Product of Subset with Intersection/Corollary

Theorem
Let $\left({G, \circ}\right)$ be an algebraic structure.

Let $X, Y, Z \subseteq G$ such that $X$ is a singleton.

Then:
 * $X \circ \left({Y \cap Z}\right) = \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$
 * $\left({Y \cap Z}\right) \circ X = \left({Y \circ X}\right) \cap \left({Z \circ X}\right)$

where $X \circ Y$ denotes the subset product of $X$ and $Y$.

Proof
Let $X = \left\{{x}\right\}$.

We have from Product of Subset with Intersection that:
 * $X \circ \left({Y \cap Z}\right) \subseteq \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$

Let $g \in \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$.

Then by definition of subset product:
 * $\exists y \in Y, z \in Z: g = x \circ y = x \circ z$

By the Cancellation Laws it follows that $y = z$.

But:
 * $g = x \circ y \implies g \in X \circ Y$
 * $g = x \circ z \implies g \in X \circ Z$

and so as $x \circ y = x \circ z$ it follows that $g \in X \circ \left({Y \cap Z}\right)$.

Thus:
 * $X \circ \left({Y \cap Z}\right) \supseteq \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$

It follows by definition of set equality that:
 * $X \circ \left({Y \cap Z}\right) = \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$

By similar reasoning:
 * $\left({Y \cap Z}\right) \circ X = \left({Y \circ X}\right) \cap \left({Z \circ X}\right)$

Hence the result.