Strictly Well-Founded Relation determines Strictly Minimal Elements/Lemma

Theorem
Let $A$ be a non-empty class.

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

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

Proof
The general strategy of the proof is as follows:

We will recursively define a certain subset, $a$, of $A$.

We will use the fact that $\prec$ is a foundational relation to choose a minimal element, $m$, of $a$.

Then we will prove that $m$ is in fact a minimal element of $A$.

For each $x \in A$, let $\map {\prec^{-1} } x$ denote the preimage under $\prec$ of $x$ in $A$.

For each class $C$, let $\map R C$ denote the set of elements of $C$ of minimal rank, and let $\map R \O = \O$.

That is:

For a given class $C$, let $\alpha_C$ be the smallest ordinal such that:
 * $C \cap \map V {\alpha_C} \ne \O$

where $V$ is the von Neumann hierarchy.

Then let $\map R C$ denote the set $C \cap \map V {\alpha_C}$.

Let $F$ be a function defined recursively:


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

Lemma
$\map F n$ is a set for each $n \in \omega$.

Proof
Proceed by induction:

$\map R A$ is a set, so $\map F 0$ is a set.

Suppose that $\map R {\map F n}$ is a set.

We know that for each $y \in \map F n$, $\map R {\map {\prec^{-1} } y}$ is a set, so by the Axiom of Unions, $\map F {n + 1}$ is a set.

Let $\displaystyle a = \bigcup_{n \mathop \in \omega} \map F n$.

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

Since $\map F n \subseteq A$ for each $n \in \omega$, $a \subseteq A$.

By Non-Empty Class has Element of Least Rank, $\map F 0 \ne \O$, so $a \ne \O$.

Since $\prec$ is foundational, $a$ has a $\prec$-minimal element $m$.

that $m$ is not a $\prec$-minimal element of $A$.

Then, by Characterization of Minimal Element,
 * $\map {\prec^{-1} } m \ne \O$

By Non-Empty Class has Element of Least Rank, $\map {\prec^{-1} } m$ has an element $w$ of least rank.

Therefore, $a \cap \map {\prec^{-1} } m \ne \O$, contradicting the fact that $m$ is $\prec$-minimal in $a$.

Thus we conclude that $m$ is a $\prec$-minimal element of $A$.

Also see

 * Well-Founded Proper Relational Structure Determines Minimal Elements‎
 * Proper Well-Ordering Determines Smallest Elements

These are weaker results that do not require the Axiom of Foundation.