Image of Set Difference under Relation

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

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

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

where:
 * $\setminus$ denotes set difference
 * $\RR \sqbrk A$ denotes image of $A$ under $\RR$.

Also see
Note that equality does not hold in general.

See Difference of Images under Mapping not necessarily equal to Image of Difference for an example of a mapping (which is of course a relation) for which it does not.