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


 * $A \subseteq S \implies A \subseteq \left({\mathcal R^{-1} \circ \mathcal R}\right) \left({A}\right)$
 * $B \subseteq T \implies \left({\mathcal R \circ \mathcal R^{-1}}\right) \left({B}\right) \subseteq B$

Proof
Suppose $A \subseteq S$.

We have:

So by definition of composition of relations:
 * $A \subseteq S \implies A \subseteq \left({\mathcal R^{-1} \circ \mathcal R}\right) \left({A}\right)$

Let $B \subseteq T$.

Then:

So:
 * $B \subseteq T \implies \left({\mathcal R \circ \mathcal R^{-1}}\right) \left({B}\right) \subseteq B$