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 suppose that $B \subseteq A$.

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

Proof
For each $x \in A$, let $\prec^{-1} \left({ x }\right)$ denote the $\prec$-initial segment of $x$ in $A$.

Let $F$ be a function defined recursively:


 * $F \left({0}\right) = \left\{ x \in B : \operatorname{rank} \left({ x }\right) = \operatorname{minrank} \left({ B }\right) \right\}$
 * $\displaystyle F \left({n+1}\right) =

\bigcup_{y \mathop\in F\left({n}\right)} \left\{{x \in B: x \prec y \text{ and }\operatorname{rank} \left({ x }\right) \le \operatorname{minrank} \left({ F\left({n}\right) \cap \prec^{-1} \left({ y }\right)}\right) }\right\}$,
 * where for any non-empty class $C$, $\operatorname{minrank}\left({C}\right)$ is the rank of an element of $C$ of least rank and $\operatorname{minrank}\left({\varnothing}\right)$ is taken to be $0$.

Lemma
$F\left({n}\right)$ is of bounded rank for each $n \in \omega$.

Proof
The rank of elements of $F\left({0}\right)$ is bounded above by $\operatorname{minrank} \left({ B }\right)$.

If the rank of elements of $F\left({n}\right)$ is bounded above by $m$, then

$\operatorname{minrank} \left({ F\left({n}\right) \cap \prec^{-1} \left({ y }\right)}\right) \le m$, so the ranks of elements of $F\left({n+1}\right)$ is bounded above by $m$.

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

By the Axiom of Union, $b$ is a set.

Since $F\left({n}\right)\subseteq B$ for each $n \in \omega$, $b \subseteq B$.

By Non-Empty Class has Element of Least Rank, $F(0) \ne \varnothing$, so $b \ne \varnothing$.

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

Then, by Characterization of Minimal Element,


 * $\forall x \in B: B \cap \prec^{-1} \left({ x }\right) \ne \varnothing$.

Since $b \subseteq B$,
 * $\forall x \in b: B \cap \prec^{-1} \left({ x }\right) \ne \varnothing$.

Let $x$ be any element of $b$.

By Non-Empty Class has Element of Least Rank, $B \cap \prec^{-1} \left({ x }\right)$ has an element $w$ of least rank.

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