Theorem
Intersection is commutative
$ A\cap B = B\cap A $
where $ A $ and $ B $ are sets.
Proof
$ \begin{align} A\cap B &\triangleq \{x\in\mathcal S:\;x\in A\;\mbox{and}\; x\in B\}\\ &= \{x\in\mathcal S:\;x\in B\;\mbox{and}\; x\in A\}\\ &= B\cap A\\ \blacksquare \end{align} $
