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: $(A \cap B^{C}) \cup (A^{C} \cap B) = (A \cap B^{C} \cup A^{C}) \cup (A \cap B^{C} \cup B^{C})$ 
 A: since $A \cap B \subseteq A \cup B$
$$
A \cap B = (A \cap B) \cap (A \cup B) \\
= (A \cap B) \cap ( (A \cap B^{C}) \cup (A^{C} \cap B)) \\
=  ((A \cap B) \cap ( (A \cap B^{C}))\cup ((A \cap B) \cap ( (A^{C} \cap B)) \\
= \varnothing
$$
A: Suppose $ x\in A \cap B $. Then $x \in A$ and $ x \in B $. 
But then $ x \not \in (A \cap B^{C})  $ and $ x \not \in (A^{C} \cap B) $ which forces $ x \not \in  (A \cap B^{C}) \cup (A^{C} \cap B)$ and hence $ x\not \in A \cup B $. But then $x \not \in A$ and $ x \not \in B $ which is more than a contradiction. 
A: Here you can see that $\color{red}{(A- B) ∪ (B -A) = A\Delta B = (A ∪ B) - (A ∩ B)}$. So:  
$$A\cup B=(A \cap B^{C}) \cup (A^{C} \cap B) = \color{lime}{(A-B)\cup(B-A)=A\Delta B=(A\cup B)-(A\cap B)}.$$ This means $A \cap B = \emptyset$
A: I'll add an answer expanding your idea, since that works too!
Only, you made a few mistakes.
It should be:  
$(A \cap B^{C}) \cup (A^{C} \cap B) = ((A \cap B^{C}) \cup A^{C}) \cap ((A \cap B^{C}) \cup B)$   
Then, 
$=((A\cup A^{C}) \cap (B^{C}\cup A^{C})) \cap ((A\cup B) \cap (B^{C}\cup B))=$
$=(B^{C}\cup A^{C}) \cap (A\cup B)$  
We know this to be $A\cup B$, so we can conclude $A \cap B = \emptyset$ (for instance by drawing a Venn diagram, or by further manipulation)
A: 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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\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
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\newcommand{\then}{\Rightarrow}
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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.
