Reflexive Reduction of Ordering is Strict Ordering/Proof 2

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Theorem

Let $\RR$ be an ordering on a set $S$.

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


Then $\RR^\ne$ is a strict ordering on $S$.


Proof

By definition, an ordering is both antisymmetric and transitive.

The result then follows from Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering.

$\blacksquare$