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 $B$ has at least one element $x$.

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

By Relational Closure Exists for Set-Like Relation:
 * $\left\{ x \right\}$ has a $\prec$-relational closure.

This $\prec$-relational closure shall be denoted $y$.

By Intersection is Subset:
 * $y \cap B \subseteq A$

As $x \in y$ and $x \in B$:
 * $y \cap B \ne \varnothing$

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 $(1)$.

Therefore:
 * $w \not \prec x$

and so $B$ has a $\prec$-minimal element.