Product of Subset with Intersection/Corollary

Theorem
Let $\struct {G, \circ}$ be a group.

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

Then:
 * $X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
 * $\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$

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

Proof
Let $X = \set x$.

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

Let $g \in \paren {X \circ Y} \cap \paren {X \circ Z}$.

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 $g \in X \circ \paren {Y \cap Z}$.

Thus:
 * $X \circ \paren {Y \cap Z} \supseteq \paren {X \circ Y} \cap \paren {X \circ Z}$

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

By similar reasoning:
 * $\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$

Hence the result.