Subset Product within Commutative Structure is Commutative

Theorem
Let $$\left({S, \circ}\right)$$ be a groupoid.

If $$\circ$$ is commutative, then the operation $\circ_{\mathcal P}$ induced on the power set of $$S$$ is also commutative.

Proof
Let $$\left({S, \circ}\right)$$ be a groupoid in which $$\circ$$ is commutative.

Let $$X, Y \in \mathcal P \left({S}\right)$$.

Then:


 * $$X \circ_{\mathcal P} Y = \left\{{x \circ y: x \in X, y \in Y}\right\}$$


 * $$Y \circ_{\mathcal P} X = \left\{{y \circ x: x \in X, y \in Y}\right\}$$

... from which it follows that $$\circ_{\mathcal P}$$ is commutative on $$\mathcal P \left({S}\right)$$.