Image of Subset under Relation is Subset of Image/Corollary 1

Corollary to Image of Subset under Relation is Subset of Image
Let $S$ and $T$ be sets.

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

Let $C, D \subseteq T$.

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

where $\RR^{-1} \sqbrk C$ is the preimage of $C$ under $\RR$.

Proof
We have that $\RR^{-1}$ is itself a relation, by definition of inverse relation.

The result follows directly from Image of Subset under Relation is Subset of Image.