Prove or disprove $ \exists x \in S [ p(x) \land q(x)] \iff [\exists x \in S p(x)] \land [\exists x \in S q(x)] $ I proved it as follows. Can anybody tell me if it is correct or wrong?
Assume $ \exists x \in S [ p(x) \land q(x)] $ ,
Let $ x_0 \in S, st [ p(x_0) \land q(x_0)]$
$p(x_0)$ and $ q(x_0) $
Let x =$x_0$
p(x) and q(x)
$[\exists x \in S, p(x)]$ and  $[\exists x \in S, q(x)] $
$[\exists x \in S, p(x)] \land  [\exists x \in S, q(x)] $
$ \therefore \exists x \in S [ p(x) \land q(x)] \implies [\exists x \in S p(x)] \land  [\exists x \in S q(x)] $ ---------------------------1
Assume $[\exists x \in S p(x)] \land  [\exists x \in S q(x)] $ 
Let $x_0 \in S$
p($x_0$)
q($x_0$)
$[ p(x_0) \land q(x_0)] $
Let x=$x_0$
$[ p(x) \land q(x)] $
$ \therefore [\exists x \in S p(x)] \land  [\exists x \in S q(x)] \implies \exists x \in S [ p(x) \land q(x)] $  ---------------------2
By 1 and 2,
$ \exists x \in S [ p(x) \land q(x)] \iff [\exists x \in S p(x)] \land  [\exists x \in S q(x)] $
 A: The equivalence is false.  It is true that ${\rm LHS}\Rightarrow{\rm RHS}$, and your proof is accurate.
However, the other direction is false.  For example, take $S=\Bbb Z$ and $p(x)={}$"$x$ is even" and $q(x)={}$"$x$ is odd".
The error in your attempt at proving the converse is right at the beginning.  You have
$$[\exists x \in S p(x)] \land  [\exists x \in S q(x)]\ ,$$
which means that there is some $x$, say $x=x_1$, for which $p(x)$ is true and some $x$, say $x=x_2$, for which $q(x)$ is true.  However there is no reason why $x_1$ should equal $x_2$, that is, no reason why $x_1=x_2$.  By calling both of them $x_0$ you have assumed that they are the same.
It may be helpful to think of this in terms of ordinary language.  If you use the word "some" twice, it does not have to refer to the same object both times.  For example, if you say "I met somebody for lunch today" and your friend says "Well I met somebody for lunch today as well", that doesn't mean you both met the same person for lunch.  Ordinary language is not always identical with mathematical language, but in this case the issue is the same.
