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 $\{ p \} \ne \varnothing$ and $\{ p \} \subseteq A$.

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

Since $x \in \{ p \}$, $x = p$.

Then $p \not\mathrel{\mathcal R} p$.

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