Intersection is Subset of Union of Intersections with Complements

Theorem
Let $R, S, T$ be sets.

Then:
 * $S \cap T \subseteq \left({R \cap S}\right) \cup \left({\overline R \cap T}\right)$

where $\overline R$ denotes the complement of $R$.

Proof
Let $x \in S \cap T$.

Then by definition of set intersection, $x \in S \land x \in T$.

From Conjunction therefore Disjunction of Conjunctions with Complements, it follows that:


 * $\left({x \in S \land \psi}\right) \lor \left({x \in T \land \neg \psi}\right)$

where $\psi$ is any arbitrary statement.

Let $\psi$ be the statement $x \in R$.

Thus:
 * $\left({x \in S \land x \in T}\right) \implies \left({x \in S \land x \in R}\right) \lor \left({x \in T \land x \notin R}\right)$

By definition of subset, set intersection, set union and set complement, it follows that:


 * $S \cap T \subseteq \left({R \cap S}\right) \cup \left({\overline R \cap T}\right)$

Illustration by Venn Diagram
A Venn diagram illustrating this result is given below:


 * IntersectionSubsetUnIntsComps.png

The red field marks $R \cap S$.

The blue field marks $\overline R \cap T$.

The purple field marks $S \cap T$, where both the red and blue are seen to intersect with the purple field.