Product of Subset with Intersection/Proof 1

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

Let $X, Y, Z \subseteq G$.

Then:

where $X \circ Y$ denotes the subset product of $X$ and $Y$.
 * $X \circ \left({Y \cap Z}\right) \subseteq \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$
 * $\left({Y \cap Z}\right) \circ X \subseteq \left({Y \circ X}\right) \cap \left({Z \circ X}\right)$

Proof
Let $x \in X, t \in Y \cap Z$.

By the definition of intersection, $t \in Y$ and $t \in Z$.

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

We have $x \circ t \in X \circ \left({Y \cap Z}\right)$ by definition of subset product.

As $t \in Y$ and $t \in Z$, we also have $x \circ t \in X \circ Y$ and $x \circ t \in X \circ Z$.

The result follows.

Similarly, consider $\left({Y \cap Z}\right) \circ X$.

Then we have $t \circ x \in \left({Y \cap Z}\right) \circ X$ by definition of subset product.

As $t \in Y$ and $t \in Z$, we also have $t \circ x \in Y \circ X$ and $t \circ x \in Z \circ X$.

Again, the result follows.