Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering

Theorem
Let $S$ be a set.

Let $\preceq$ be a transitive, antisymmetric relation on $S$.

Let $\prec$ be the reflexive reduction of $\preceq$.

Then $\prec$ is a strict ordering.

Proof
We will show that $\prec$ is antireflexive and transitive, meaning that it is a strict ordering.

Antireflexive
Follows from Reflexive Reduction is Antireflexive.

Transitive
Let $a,b,c \in S$.

Suppose that $a \prec b$ and $b \prec c$.

By the definition of reflexive reduction,


 * $a \preceq b \qquad a \ne b$
 * $b \preceq c \qquad b \ne c$.

Since $\preceq$ is transitive,


 * $a \preceq c$.

Suppose, with a view to deriving a contradiction, that $a = c$.

Then since $a \preceq b$ we must have $c \preceq b$.

Since $c \preceq b$, $b \preceq c$, and $\preceq$ is antisymmetric, $b = c$.

But this contradicts the fact that $b \ne c$.

We therefore conclude that $a ≠ c$.

Since we already know that $a \preceq c$, by the definition of $\prec$, $a \prec c$.