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?

Thanks in advance.


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 $x\in A$.


There are no two cases.

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.


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