$$\exists xP(x) \wedge Q(x) \wedge S(x)$$
In this formular the first x is bound. The second and third x are free.
Because of this, I can subsitute them (2nd and 3rd) with y
$$\exists xP(x) \wedge Q(y) \wedge S(y)$$
But can I also substitute them in this way?
$$\exists xP(y) \wedge Q(y1) \wedge S(y2)$$
Because as far as I understand the free variable can be any element from the "universal set", but are they the same?