Well-Founded Proper Relational Structure Determines Minimal Elements

Theorem
Let $A$ and $B$ be classes.

Let $\left({ A, \prec }\right)$ be a proper relational structure.

Let $\prec$ be a foundational relation.

Suppose $B \subset A$ and $B \ne \varnothing$.

Then $B$ has a $\prec$-minimal element.

Proof
$B$ is not empty, so it has at least one element $x$.

By Singleton of Element is Subset:
 * $\left\{{ x }\right\} \subseteq B$

Since $\left\{ x \right\} \subseteq B$, it follows that $\left\{ x \right\}$ has a $\prec$-relational closure by Relational Closure Exists for Set-Like Relation. The $\prec$-relational closure shall be denoted $y$.


 * $y \cap B \subseteq A$ by Intersection is Subset


 * $y \cap B \ne \varnothing$ since $x \in y$ and $x \in B$

By the definition of foundational relation:


 * $(1): quad \exists x \in \left({ y \cap B }\right): \forall w \in \left({ y \cap B }\right): w \not \prec x$

Suppose that $w \in B$ and $w \prec x$.

Since $x \in y$, it follows that $w \in y$, so $w \in \left({ B \cap y }\right)$ by the definition of intersection. This contradicts $(1)$, so $w \not \prec x$ and $B$ has a minimal element.