Image of Set Difference under Relation

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

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

Then:
 * $\mathcal R \sqbrk A \setminus \mathcal R \sqbrk B \subseteq \mathcal R \sqbrk {A \setminus B}$

where:
 * $\setminus$ denotes set difference
 * $\mathcal R \left[{A}\right]$ denotes image of $A$ under $\mathcal R$.

Also see
Note that equality does not hold in general.

See the note on Image of Set Difference under Mapping for an example of a mapping (which is of course a relation) for which it does not.