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.