Product of Subset with Intersection/Proof 2

Theorem

Let $\struct {G, \circ}$ be an algebraic structure.

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

Then:

$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$

$\paren {Y \cap Z} \circ X \subseteq \paren {Y \circ X} \cap \paren {Z \circ X}$

where $X \circ Y$ denotes the subset product of $X$ and $Y$.

Consider the relation $\RR \subseteq G \times G$ defined as:

$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$

Then:

$\forall S \subseteq G: X \circ S = \RR \sqbrk S$

Then:

$\ds X \circ \paren {Y \cap Z}$

$=$

$\ds \RR \sqbrk {Y \cap Z}$

$\ds $

$\subseteq$

$\ds \RR \sqbrk Y \cap \RR \sqbrk Z$

$\ds $

$=$

$\ds \paren {X \circ Y} \cap \paren {X \circ Z}$

Next, consider the relation $\RR \subseteq G \times G$ defined as:

$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists h \in X$

Then:

$\forall S \subseteq G: S \circ X = \RR \sqbrk S$

Then:

$\ds \paren {Y \cap Z} \circ X$

$=$

$\ds \RR \sqbrk {Y \cap Z}$

$\ds $

$\subseteq$

$\ds \RR \sqbrk Y \cap \RR \sqbrk Z$

$\ds $

$=$

$\ds \paren {Y \circ X} \cap \paren {Z \circ X}$

$\blacksquare$