Reflexive Reduction of Well-Founded Relation is Strictly Well-Founded Relation

Theorem
Let $\struct {S, \RR}$ be a relational structure.

Let $\RR$ be a well-founded relation on $S$.

Let $\RR^{\ne}$ be the reflexive reduction of $\RR$.

Then $\RR^{\ne}$ is a strictly well-founded relation.

Proof
Let $T$ be a non-empty subset of $S$.

Since $\RR$ is a well-founded relation, $T$ has a minimal element with respect to the relation $\RR$.

That is, there is an element $m \in T$ such that $\forall x \in T: \paren {\tuple {x, m} \notin \RR} \lor \paren {x = m}$.

Let $x \in T$.

Then $\tuple {x, m} \notin \RR$ or $x = m$.

By the definition of reflexive reduction, $\RR^{\ne}$ is a subset of $\RR$.

Thus:
 * if $\tuple {x, m} \notin \RR$, it follows that $\tuple {x, m} \notin \RR^{\ne}$.


 * if $x = m$ then by Reflexive Reduction is Antireflexive, $\tuple {x, m} \notin \RR^{\ne}$.

As $x$ is arbitrary, this holds for all $x \in T$.

Thus $m$ is a strictly minimal element under $\RR^{\ne}$ in $T$.

As $T$ is arbitrary, this condition holds for all non-empty subsets of $S$.

That is, every non-empty subset of $S$ has a strictly minimal element under $\RR^{\ne}$.

Hence, by definition, $\RR^{\ne}$ is a strictly well-founded relation on $S$.