Prove that:
$(A-B)\cup (B-A)=(A\cup B)-(A\cap B)$
My Attempt:
$x\in (A-B)\cup (B-A)$
so $x\in (A-B)\vee x\in (B-A)$
$(x\in A \wedge x\notin B)\vee (x\in B\wedge x\notin A)$
I suppose I should try different cases from here? I haven't been able to make progress after this step. Thanks for your help!
Looks like you're on the right track. You could make two cases now based on your "or" statement.
Case $1$: $x \in A$ and $x\notin B$. Then obviously $x\notin A\cap B$.
Case $2$: $x \in B$ and $x\notin A$. Then again $x \notin A\cap B$.
In either case (that is, whether we start with $x\in A$ or $x\in B \equiv x \in A\cup B$) we know $x\notin A\cap B$ so $x\in (A \ \cup B)\setminus (A\cap B)$ and you can conclude that $$(A-B)\cup (B-A)\subseteq (A\cup B)-(A\cap B)$$ Can you complete the second half of the proof?