Is the following set operation true? Prove the following or else find a counter example:
For all sets $A$, $B$, and $C$, $$((A \cup B) − C) \cup (A \cap B) = ((A − B) \cup (B − A)) − C$$
For the life of me, I can't figure out if its true or not. I have tried drawing the Venn diagrams but I'm not too good with them and can't quite figure out if diagrams representing either side of above equation look alike.Help appreciated.
 A: It might help to let $C = \emptyset$, since it seems that's allowed. I'll assume here that $C = \emptyset$.
Just thinking about the right side, the set $(A - B) \cup (B - A)$ has a special name: it's the symmetric difference of $A$ and $B$. It's the set of everything that's in exactly one of $A$ or $B$. In particular, it doesn't contain anything in $A \cap B$.
But if you notice, the left side certainly contains everything in $A \cap B$, doesn't it? So, can you cook up a situation that takes advantage of this fact? 
A: Notice that in the Right set $((A-B)\cup (B-A)) - C$ nothing in $A\cap B$ is included. However such elements are obviously included in the left set, since they are one part of the union.
Thus any choice of sets A, B and C such that $A\cap B\neq \emptyset$ is a good answer.
Ex: $A={1,2}, B={2,3} C={4}$ or maybe if you want something simple $A=B=\{1\}$. 
A: Let $A=\{1,2,3\}$,
$B=\{2,3,4\}$ and
$C=\{0\}$
Then $$[(A\cup B)-C]\cup (A\cap B)=[\{1,2,3,4\}-\{0\}]\cup \{2,3\}$$
$$=\{1,2,3,4\}\cup \{2,3\}$$
$$=\{1,2,3,4\}$$
But $$[(A-B)\cup(B-A)]-C=[\{1\}\cup\{4\}]-\{0\}$$
$$=\{1,4\}-\{0\}$$
$$=\{1,4\}$$
So there's a counterexample.
