Strict Ordering can be Expanded to Compare Additional Pair/Proof 1

Theorem
Let $(S, \prec)$ be a strictly ordered set.

Let $a$ and $b$ be distinct, $\prec$-incomparable elements of $S$. That is, let:
 * $a \not\prec b$ and $b \not\prec a$.

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

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

where $\preceq$ is the reflexive closure of $\prec$.

Then:


 * $\prec'^+$ is a strict ordering.
 * $\prec^+$ is the transitive closure of $\prec'$.

Proof
First, note that since $\prec$ is a strict ordering, $\preceq$ is an ordering by Reflexive Closure of Strict Ordering is Ordering.

$a$ and $b$ are $\preceq$-incomparable
Suppose that $a \preceq b$.

By the definition of reflexive closure, either $a \prec b$ or $a = b$.

Each possibility contradicts one of the premises, so $a \not\preceq b$.

For the same reasons, $b \not\preceq a$.

$\prec'^+$ is antireflexive
Let $p \in S$.

Then by the definition of strict ordering:

$p \not\prec p$.

Suppose that $p \preceq a$ and $b \preceq p$.

Then since $\preceq$ is transitive, $b \preceq a$, contradicting the fact that $a$ and $b$ are $\preceq$-incomparable.

Since neither $p \prec p$ nor $(p \preceq a) \land (b \preceq p)$ holds:
 * $p \not \prec'^+ p$.

Since this is the case for all $p \in S$, $\prec'^+$ is antireflexive.

$\prec'^+$ is transitive
Suppose that $p \prec'^+ q$ and $q \prec'^+ r$.

Then there are three possibilities:

$(1)\quad p \prec q$ and $p \prec r$

Because $\prec$ is transitive, $p \prec r$.

Thus $p \prec'^+ r$

$(2)\quad p \prec q$, $q \preceq a$, and $b \preceq r$

By Extended Transitivity, $p \prec a$.

Since $p \prec a$ and $b \prec r$, $p \prec'^+ r$.

$(3)\quad p \preceq a$, $b \preceq q$, and $q \prec r$

By Extended Transitivity, $b \prec r$, so $b \preceq r$.

Since $p \preceq a$ and $b \preceq r$, $p \prec'^+ r$.

Note that it is impossible to have $p \preceq a$, $b \preceq q$, $q \preceq a$ and $b \preceq r$:

If that were so, $b \preceq q$ and $q \preceq a$ together would imply by transitivity that $b \preceq a$.

But this contradicts the fact that $a$ and $b$ are $\preceq$-incomparable.

Thus in all cases, $p \prec'^+ q$ and $q \prec'^+ r$ imply $p \prec'^+ r$, so $\prec'^+$ is transitive.

Since $\prec'^+$ is transitive and antireflexive, it is by definition a strict ordering.

$\prec'^+$ is the transitive closure of $\prec'$
First note that $\prec'$ is a subset of $\prec'^+$:

If $p \prec' q$ then either
 * $p \prec q$ or
 * $p = a$ and $q = b$

If $p \prec q$ then $p \prec'^+ q$ by definition.

If $p = a$ and $q = b$, then $p \preceq a$ and $q \preceq b$ by the definition of $\preceq$.

Thus $p \prec'^+ q$ by the definition of $\preceq'^+$.

Let $\mathcal R$ be a transitive relation containing $\prec'$.

Let $p, q \in S$ and let $p \prec'^+ q$.

Then either:
 * $p \prec q$ or
 * $p \preceq a$ and $b \preceq q$

If $p \prec q$, then by the definition of subset, $p \mathrel{\mathcal R} q$.

Suppose instead that $p \preceq a$ and $b \preceq q$. By the definition of $\preceq'$, $a \preceq' b$.

Since Reflexive Closure is Closure Operator, it is increasing, so $\preceq \subseteq \mathcal R^=$.

Thus $p \mathrel{\mathcal R}^= a$, $a \mathrel{\mathcal R}^= b$, and $b \mathrel{\mathcal R}^= q$.

Since $\mathcal R$ is transitive, $\mathcal R^=$ is as well, by Reflexive Closure of Transitive Relation is Transitive.

Thus $p \mathrel{\mathcal R}^= q$.

Since $\preceq'^+$ is antireflexive and we assumed $p \preceq'^+ q$, we conclude that $p \ne q$.

Thus by the definition of reflexive closure:
 * $p \mathrel{\mathcal R} q$

Thus we have shown that $\preceq'^+$ is a subset of $\mathcal R$.

Since this holds when $\mathcal R$ is any transitive superset of $\preceq'$, $\preceq'^+$ is the transitive closure of $\preceq'$.