Show $(A \cup B)\setminus(A \cap B) = (A\setminus B) \cup (B\setminus A)$ Show $(A \cup B)\setminus(A \cap B) = (A\setminus B) \cup (B\setminus A)$.
What I have so far...
This is (A or B) and (A and B)' = (A and B') or (B and A')
(A or B) and (A' or B') = (A and B') or (B and A')
((A or B) and A') or ((A or B) and B') = (A and B') or (B and A')
I feel like I'm just getting farther away. There must be something simple I'm missing.
 A: As Simon's answer points out, often you want to construct a so-called "element-chasing" proof to show that $S=T$ by showing $S\subseteq T$ and $T\subseteq S$.
However, I would only use an element-chasing proof if you can't "squeeze out" the equality by using some basic set algebra. Assuming you know about what the complement of a set is, distributivity, etc., see if you can follow the proof below (more steps than usual have been added to emphasize clarity, but let me know if a step or two does not make sense):
\begin{align}
\text{LHS}&=(A\cup B)\setminus(A\cap B)\\ &= (A\cup B)\cap(A\cap B)^C\tag{definition}\\[0.5em]
&= (A\cup B)\cap(A^C\cup B^C)\tag{De Morgan}\\[0.5em]
&= [(A\cup B)\cap A^C]\cup[(A\cup B)\cap B^C]\tag{distrib.}\\[0.5em]
&= [(A\cap A^C)\cup(B\cap A^C)]\cup[(A\cap B^C)\cup(B\cap B^C)]\tag{distrib.}\\[0.5em]
&= [\varnothing\cup(B\cap A^C)]\cup[(A\cap B^C)\cup\varnothing]\tag{$S\cap S^C=\varnothing$}\\[0.5em]
&= (B\cap A^C)\cup(A\cap B^C)\tag{simplify}\\[0.5em]
&= (A\cap B^C)\cup(B\cap A^C)\tag{commutativity}\\[0.5em]
&= (A\setminus B)\cup(B\setminus A)\tag{definition}\\[0.5em]
&= \text{RHS}.
\end{align}
Are all of those steps clear? Can you see how the proof above, despite seeming somewhat long winded, is exactly a good bit simpler than cranking out two proofs (one for each direction)?
A: Hint:
In general to show that two sets, say $S$ and $T$, are equal you want to show that $S \subset T$ and $T \subset S$. In other words $x \in S \Rightarrow x \in T$ and $x \in T \Rightarrow x \in S$.
Hence what you want to show is that
$$1. \quad x \in (A \cup B)\backslash (A \cap B) \ \Rightarrow x \in (A\backslash B)\cup(B\backslash  A)$$
and 
$$2. \quad  x \in (A\backslash B)\cup(B\backslash  A) \ \Rightarrow x \in (A \cup B)\backslash (A \cap B) $$
Try writing arguments for these two statements. A Venn diagram will also help you see what is going on here.

Here's the beginning of a possible argument for statement 1:
If $x \in (A \cup B)\backslash (A \cap B)$ then  $x \in A$ or $x \in B$. However if $x \in A$ then $x \not\in B$ as $x \not\in A \cap B$. Similarly if $x \in B$ then ...
