Proving equality of sets using proof by contradiction

Question

I was wondering how can I prove the following conjecture using proof by contradiction.

If $A-B = B-A$, then it must be the case that $A=B$.

I tried to assume that $A \ne B$ and tried to show that $A-B = B-A$ still holds. However, I'm not sure how to deduce this part.

I am new to set theory and proofs, so any kind of help would be appreciated.

Answer 1

I tried to assume that $A≠B$ and tried to show that $A−B=B−A$ still holds. However, I'm not sure how to deduce this part.

Suppose there is an $a$ such that $a\in A$ and $a\notin B,$ yet $A-B = B-A$

$a\in A-B, a\notin B-A$

$A-B \ne B-A$

Contradiction!

All $a$ in $A$ are also in $B.$

And by an identical argument all $b$ in $B$ are also in $A.$

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Alternatively, and a little longer for clarity.

• Assume $A\smallsetminus B=B\smallsetminus A$.
  • Take an arbitrary $a$
    • Assume $a\in A$.
      • Assume $a\notin B$.
      • $a\in A \wedge a\notin B$ by conjunction .
      • $a\in A\smallsetminus B$ by definition of set minus.
      • $a\in B\smallsetminus A$ by the first assumption (and substitution).
      • $a\in B\wedge a\notin A$ by definition of set minus.
      • $a\in B$ by simplification.
      • Contradiction! ($a\notin B$ and $a\in B$)
    • $a\in B$ via proof by contradiction.
  • $a\in A\to a\in B$ via conditional proof
• $\forall x~(x\in A\to x\in B)$ by universal generalisation.
• $A\subseteq B$ by definition of subset
• $B\subseteq A$ similarly.
• $A=B$