Relation is Antisymmetric and Reflexive iff Intersection with Inverse equals Diagonal Relation

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

Then $\mathcal R$ is both antisymmetric and reflexive iff:
 * $\mathcal R \cap \mathcal R^{-1} = \Delta_S$

where $\Delta_S$ denotes the diagonal relation.

Necessary Condition
Let $\mathcal R$ be both antisymmetric and reflexive.

From Relation is Antisymmetric iff Intersection with Inverse is Coreflexive:
 * $\mathcal R \cap \mathcal R^{-1} \subseteq \Delta_S$

By definition of reflexive relation:
 * $\Delta_S \subseteq \mathcal R$

By Inverse of Reflexive Relation is Reflexive:
 * $\Delta_S \subseteq \mathcal R^{-1}$

By Intersection is Largest Subset:
 * $\Delta_S \subseteq \mathcal R \cap \mathcal R^{-1}$

Thus by definition of set equality:
 * $\mathcal R \cap \mathcal R^{-1} = \Delta_S$

Sufficient Condition
Let $\mathcal R$ be such that:
 * $\mathcal R \cap \mathcal R^{-1} = \Delta_S$

Then by definition of set equality:
 * $\mathcal R \cap \mathcal R^{-1} \subseteq \Delta_S$

and so by Relation is Antisymmetric iff Intersection with Inverse is Coreflexive, $\mathcal R$ is antisymmetric.

Also by definition of set equality:
 * $\Delta_S \subseteq \mathcal R \cap \mathcal R^{-1}$

By Intersection is Subset:
 * $\mathcal R \cap \mathcal R^{-1} \subseteq \mathcal R$

By Subset Relation is Transitive:
 * $\Delta_S \subseteq \mathcal R$

So, by definition, $\mathcal R$ is reflexive. }