Image of Set Difference under Relation/Corollary 2

Corollary to Image of Set Difference under Relation
Let $\mathcal R \subseteq S \times T$ be a relation.

Let $A$ be a subset of $S$.

Then:


 * $\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk A} \subseteq \mathcal R \sqbrk {\relcomp S A}$

where:
 * $\Img {\mathcal R}$ denotes the image of $\mathcal R$
 * $\mathcal R \sqbrk A$ denotes the image of $A$ under $\mathcal R$.

Proof
By definition of the image of $\mathcal R$:
 * $\Img {\mathcal R} = \mathcal R \sqbrk S$

So, when $B = S$ in Image of Set Difference under Relation: Corollary 1:
 * $\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk A} = \relcomp {\mathcal R \sqbrk S} {\mathcal R \sqbrk A}$

Hence:
 * $\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk A} \subseteq \mathcal R \sqbrk {\relcomp S A}$

means exactly the same thing as:
 * $\relcomp {\mathcal R \sqbrk S} {\mathcal R \sqbrk A} \subseteq \mathcal R \sqbrk {\relcomp S A}$

that is:
 * $\mathcal R \sqbrk S \setminus \mathcal R \sqbrk A \subseteq \mathcal R \sqbrk {S \setminus A}$