Relation is Antisymmetric iff Intersection with Inverse is Coreflexive

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

Then $\mathcal R$ is antisymmetric iff:
 * $\mathcal R \cap \mathcal R^{-1} \subseteq \Delta_S$

where:
 * $\mathcal R^{-1}$ is the inverse of $\mathcal R$
 * $\Delta_S$ is the diagonal relation on $S$.

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

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

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

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

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

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

Thus $\left({a, b}\right) = \left({a, a}\right)$ and so $\left({a, b}\right) \in \Delta_S$.

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

Sufficient Condition
Let $\mathcal R \cap \mathcal R^{-1} \subseteq \Delta_S$.

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

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

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

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

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

So by definition $\mathcal R$ is antisymmetric.