Intersection with Subgroup Product of Superset

Theorem
Let $X, Y, Z$ be subgroups of a group $\left({G, \circ}\right)$.

Let $Y \subseteq X$.

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

where $Y \circ Z$ denotes subset product.

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

Then:
 * $x \in Y \circ Z \implies x \in Z$

and:
 * $x \in Y \implies x \in X \implies x \in Y \circ X$.

So $x \in X \cap \left({Y \circ Z}\right) \implies x \in \left({Y \circ X}\right) \cap \left({Y \circ Z}\right) \implies x \in Y \left({X \cap Z}\right)$.

Now let $x \in Y \left({X \cap Z}\right)$.

Then $x \in \left({Y \circ X}\right) \cap \left({Y \circ Z}\right)$.

Now $x \in \left({Y \circ X}\right) \implies x \in X$ as $x \in Y \implies x \in X$.

So $x \in X \cap \left({Y \circ Z}\right)$.

So $X \cap \left({Y \circ Z}\right) \subseteq Y \circ \left({X \cap Z}\right)$ and $Y \circ \left({X \cap Z}\right) \subseteq X \cap \left({Y \circ Z}\right)$.

So $X \cap \left({Y \circ Z}\right) = Y \circ \left({X \cap Z}\right)$.