Applying the rules of logic. I am asked to prove that $(A \cup B)\setminus(A \cap B) = (A\setminus B) \cup (B\setminus A)$.
My approach is as follows:
Let $x \in (A \cup B)\setminus(A \cap B)$
implies that
$x \in (A \cup B)$ and
 $x \notin (A \cap B)$
implies that
$(x \in A \lor x \in B) \land
(x \notin A \land x \notin B)$.
Further if I take statements $p: x \in A$
$q: x \in B$
and $\neg p$ & $\neg q$ to be their negations.
Then using rules of logic, 
$[(p \lor q) \land (\neg p \land \neg q)] \Leftrightarrow (\neg p \land q) \land (p \land \neg q)$.
But this does not give me desired proof. 
I know that for union the "or" I used is inclusive one, means x may belong to both A and B,
so dont know how to represent this properly.
 A: If $x$ belongs to ($A$ union $B$), then $x$ belongs to $A$ or $x$ belongs to $B$.
If also $x$ does not belong to ($A$ intersection $B$), then $x$ does not belong to both, and so $x$ belongs to $A$ but not $B$, or $x$ belongs to $B$ but not $A$.
That is equivalent to saying $x$ belongs to ($A-B$) or $x$ belongs to $(B-A)$, i.e.  $x$ belongs to ($A-B$ union $B-A$)
A: You've made an mistake in your proof. The statement $x\notin (A\cap B)$ is not the same as $x \notin A \land x\notin B$. You have to use DeMorgan law here so $x\notin (A\cap B)$ is in fact $x\notin A \lor x\notin B$
You also should be aware that you should be able to work directly at sets as you would do on propositional calculus. So you have:
$$(A\cup B) \setminus (A\cap  B) = (A\cup B) \cap \overline{A \cap B} = (A\cup B) \cap (\overline A \cup \overline B) = \\ (A\cap \overline A) \cup (A\cap\overline B) \cup (B\cup\overline A) \cup (B\cup\overline B) = (A\cap\overline B)\cup(B\cup\overline A)$$
And similarily for the RHS:
$$(A\setminus B) \cup (B\setminus A) = (A \cap \overline B) \cup (B \cap \overline A)$$
