# Prove that $A \vartriangle B \subseteq C$ iff $A \cup C = B \cup C$.

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.

1. 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 $$A \cup C = B \cup C$$. $$\square$$

• A suggested shorter proof, easier (for me at least!) to grasp as a whole: $$(A \cup C) \setminus (B \cup C) = ((A \cup C) \setminus C) \setminus B = (A \setminus C) \setminus B = (A \setminus B) \setminus C,$$ and similarly $(B \cup C) \setminus (A \cup C) = (B \setminus A) \setminus C,$ therefore: $$(A \cup C) \vartriangle (B \cup C) = (A \vartriangle B) \setminus C,$$ whence the result. Apr 7, 2020 at 14:59

• You're welcome, @Iyeeke. I mean starting with $A \vartriangle B\subseteq C$ then using iff statements to get intermediate results until, eventually, you get $A\cup C=B\cup C$. I'm thinking of how to do it myself, so I might update this answer soon. Apr 7, 2020 at 14:14