Image of Set Difference under Relation/Corollary 2

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

Let $A$ and $B$ be subsets of $S$.

Then:


 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) \subseteq \mathcal R \left({\complement_S \left({A}\right)}\right)$

where $\operatorname{Im} \left({\mathcal R}\right)$ denotes the image of $\mathcal R$.

Proof
By definition of the image of $\mathcal R$:
 * $\operatorname{Im} \left({\mathcal R}\right) = \mathcal R \left({S}\right)$

So, when $B = S$ in Image of Set Difference: Corollary 1:
 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) = \complement_{\mathcal R \left({S}\right)} \left({\mathcal R \left({A}\right)}\right)$

Hence:
 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) \subseteq \mathcal R \left({\complement_S \left({A}\right)}\right)$

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

that is:
 * $\mathcal R \left({S}\right) \setminus \mathcal R \left({A}\right) \subseteq \mathcal R \left({S \setminus A}\right)$