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$$\exists xP(x) \wedge Q(x) \wedge S(x)$$

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?

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    No, in general you _can't_ rename _free_ variables at will and expect the meaning of the formula to be unchanged -- because the "meaning" of the formula consists of how its truth variables depends on the values of the free variables, and depending on $y$ is a different meaning than depending on $x$. – 2017-01-09
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    Another matter is whether $\exists x\,P(x) \land Q(x) \land S(x)$ really means $(\exists x\,P(x)) \land Q(x) \land S(x)$ rather than $\exists x\,(P(x) \land Q(x) \land S(x))$. Different authors use different conventions for when parentheses can be omitted. – 2017-01-09
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    Okay thanks so far. In my case it means $$(\exists xP(x)) \wedge Q(x) \wedge S(x)$$ – 2017-01-09
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    It is actually the *other* variable you can rename. $(\exists y P(y)) \land Q(x) \land S(x)$ is equivalent. – 2017-01-09
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    +1 for using latex on your first post. It is appreciated. – 2017-01-09

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