Show that the set $A \cap B = \emptyset$ Let  $A$ and $B$ be two sets for which the following applies: $A \cup B = (A \cap B^{C}) \cup (A^{C} \cap B)$. Show that $A \cap B = \emptyset$.
How?! I am seriously stuck. One thought I had is to distribute the right part, so that: $(A \cap B^{C}) \cup (A^{C} \cap B) = (A \cap B^{C} \cup A^{C}) \cup (A \cap B^{C} \cup B^{C})$ 

Just to give an answer in a different style, expanding the definitions and using the laws of logic results in the following straightforward (albeit perhaps a bit tedious) calculation.$
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We start by simplifying the assumption, since it looks like it can use some:
$$\calc
  A \cup B \;=\; (A \cap B^{C}) \cup (A^{C} \cap B)
\op\equiv\hints{definitions of $\;=,\cup,\cap,{}^C\;$}\hint{-- going from set level to logic level}
  \langle \forall x :: x \in A \lor x \in B \;\equiv\; (x \in A \land x \not\in B) \lor (x \not\in A \land x \in B) \rangle
\op\equiv\hints{logic: in RHS, distribute $\;\lor\;$ over $\;\land\;$, three times}\hint{-- this looks like the only way forward at this point}
  \langle \forall x :: x \in A \lor x \in B \;\equiv\; (x \in A \lor x \not\in A) \land (x \not\in B \lor x \not\in A) \land (x \in A \lor x \in B) \land (x \not\in B \lor x \in B) \rangle
\op\equiv\hint{logic: simplify using $\;P \lor \lnot P \;\equiv\; \true\;$, twice}
  \langle \forall x :: x \in A \lor x \in B \;\equiv\; (x \not\in B \lor x \not\in A) \land (x \in A \lor x \in B) \rangle
\op\equiv\hints{logic: simplify, using the fact that $\;P \;\equiv\; P \land Q\;$ and}\hint{$\;\lnot P \lor Q\;$ are two ways of writing $\;P \then Q\;$}
  \langle \forall x :: \lnot(x \in A \lor x \in B) \lor x \not\in B \lor x \not\in A \rangle
\op\equiv\hints{logic: use negation of $\;x \not\in A\;$ on other side of $\;\lor\;$;}\hint{same for $\;x \in B\;$; simplify}
  \langle \forall x :: x \not\in B \lor x \not\in A \rangle
\op\equiv\hint{logic: DeMorgan}
  \langle \forall x :: \lnot(x \in B \land x \in A) \rangle
\op\equiv\hint{definitions of $\;\emptyset,\cap\;$}
  B \cap A = \emptyset
\endcalc$$
So we've shown that the two statements are actually equivalent.
