Preimage of Intersection under Relation

Theorem
Let $S$ and $T$ be sets.

Let $\RR \subseteq S \times T$ be a relation.

Let $C$ and $D$ be subsets of $T$.

Then:
 * $\RR^{-1} \sqbrk {C \cap D} \subseteq \RR^{-1} \sqbrk C \cap \RR^{-1} \sqbrk D$

Also see

 * Image of Intersection under Relation
 * Image of Union under Relation
 * Preimage of Union under Relation

Proof
This follows from Image of Intersection under Relation, and the fact that $\RR^{-1}$ is itself a relation, and therefore obeys the same rules.