I'm trying to show the proof of the statement in the title. What I have so far is the following:
Right side is equivalent to $(x \in A \cup B) \wedge \neg(x \in A \cap B)$.
From here I distribute the 'not' to the parentheses $(A \cap B)$. ($x \in A \cup B) \wedge [\neg(x \in A) \vee \neg(x \in B)]$ where the 'and' in the $A \cap B$ parentheses has become an 'or' after being operated on by the 'not' (inverse?) operator.
From here I'm stuck. My last step is separation of the two sets in the second parentheses, which looks like $(A \cup B)$ and $\neg (A) \cup \neg(B)$. I would appreciate any assistance in determining where to go from here.
I apologize for my poor formatting. This is my first time on this website.