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.