# 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:

$\mathcal R \sqbrk S \setminus \mathcal R \sqbrk A \subseteq \mathcal R \sqbrk {S \setminus A}$

$\blacksquare$