This is a exercise from How To Prove It by Velleman. Exercise 3.5.24 to be exact. The solutions I can find online involve using 4 cases. My proof involves 2 so I was wondering if I was skipping some steps.
Proof: Let x be an arbitrary element of $(A \triangle C) \cap (B \triangle C)$. It follows that $x \in A \triangle C$ and $x \in B \triangle C$.
We will consider 2 cases.
Case 1: $x \in C$. Since $x \in A \triangle C$, it follows that $x \notin A$. Also since $x \in B \triangle C$, it follows that $x \notin B$. This means $x \in C \setminus (A \cap B)$ thus $x \in (A \cap B) \triangle C$.
Case 2: $x \notin C$. Since $x \in A \triangle C$, it follows that $x \in A$. Also since $x \in B \triangle C$, it follows that $x \in B$. This means $x \in (A \cap B) \setminus C$ thus $x \in (A \cap B) \triangle C$.
Both cases gave us $x \in (A \cap B) \triangle C$ and since x was arbitrary $(A \triangle C) \cap (B \triangle C) \subseteq (A \cap B) \triangle C$
Is this proof a valid one? Any feedback would be greatly appreciated.
Thanks in advance.