Product of Subset with Intersection

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

Note
It is not always the case that:
 * $X \circ \left({Y \cap Z}\right) \supseteq \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$
 * $\left({Y \cap Z}\right) \circ X \supseteq \left({Y \circ X}\right) \cap \left({Z \circ X}\right)$

so this result can not be expressed as an equality.

Example
Let $a \in G$ such that $a \ne a^{-1}$.

Let $X = \left\{{a, a^{-1}}\right\}, Y = \left\{{a}\right\}, Z = \left\{{a^{-1}}\right\}$.

Then: so clearly $X \circ \left({Y \cap Z}\right) \ne \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$.
 * $X \circ \left({Y \cap Z}\right) = X \circ \varnothing = \varnothing$
 * $\left({X \circ Y}\right) \cap \left({X \circ Z}\right) = \left\{{a^2, e}\right\} \cap \left\{{e, a^{-2}}\right\} \ne \varnothing$

But see Product of Subset with Intersection/Corollary where equality holds when $X$ is a singleton.