Relation is Symmetric and Antisymmetric iff Coreflexive

Theorem
Let $S$ be a set.

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

Then:
 * $\mathcal R$ is both symmetric and antisymmetric

iff:
 * $\mathcal R \subseteq \Delta_S$

where $\Delta_S$ is the diagonal relation.

Necessary Condition
Let $\mathcal R$ be both symmetric and antisymmetric

Suppose $\mathcal R \not \subseteq \Delta_S$.

Then:
 * $\exists \left({x, y}\right) \in \mathcal R: x \ne y$

But then as $\mathcal R$ is symmetric:
 * $\left({y, x}\right) \in \mathcal R$

So we have:
 * $\left({x, y}\right) \in \mathcal R$

and:
 * $\left({y, x}\right) \in \mathcal R$

where $x \ne y$.

Thus $\mathcal R$ is not antisymmetric, contrary to hypothesis.

So our assumption that $\mathcal R \not \subseteq \Delta_S$ is false.

That is:
 * $\mathcal R \subseteq \Delta_S$

Sufficient Condition
Let $\mathcal R \subseteq \Delta_S$.

Let $\left({x, y}\right) \in \mathcal R$.

and so by definition $\mathcal R$ is symmetric.

Let $\left({x, y}\right) \in \mathcal R$ and $\left({y, x}\right) \in \mathcal R$.

From the above, this can happen only when $x = y$.

That is, by definition, $\mathcal R$ is antisymmetric.

So $\mathcal R$ is both symmetric and antisymmetric.

Hence the result.