Relation is Antisymmetric iff Intersection with Inverse is Coreflexive

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

Then:
 * $\mathcal R$ is antisymmetric


 * $\mathcal R \cap \mathcal R^{-1}$ is coreflexive
 * $\mathcal R \cap \mathcal R^{-1}$ is coreflexive

where $\mathcal R^{-1}$ is the inverse of $\mathcal R$.

Necessary Condition
Let $\mathcal R$ be an antisymmetric relation.

Let $\left({a, b}\right) \in \mathcal R \cap \mathcal R^{-1}$.

That means:
 * $\left({a, b}\right) \in \mathcal R$

and
 * $\left({a, b}\right) \in \mathcal R^{-1}$

which means, by definition of inverse relation:
 * $\left({b, a}\right) \in \mathcal R$

But as $\mathcal R$ is antisymmetric, that means $a = b$.

Thus:
 * $\left({a, b}\right) = \left({a, a}\right)$

and so:
 * $\left({a, b}\right) \in \Delta_S$

where $\Delta_S$ is the diagonal relation.

Thus from the definition of subset:
 * $\mathcal R \cap \mathcal R^{-1} \subseteq \Delta_S$

Hence, by definition, $\mathcal R$ is coreflexive.

Sufficient Condition
Let $\mathcal R \cap \mathcal R^{-1}$ be coreflexive.

Hence by definition of coreflexive:
 * $\mathcal R \cap \mathcal R^{-1} \subseteq \Delta_S$

Let $\left({a, b}\right) \in \mathcal R$ and $\left({b, a}\right) \in \mathcal R$.

That is, by definition of inverse relation:
 * $\left({a, b}\right) \in \mathcal R$

and
 * $\left({a, b}\right) \in \mathcal R^{-1}$

That is:
 * $\left({a, b}\right) \in \mathcal R \cap \mathcal R^{-1}$

But as $\mathcal R \cap \mathcal R^{-1} \subseteq \Delta_S$ it follows that $a = b$.

So by definition $\mathcal R$ is antisymmetric.