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 (strict) $\prec$-initial segment of $x$ in $A$.

For each class $C$, let $R\left({C}\right)$ denote the set of elements of $C$ of minimal rank, and let $R\left({\varnothing}\right) = \varnothing$. Note that this is a set because each level of the Von Neumann Hierarchy is a set.

Let $F$ be a function defined recursively:


 * $F \left({0}\right) = R\left({B}\right)$
 * $\displaystyle F \left({n+1}\right) =

\bigcup_{y \mathop\in F\left({n}\right)} R\left({B \cap \prec^{-1}\left({y}\right)}\right)$,

Lemma
$F \left({n}\right)$ is a set for each $n \in \omega$.

Proof
Proceed by induction:

$R\left({B}\right)$ is a set, so $F \left({0}\right)$ is a set.

Suppose that $R\left({F \left({n}\right)}\right)$ is a set.

We know that for each $y \in F\left({n}\right)$, $R\left({B \cap \prec^{-1}\left({y}\right)}\right)$ is a set, so by the Axiom of Union, $F\left({n+1}\right)$ is a set.

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

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.