Intersection with Subgroup Product of Superset

Theorem
Let $X, Y, Z$ be subgroups of a group $\struct {G, \circ}$.

Let $Y \subseteq X$.

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

where $Y \circ Z$ denotes subset product.

Proof
By definition of set equality, it suffices to prove:
 * $X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$

and:
 * $Y \circ \paren {X \cap Z} \subseteq X \cap \paren {Y \circ Z}$

$X \cap \paren {Y \circ Z}$ is contained in $Y \circ \paren {X \cap Z}$
Let $s \in X \cap \paren {Y \circ Z}$.

Then:

So we have:

Thus by definition of set intersection:
 * $z \in X \cap Z$

So:
 * $s = y \circ z \in Y \circ \paren {X \cap Z}$

By definition of subset:
 * $X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$

$Y \circ \paren {X \cap Z}$ is contained in $X \cap \paren {Y \circ Z}$
Then:

We have established that:
 * $x \in Y \circ \paren {X \cap Z} \iff x \in X \cap \paren {Y \circ Z}$

From the definition of set equality:
 * $Y \circ \paren {X \cap Z} = X \cap \paren {Y \circ Z}$