, let me know in the comments and I will edit my question later. In either case, I hope I was able to help!
A: 
Example:
$$\phi_1 := \forall x\; P(x )\rightarrow \exists y\; T(y) \\
\phi_2 := \color{green}{\Big(}\forall x\;\exists y\; P(x) \color{green}{\Big)}\rightarrow  T(y) \\$$

$$\phi_3 := \forall x\;\exists y \Big( P(x )\rightarrow  T(y) \Big)\\
\phi_4 := \exists x\;\exists y\;\Big( P(x )\rightarrow  T(y)\Big)$$
Formulae $1$ and $4$ are equivalent, whereas neither formula $2$ nor $3$ are equivalent to any of the others.
Therefore, non-collision is not a sufficient condition for a formula to be equivalent to the new one created by simply shifting its quantifiers to the front.

Addendum

can you provide a counterexample where $\phi_1$ and $\phi_3$ are not equals? or an explanation

Consider the set of integers. Let
 $P(x) := x$ is even, and
 $T(x) := x$ is smaller than itself.
For $x=2,$ no integer can make $\Big( P(x )\rightarrow  T(y) \Big)$ true; thus, $\phi_3$ is false.
On the other hand, $\phi_1$ is vacuously true.
