Commutativity of intersection

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A\cap B=B\cap A

Note

This is somewhere between a theorem and a definition because at some point you have to accept "and" is commutative or something.

Proof

\implies

x\in A\cap B\implies x\in A\text{ and }x\in B\implies x\in B\text{ and }x\in A\implies x\in B\cap A, thus by the implies and subset relation we see A\cap B\subset B\cap A

\impliedby

By the exact same procedure we see B\cap A\subset A\cap B

Thus we conclude A\cap B=B\cap A

[Expand]Set Theory