Image of Set Difference under Relation/Corollary 1

Theorem
Let $\mathcal R \subseteq S \times T$ be a relation.

Let $A \subseteq B \subseteq S$.

Then:
 * $\complement_{\mathcal R \left({B}\right)} \left({\mathcal R \left({A}\right)}\right) \subseteq \mathcal R \left({\complement_{B} \left({A}\right)}\right)$

where:
 * $\mathcal R \left({B}\right)$ denotes the image of $B$ under $\mathcal R$
 * $\complement$ (in this context) denotes relative complement.

Proof
We have that $A \subseteq B$.

Then by definition of relative complement:
 * $\complement_B \left({A}\right) = B \setminus A$
 * $\complement_{\mathcal R \left({B}\right)} \left({\mathcal R \left({A}\right)}\right) = \mathcal R \left({B}\right) \setminus \mathcal R \left({A}\right)$

Hence, when $A \subseteq B$:
 * $\complement_{\mathcal R \left({B}\right)} \left({\mathcal R \left({A}\right)}\right) \subseteq \mathcal R \left({\complement_B \left({A}\right)}\right)$

means exactly the same thing as:
 * $\mathcal R \left({B}\right) \setminus \mathcal R \left({A}\right) \subseteq \mathcal R \left({B \setminus A}\right)$