Order-Extension Principle/Proof 2

Theorem
Let $S$ be a set.

Let $\preceq$ be an ordering on $S$.

Then there exists a total ordering $\le$ on $S$ such that:
 * $\forall a, b \in S: \left({a \preceq b \implies a \le b}\right)$

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

By Reflexive Reduction of Ordering is Strict Ordering, $\prec$ is a strict ordering.

By the strict form of the Order-Extension Principle, there exists a strict total ordering $<$ on $S$ such that:


 * $\forall a, b \in S: \left({a \prec b \implies a < b}\right)$

Let $\le$ be the reflexive closure of $<$.

Let $a, b \in S$.

If $a \preceq b$, then by Law of Excluded Middle either $a \prec b$ or $a = b$.

If $a = b$, then by the definition of reflexive closure, $a \le b$.

If $a \prec b$, then by the choice of $<$, $a < b$ so $a \le b$.

Thus for all $a, b \in S$, $a \preceq b \implies a < b$.

By Reflexive Closure of Strict Total Ordering is Total Ordering, $\le$ is a total ordering.