One-to-Many Image of Set Difference/Corollary 2

Theorem
Let $\mathcal R \subseteq S \times T$ be a relation which is one-to-many.

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

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

where $\complement$ (in this context) denotes relative complement.

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 One-to-Many 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:
 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) = \mathcal R \left({\complement_S \left({A}\right)}\right)$

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

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

Hence the result from One-to-Many Image of Set Difference.