Subset Product within Semigroup is Associative

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

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

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

If $$\circ$$ is associative, then:


 * $$x \left({y S}\right) = \left({x y}\right) S$$
 * $$x \left({S y}\right) = \left({x S}\right) y$$
 * $$\left({S x}\right) y = S \left({x y}\right)$$

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

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

Then:


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


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

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

Proof of Corollary
Follows directly from the definition of Subset Product with Singleton:


 * $$x \left({y S}\right) = \left\{{x}\right\} \left({\left\{{y}\right\} S}\right)$$

and so on.