Using elementary set theory to show the union of two sets is equal to the set difference of two sets I am supposed to prove (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B)
So far I have:  
(A \ B) ∪ (B \ A) = x ∈ (A ∪ B) and x ∉ (A ∩ B) [by the definition of set difference]
= (x ∈ A or x ∈ B) and x ∉ (A ∩ B) [by the definition of union]
= (x ∈ A or x ∈ B) and x ∈ (A ∩ B)^c [by the definition of complement]
= (x ∈ A or x ∈ B) and (x ∈ A^c or x ∈ B^c) [by the definition of distribution]
I can't figure out how to further manipulate the right hand side. I have also tried manipulating the left hand side:
(A ∪ B) \ (B ∩ A) = (A \ B) ∪ (B \ A)  
= x ∈ (A \ B) or x ∈ (B \ A)  
= (x ∈ A and x ∉ B) or (x ∈ B and x ∉ A)
 A: You are, in fact, almost there:


*

*($x\in A$ or $x\in B$) and ($x \in A^c$ or $x \in B^c$)

*($x \in A$ or $x\in B$) and ($x \notin A$ or $x \notin B$)

*($x \in A$ or $x\in B$) and $x \notin A \cap B$


and the last expression is just the definition of $(A \cup B) \setminus (A \cap B)$. The last step used De Morgan's law for negation, but it's not hard to verify its truth directly.
A: $(A\setminus B)\cup (B\setminus A) = $
$(A\cap B^c)\cup(B\cap A^c)=[(A\cap B^c)^c\cap(B\cap A^c)^c]^c$[From De Morgan's Law]
$(A^c\cup B)\cap(B^c\cup A)]^c=[((A^c\cup B)\cap B^c) \cup(A^c\cup B)\cap A)]^c=$
$[(A^c\cap B^c)\ \cup(A\cap B)]^c=(A^c\cap B^c)^c \cap(A\cap B)^c=$
$(A\cup B)\cap(A\cap B)^c\equiv (A\cup B)\setminus(A\cap B)$
A: Use Distribution, Idempotent Tautology, and DeMorgan's Law.  (Also commutation)
$\begin{align}
 &x\in (A\setminus B)\cup (B\setminus A)
\\ \iff & x\in (A\setminus B) \lor x\in (B\setminus A)
\\ \iff & (x\in A\land x\not\in B)\lor (x\in B\land x\not\in A)
\\ \iff & (x\in A\lor x\in B) \land (x\in A\lor x\not\in A) \land (x\not\in B \lor x\in B) \land (x\not\in B \lor x\not\in A)
\\ \iff & (x\in A\lor x\in B) \land (x\not\in B \lor x\not\in A)
\\ \iff & x\in (A\cup B) \land \neg(x\in B \land x\in A)
\\ \iff & x\in (A\cup B) \land x\not\in (B\cap A)
\\ \iff & x\in (A\cup B) \setminus (B\cap A)
\\ \iff & x\in (A\cup B) \setminus (A\cap B)
\\[2ex] \therefore & (A\setminus B)\cup (B\setminus A) \quad\equiv\quad (A\cup B)\setminus(A\cap B)
\end{align}$
