Ordering can be Expanded to compare Additional Pair

Theorem
Let $S$ be a set,

let $\le$ be an ordering on $S$,

let $a$ and $b$ be elements of $S$ such that $a \not\le b$ and $b \not\le a$,

let $\le' = \le \cup \left\{ {\left({a,b}\right)} \right\}$, and

let $\le'^-$ be the transitive closure of $\le'$.

Then $\le'^-$ is an ordering.

Proof
The reflexive property of $\le'^-$ follows trivially from the reflexive property of $\le$.

The transitive property of $\le'^-$ follows trivially from the definition of transitive closure.

$\le'^-$ is antisymmetric:

Suppose $x \le'^- y$ and $y \le'^- x$, and $y \ne x$.

By the definition of transitive closure: $x_0=x, x_1, \ldots, x_n = y$ and some $y_0=y, y_1, \ldots, y_m = x$


 * $x_0 \le' x_1 \le' \cdots \le' x_n$ and $y_0 \le' y_1 \le' \cdots \le' y_m$.
 * In the statement $x_0 \le' x_1 \le' \cdots \le' x_n$, only one of the relations can usefully be $a\le' b$, because any segment of the form $a \le' b \le' \cdots \le' a \le' b$ can be collapsed to $a \le' b$.
 * Thus the statement can be put in the form
 * $x_0 \le x_1 \le \cdots \le x_n$ or the form
 * $x_0 \le \cdots \le x_j \le a \le' b \le x_{j+3} \le \cdots \le x_n$.
 * In the first case, transitivity of $\le$ implies $x_0 \le x_n$, or $x \le y$.
 * In the second case, transitivity of $\le$ implies
 * $x_0 \le a \le' b \le x_n$, that is, $x \le a$ and $b \le y$.
 * Similarly, we see that the statement $y_0 \le' y_1 \le' \cdots \le' y_m$ implies that
 * $y \le x$ or both $y \le a$ and $b \le x$.
 * If $x \le y$ and $y \le x$, then $x=y$ by antisymmetry of $\le$, contradicting the assumption that $x \ne y$.
 * If, however, $x \le a$ and $b \le y$, we must have $x \le'^- a$ and $b \le'^- y$.
 * Since $y \le'^- x$ and $\le'^-$ is transitive, $b \le'^- x$, so $b \le'^- a$. But $a \le'^- b$, so $a=b$.
 * This contradicts the assumption that $a$ and $b$ are not $\le$-comparable.
 * The same contradiction occurs if we suppose that $y \le a$ and $b \le x$,
 * Thus assuming $\le'^-$ is not anti-symmetric leads to a contradiction, so
 * $\le'^-$ is anti-symmetric.