Preimage of Set Difference under Relation

Theorem

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

Let $C$ and $D$ be subsets of $T$.

Then:

$\RR^{-1} \sqbrk C \setminus \RR^{-1} \sqbrk D \subseteq \RR^{-1} \sqbrk {C \setminus D}$

where:

$\setminus$ denotes set difference

$\RR^{-1} \sqbrk C$ denotes the preimage of $C$ under $\RR$.

Proof

We have that $\RR^{-1}$ is itself a relation

The result then follows from Image of Set Difference under Relation.

$\blacksquare$