Show $A\setminus (A\cap B)=(A\cup B)\setminus B$ I'm having problems with this; the strategy is to show the left side is in the right side, and vice versa. Thus far I can only show that both sides of the equality are in A\B. Can I use "the empty set = A and not A" as a fact? Because then it works. Or just show me how to do it!
 A: Yes, to show "X= Y"  you need to show "X is a subset of Y" and "Y is a subset of X".  And the standard way to show "X is a subset of Y" is to start "let x be a member of X" and use the definitions of X and Y to conclude "therefore x is a member of Y".  
Here, if x is a member of $A\setminus(A\cap B)$ then x must be in A but not in $A\cap B$ so not in B.  Since x is in A it is in $A\cup B$.  And x is not in B so it is still in $A\cup B\setminus B$.  Therefore x is in $A\cup B\setminus B$.
If x is in $A\cup B\setminus B$ then x is in $A\cup B$ but not in B.  That means that x is in A but not in B.  So x is not in $A\cap B$ and so is in $A\setminus(A\cap B)$.
A: 
Can I use "the empty set = A and not A" as a fact? 

Yes.
$$\begin{align}A\setminus(A\cap B)
  &= A\cap(A\cap B)^\complement
\\&~\vdots
\\&= (A\cap A^\complement)\cup(A\cap B^\complement)
\\&= \emptyset\cup(A\cap B^\complement)
\\&= A\cap B^\complement &&=A\setminus B
\\&= (A\cap B^\complement)\cap\emptyset
\\&~\vdots
\\&= (A\cap B)\cap B^\complement
\\[3ex]\therefore\quad A\setminus (A\cap B)&= (A\cap B)\setminus B
\end{align} $$
