This is an exercise from Velleman's "How To Prove It". The end of chapter questions have escalated in difficulty, so I just want to make sure that I am on the right track.
- Suppose $A$, $B$, and $C$ are sets. Prove that $A \vartriangle B \subseteq C$ iff $A \cup C = B \cup C$.
Proof: Suppose $A \vartriangle B \subseteq C$. Let $x$ be arbitrary. Suppose $x \in A \cup C$, then either $x \in A$ or $x \in C$. We consider these two cases:
Case 1. $x \in A$. Suppose $x \notin B \cup C$. So $x \notin B$ and $x \notin C$. Since $x \in A$ and $x \notin B$, $x \in A\setminus B$. It follows that $x \in A \setminus B \cup B \setminus A$, so $x \in A \vartriangle B$. Since $A \vartriangle B \subseteq C$ and $x \in A \vartriangle B $, $x \in C$. But then we have $x \in C$ and $x \notin C$, which is a contradiction. Thus, $x \in B \cup C$
Case 2. $x \in C$. It immediately follows that $x \in B \cup C$.
In every case, we have shown that $x \in B \cup C$. The proof of $x \in B \cup C \implies x \in A \cup C$ will be similar, but with the roles of $A$ and $B$ switched. Therefore, $A \cup C = B \cup C$.
Now suppose $A \cup C = B \cup C$. Let $x \in A \vartriangle B$ be arbitrary. Then $x \in A \setminus B \cup B \setminus A$, which means that $x \in A \setminus B$ or $x \in B \setminus A$. We consider these two cases:
Case 1. $x \in A \setminus B$. Then $x \in A$ and $x \notin B$. Suppose $x \notin C$. Then since $x \notin B$ and $x \notin C$, $x \notin B \cup C$. Since $x \in A$, $x \in A \cup C$. Then since $A \cup C = B \cup C$, $x \in B \cup C$. But then we have $x \in B \cup C$ and $x \notin B \cup C$, which is a contradiction. Thus, $x \in C$.
Case 2. $x \in B \setminus A$. By similar reasoning as case 1 with $A$ and $B$ switched, we also find that $x \in C$.
In every case, we have shown that $x \in C$. Since $x$ was arbitrary, it follows that $A \vartriangle B \subseteq C$. Therefore, $A \vartriangle B \subseteq C$ iff $A \cup C = B \cup C$. $\square$
Best Answer
Your proof is fine. I'll go a little further and compliment the way you explained the result in detail.
Having said that, I believe that it can be proven using "iff" statements all the way.