$(\exists x)[A(x) \wedge B(x)] \rightarrow (\exists x)A(x) \wedge (\exists x)B(x)$
I realize that this is an equivalence but I am wondering how I might prove this.
I feel like I would use existential instantiation on $(\exists x)[A(x) \wedge B(x)]$ to get $A(x) \wedge B(x)$ and then I would use existential generalization on $A(x) \wedge B(x)$ to get $(\exists x)A(x) \wedge (\exists x)B(x)$
is this correct?
You can use simplification (sometimes called conjunction elimination):
1. $A(x) ∧ B(x) \rightarrow A(x)$
2. $A(x) ∧ B(x) \rightarrow B(x)$.
From 1 and 2, we have $\exists x(A(x) ∧ B(x)) \rightarrow \exists x(A(x)) ∧ \exists x(B(x))$.
The converse of the above is not true, because if $\exists x_1$ such that $A(x_1)$ and $\exists x_2$ such that $B(x_2)$ there is not enough information to conclude that $x_1 = x_2$. In other words, we cannot conclude that $\exists x$ such that both $A(x)$ and $B(x)$.