Prove $(A\cup C=B\cup C)\implies(A\mathbin{\Delta} B\subseteq C)$

Question

My full problem is: Suppose $A$, $B$, and $C$ are sets. Prove: $A\mathbin{\Delta}B \subseteq C\text{ iff }A\cup C=B\cup C$.

I've proved my case one, which is that $(A\mathbin{\Delta} B\subseteq C)\implies (A\cup C=B\cup C)$, but I've been trying and trying to prove the second case and I keep having difficulties. I've tried to start with assuming that there is an arbitrary $x$ that is in $(A\cap B)$ to get a contradiction, which hasn't worked so far. Any guidance would be wonderful.

Answer 1

$\Leftarrow$: Note that $A\triangle B = (A\setminus B) \cup (B\setminus A)$.   Suppose $x\in A\triangle B$ — say, $x \in A\setminus B$, without loss of generality.   Then $x\in A \subseteq A\cup C = B\cup C$.   But $x\notin B$, so $x\in C$.

Answer 2

$A\cup C = B\cup C$ means: any element in $A$ is in the union $B\cup C$, including those of $A$ not in $B$.   That is: $A\setminus B\subseteq C$.   And by symmetry this means $B\setminus A\subseteq C$, and consequently:

$$\therefore A\cup C=B\cup C \implies A\triangle B \subseteq C$$

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$$\dfrac{\dfrac{A\setminus B \subseteq A\cup C \;,\; A\cup C = B\cup C}{A\setminus B \subseteq C} \;,\; \dfrac{B\setminus A \subseteq B\cup C \;,\; A\cup C = B\cup C}{B\setminus A \subseteq C}}{A\triangle B\subseteq C}$$