Strictly Well-Founded Relation determines Strictly Minimal Elements/Lemma

Lemma
Let $A$ be a nonempty class.

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

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

Proof
For each $x \in A$, let $\prec^{-1} \left({x}\right)$ denote the preimage under $\prec$ 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 $R \left({\varnothing}\right) = \varnothing$.

That is:

For a given class $C$, let $\alpha_C$ be the smallest ordinal such that $C \cap V \left({\alpha_C}\right) ≠ \varnothing$,

where $V$ is the von Neumann hierarchy.

Then let $R \left({C}\right)$ denote the set $C \cap V \left({\alpha_C}\right)$.

Let $F$ be a function defined recursively:


 * $F \left({0}\right) = R \left({A}\right)$
 * $\displaystyle F \left({n+1}\right) = \bigcup_{y \mathop \in F \left({n}\right)} R \left({\prec^{-1} \left({y}\right)}\right)$

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

Proof
Proceed by induction:

$R \left({A}\right)$ 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 $y \in F \left({n}\right)$, $R \left({\prec^{-1} \left({y}\right)}\right)$ is a set, so by the Axiom of Union, $F \left({n+1}\right)$ is a set.

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

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

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

By Non-Empty Class has Element of Least Rank, $F \left({0}\right) \ne \varnothing$, so $a \ne \varnothing$.

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

Then, by Characterization of Minimal Element,
 * $\forall x \in A: \prec^{-1} \left({x}\right) \ne \varnothing$

Since $a \subseteq A$:
 * $\forall x \in a: \prec^{-1} \left({x}\right) \ne \varnothing$

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

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

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

Also see
and weaker results that don't require the Axiom of Foundation.
 * Well-Founded Proper Relational Structure Determines Minimal Elements‎
 * Proper Well-Ordering Determines Minimal Elements,