Preimage of Set Difference under Relation

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Theorem

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

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


Then:

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

where:

$\setminus$ denotes set difference
$\mathcal R^{-1} \sqbrk C$ denotes the preimage of $C$ under $\mathcal R$.


Proof

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

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

$\blacksquare$