Prove $(A \triangle B) \cap (B\triangle C) \cap (C\triangle A) = \emptyset$ This can be proved by assuming that there exists some $x \in (A \triangle B) \cap (B\triangle C) \cap (C\triangle A) $ and then deriving a contradiction by considering each of the cases that arise. 
[$X\triangle Y$ is the symmetric difference of $X$ and $Y$]
Now, this proof ultimately breaks down into six cases which makes it a bit long considering the simple goal. So I was wondering if there was a better way of doing this?
My six case proof goes something like this: 
Suppose to the contrary that $x \in (A \triangle B) \cap (B\triangle C) \cap (C\triangle A) $. Then $x \in (A \triangle B)$ and $ x \in (B \triangle C)$ and $x \in (C\triangle A)$. Since $x \in (C \triangle A)$, $x \in A\backslash C$ or $x \in C\backslash A$.
Case 1: $x \in A\backslash C$. Since $x \in (B \triangle C)$, $x \in B\backslash C$ or $x \in C\backslash B$. Case 1.1: $x\in B\backslash C$... 
Case 2: $x \in C\backslash A$. Since $x \in (B \triangle C)$, $x \in B\backslash C$ or $x \in C\backslash B$. Case 2.1: $x\in B\backslash C$...

It's possible to make only one case. 
Because
$$x\in (A \triangle B)\iff (x\in A \wedge x\notin B ) \vee (x\notin A \wedge x\in B)$$
Then we have
$$x\in (A\triangle B) \cap (B\triangle C) \cap (C\triangle A)$$
using the definition above we see that $x$ can't be in $A,B$ or $C$. Hence the intersection is empty.
