Ordering can be Expanded to compare Additional Pair/Proof 2

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

Let $\prec' = {\prec} \cup \left\{ {\left({a, b}\right)} \right\}$.

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

Define a relation $\prec'_2$ by letting $p \prec'_2 q$ :
 * $p \prec q$ or
 * $p \preceq a$ and $b \preceq q$

By Strict Ordering can be Expanded to Compare Additional Pair, $\prec'_2$ is a strict ordering and is the transitive closure of $\prec'$.

Then the reflexive closure of $\prec'_2$, ${\prec'_2}^=$ is the transitive reflexive closure of $\prec'$.

From Equivalence of Definitions of Reflexive Transitive Closure, ${\prec'_2}^=$ is the transitive closure of the reflexive closure of $\prec'$.

The reflexive closure of $\prec'$ is $\preceq'$, so ${\prec'_2}^=$ is the transitive closure of $\preceq'$.

But ${\prec'_2}^=$ is clearly the relation defined by the two conditions in the theorem statement.