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.