Strictly Well-Founded Relation determines Strictly Minimal Elements

Theorem
Let $A$ be a class.

Let $\prec$ be a foundational relation on $A$.

Let $B$ be a nonempty class and let $B \subseteq A$.

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

Proof
Let $F$ be a function defined recursively:


 * $F\left({0}\right) = \left\{ x \in B : \forall y \in B: \operatorname{rank} \left({ x }\right) \le \operatorname{rank} \left({ y }\right) \right\}$


 * $F\left({n+1}\right) = \left\{ x \in B : \exists y \in F\left({n}\right): \left({ x \prec y \land \forall z \in F\left({n}\right): \left({ z \prec y \implies \operatorname{rank} \left({ x }\right) \le \operatorname{rank} \left({ z }\right) }\right) }\right) \right\}$

$F\left({0}\right)$ consists of the elements of $B$ of minimal rank.

$F\left({n+1}\right)$ consists of the elements of $B$ that strictly precede the elements of $F\left({n}\right)$ and are of minimal rank.

Since $F\left({n}\right)$ are all of bounded rank, it follows that $F\left({n}\right)$ are sets for all $n \in \mathbb N$.

Set $\displaystyle b = \bigcup_{n \in \omega} F\left({n}\right)$.

It follows that $b \subseteq B$ and $b \ne \varnothing$.

Suppose $B$ has no $\prec$-minimal element.

Then, by de Morgan’s Laws, $\forall x \in B: \exists y \in B: y \prec x$.

As a special case, $\forall x \in b: \exists y \in B: y \prec x$ by the fact that $b \subseteq B$.

Let $\prec^{-1} \left({ x }\right)$ denote the $\prec$-initial segment of $x$.


 * $B \cap \prec^{-1} \left({ x }\right) \ne \varnothing$, so it follows that it has an element $y$ of minimal rank.

Therefore, $b \cap \prec^{-1} \left({x}\right) \ne \varnothing$, contradicting the fact that $\prec$ is foundational.