Ordering can be Expanded to compare Additional Pair/Proof 2

Theorem
Let $(S, \preceq)$ be an ordered set.

Let $a$ and $b$ be non-comparable elements of $S$. That is, let:
 * $a \not\preceq b$ and $b \not\preceq a$.

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

Let $\preceq'^+$ be the transitive closure of $\preceq'$.

Then:

$\preceq'^+$ is an ordering.

$\preceq'^+$ can be defined by letting $p \preceq'^+ q$ iff:
 * $p \preceq q$ or
 * $p \preceq a$ and $b \preceq q$.

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 $p \prec'_2 q$ iff:
 * $p \prec q$ or
 * $p \preceq a$ and $b \preceq q$

By Strict Ordering can be Expanded to Compare Additional Pair/Proof 1, $\prec'_2$ is a strict ordering and is the transitive closure of $\prec'$.

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