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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.

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Feb 15 '16 at 2:29
  • $\begingroup$ Please check that my edits did not change the meaning of what you intended to write. $\endgroup$ – N. F. Taussig Feb 15 '16 at 2:35
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It usually helps me to first just use definitions to "break up" each side of the equation and then use some basic rules of logic to deduce the desired result. You have the right idea, but I think you might not be seeing all the pieces you can use. To demonstrate, I'll do one side:

$(A - B) \cup (B - A) \subseteq (A \cup B) - (A \cap B)$:

Let $x \in (A - B) \cup (B - A)$. Then $x \in A$ and not in $B$ or $x \in B$ and not in $A$. But then $x \in A$ or $x \in B$, so $x \in (A \cup B)$. Similarly, $x \notin A$ or $x \notin B$. By Demorgan's laws, $x \notin A \lor x \notin B = \neg x \in A \lor \neg x \in B \implies \neg (x \in A \land x \in B)=x \notin A \cap B$. Putting this together, $x \in (A \cup B)-A \cap B$.

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