free variable in conversion to prenex form when convert below formula to RPF form (rectified and prenex form), how to handle the free variable of Z in both the subformulas ($\exists{x}P(x,z)$ ,$\forall{x}Q(x,z)$  ? I think it is "different" in the subformulas although it is free in both .
$(\exists{x}P(x,z))\vee(\forall{x}Q(x,z))$ . 
Which RPF is the right one  and why ?
$\exists{x}\forall{y}(P(x,z)\vee{Q(x,z)})$
$\exists{x}\forall{y}(P(x,z)\vee{Q(y,w)})$
---another question is ----
can we convert the free the variable to "universal quantified" ? 
If not, why ?
--update --
I just read a few posts about the "free variable and universal quantifier".
Looks like "free variable" could be explicitly taken as universal quantified.
If so, the RPF will be like the "second one". Am i right ?
 A: $(\exists{x}P(x,z))\lor(\forall{x}Q(x,z)) \Leftrightarrow$
$(\exists{x}P(x,z))\lor(\forall{y}Q(y,z))\Leftrightarrow$
$\exists{x}\forall{y}(P(x,z)\lor Q(y,z))$
So it is neither of the ones you indicated.
It is not 
$\exists{x}\forall{y}(P(x,z)\vee{Q(x,z)})$
because the $\forall y$ in this expression is not quantifying anything, and the $\exists x$ has captured the $x$ in the $Q(x,z)$, which it shouldn't have, since that one was universally quantified, and quantified separately from the $x$ in $P(x,z)$. 
It is also not:
$\exists{x}\forall{y}(P(x,z)\vee{Q(y,w)})$
because the variable $z$ is free, meaning that once you quantify it, it should point to the same object in both $P(x,z)$ and $Q(x,z)$. So you don't want to change that to a $w$. 
To further explain this: Think of how expressions are syntactically built up from smaller expressions.  Now that we have an expression with a $z$, we can quantify it ... but in doing so the whole expression will need to be quantified.  If you want to end up with something like your second statement, then you need to have $\exists{x}P(x,z)$ as one expression, and $\forall{x}Q(x,w)$ as another, and where you can quantify each of these statements, and then possibly put them together. But in $\exists{x}P(x,z)\land \forall{x}Q(y,z)$ we have already combined the $\exists{x}P(x,z)$ and the $\exists{x}P(x,z)$, and once we have done so, the $z$ should refer to the same object, whether you later quantify it existentially or universally.
Also: right now the expression is a formula with a free variable $z$ ... yes, for the sake of certain analyses you can treat it like a universal ... but you really don't want to add an explicit universal to the expression, because the formula could be used in a larger statement where the $z$ is existentially quantified. So: in this conversion process: if you have a free variable, keep it a free variable!
