# Prove that if $A \bigtriangleup B\subseteq A$ then $B \subseteq A.$

This is Velleman's exercise 3.5.5 (And NO! not a duplicate of Prove that if $A \mathop \triangle B \subseteq A$ then $B\subseteq A$! My question is different):

Prove that if $$A \bigtriangleup B\subseteq A$$ then $$B \subseteq A.$$

Since in the definition of a symmetric difference we have disjunction, shouldn't we prove this statement by cases?

So here's my proof of it:

Proof. Let $$x$$ be an arbitrary element of $$B$$. Now suppose $$x \not\in A$$. From $$x \in B$$ and $$x \not\in A$$, we get $$x \in (B\setminus A)$$. We now consider two cases.

Case 1. $$x \in (A\setminus B)$$. Then by $$A \bigtriangleup B \subseteq A$$, we have $$x \in A$$ which is a contradiction.

Case 2. $$x \not\in (A\setminus B)$$. Since $$x \in (B\setminus A)$$ and $$A \bigtriangleup B \subseteq A$$, $$x \in A$$ which is also a contradiction.

Since by both cases we reached a contradiction then $$x \in A$$ and since $$x$$ was arbitrary, $$B \subseteq A$$.

In other words, in proof by cases (when we have disjunction in the given/hypotheses/premises) when we also use a contradiction, do we need to reach a contradiction for all the cases or just one will be enough?

We would need to reach a contradiction for each premise. However, in your proof, there was no need to split between cases because $$x\in B\setminus A \subseteq A\ \triangle\ B\subseteq A$$ implies $x\in A$.
You assume $x \in B$ and $x \not \in A$. This means $x \in B \setminus A$. Thus, $x \in A \bigtriangleup B.$ As $$A \bigtriangleup B\subseteq A$$ therefore $x \in A.$ This a a contradiction to the assumption that $x \not\in A.$
Hence, $A \subseteq B.$
Note: When you assume that $x \not \in A$, then this means that $x \not \in A \setminus B.$ So there is no need of splitting into cases.