Strictly Well-Founded Relation is Antireflexive

Theorem
Let $\mathcal R$ be a foundational relation on a set or class $A$.

Then $\mathcal R$ is antireflexive.

Proof
Let $p \in A$.

Then $\left\{{p}\right\} \ne \varnothing$ and $\left\{{p}\right\} \subseteq A$.

Thus, by the definition of foundational relation:
 * $\exists x \in \left\{{p}\right\}: \forall y \in \left\{{p}\right\}: \neg \left({y \mathrel{\mathcal R} x}\right)$

Since $x \in \left\{{p}\right\}$, it must be that $x = p$.

It follows that $p \not\mathrel{\mathcal R} p$.

Since this holds for all $p \in A$, $\mathcal R$ is antireflexive.