Preimage of Set Difference under Relation

Theorem
Let $\mathcal R \subseteq S \times T$ be a relation. Let $C$ and $D$ be subsets of $T$.

Then:
 * $\mathcal R^{-1} \left({C}\right) \setminus \mathcal R^{-1} \left({D}\right) \subseteq \mathcal R^{-1} \left({C \setminus D}\right)$

where $\setminus$ denotes set difference.

Proof
This follows from Image of Set Difference, and the fact that $\mathcal R^{-1}$ is itself a relation, and therefore obeys the same rules.