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
By definition of set equality, it suffices to prove:
 * $X \cap \left({Y \circ Z}\right) \subseteq Y \circ \left({X \cap Z}\right)$

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

$X \cap \left({Y \circ Z}\right)$ is contained in $Y \circ \left({X \cap Z}\right)$
Suppose that $s \in X \cap \left({Y \circ Z}\right)$.

Hence $s \in Y \circ \left({X \cap Z}\right)$.

It follows by definition of subset that $X \cap \left({Y \circ Z}\right) \subseteq Y \circ \left({X \cap Z}\right)$.

$Y \circ \left({X \cap Z}\right)$ is contained in $X \cap \left({Y \circ Z}\right)$
Then:

We have established that $x \in Y \circ \left({X \cap Z}\right) \iff x \in X \cap \left({Y \circ Z}\right)$.

From the definition of set equality, it follows that $Y \circ \left({X \cap Z}\right) = X \cap \left({Y \circ Z}\right)$.