Does $\exists x$ distributes over conjunction when one of the sentences is an invariance? I know $\exists x(P(x) \land Q(x))$ isn't the same as $\exists x P(x) \land \exists x Q(x)$. This is because the first sentence means that the same object makes P(x) and Q(x) true, and the second sentence allows for different elements to make P(x) and Q(x) true.
But if Q means, for instance, "the pizza is cold", is $\exists x P(x) \land \exists x Q$ equivalent to $\exists x(P(x) \land Q)$, since the bound variable $x$ is irrelevant to "the pizza is cold"?
I think the question amount to whether or not "there is some x such that the pizza is cold" is a valid sentence in logic.
 A: Yes, if $x$ does not occur freely in proposition $Q$, then these statements are equivalent.$$\exists x~(P(x)\land Q)\iff(\exists x~P(x))\land(\exists x~Q)$$
If there is some $x$ such that $P(x)\land Q$, then there is some $x$ such that $P(x)$ and there is some $x$ such that $Q$.
If there is some $x$ such that $P(x)$ and some $x$ such that $Q$ then, because $Q$ is invariant with respect to the identity of the term $x$, there is some $x$ where $P(x)\land Q$.
(Normally you cannot guarantee that the two existentials might apply to an identical value for the term.   Here the invariance assures that it can.)
A: Yes, they are equivalent. There are actually two general equivalences from which you can establish the equivalence you are looking for. Where $Q$ is a formula that does not contain $x$ as a free variable, we have:
Prenex Law
$\exists x (P(x) \land Q) \Leftrightarrow \exists x \ P(x) \land Q$
Null Quantification
$\exists x \ Q \Leftrightarrow Q$
Therefore:
$\exists x (P(x) \land Q) \Leftrightarrow \exists x \ P(x) \land Q \Leftrightarrow \exists x \ P(x) \land \exists x \ Q$
