Prove for every two sets $A$ and $B$ that $A-B$, ${B-A}$ and $A \cap B$ are pairwise disjoint. 
Prove for every two sets $A$ and $B$ that $A-B$, ${B-A}$ and $A \cap B$ are pairwise disjoint.  

I'm really stuck on this one. I know pairwise disjoint means no two elements in $A$ and $B$ are the same. $A \cap B$ is the empty set, but how would I prove this?
 A: As you say two sets $A$ and $B$ are disjoint if $A \cap B = \emptyset$.  But note that you want to show that $(A - B) \cap (B-A) = \emptyset$ and $(A - B) \cap (A \cap B) = \emptyset$ and $(B-A) \cap (A \cap B) = \emptyset$ as John points out in the comments.   
First, can you see why $(A - B) \cap (B -A)$ should be empty?  This is the set of all elements of $A$ that are not elements of $B$ while simultaneously being elements of $B$ but not elements of $A$.  It would be weird if something satisfied both of these properties, right?  Since we are trying to prove something about the emptyset, it may be best to do a proof by contradiction.  So suppose $(A-B) \cap (B-A) \neq \emptyset$, i.e. there is some $x \in (A-B) \cap (B-A)$.  Then $$x \in (A-B) \cap (B-A) \iff x \in (A - B) \wedge x \in (B - A)$$
$$\iff (x \in A \wedge x \notin B) \wedge (x \in B \wedge x \notin A)$$
$$\iff x \in A \wedge x \notin A \wedge x \in B \wedge x \notin B$$
So we have $x \in A \wedge x \notin A$, but this can definitely not happen with any element $x$; we have arrived at a contradiction, and our claim is false.  Hence, $(A-B) \cap (B-A) = \emptyset$.
Now, for $(A-B) \cap (A \cap B)$, again try to think what this means.  For something to be an element of this set it would have to be in $A$ but not $B$ but also in $A$ and $B$.  This seems a bit fishy, huh?
We can proceed by contradiction like above, so suppose $(A-B) \cap (A \cap B) \neq \emptyset$ and let $x \in (A-B) \cap (A \cap B)$.  Then, 
$$x \in (A-B) \cap (A \cap B) \iff (x \in A \wedge x \notin B) \wedge (x \in A \wedge x \in B)  $$
$$\iff x \in A \wedge  x \notin B \wedge x \in B$$
But now, we have $x \in B \wedge x \notin B$, again a contradiction.  Thus, $(A-B) \cap (A \cap B) = \emptyset$.
I bet you can do $(B-A) \cap (A \cap B)$ since it is very similar to the previous one.
