Ordering can be Expanded to compare Additional Pair

Theorem
Let $S$ be a set,

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

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

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

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

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

Proof 1
Define a relation $\le$ by letting $p \le q$ iff:
 * $p \preceq q$ or
 * $p \preceq a$ and $b \preceq q$

$\le$ is reflexive
For any $p \in S$, $p \preceq p$ because $\preceq$ is reflexive.

Thus $p \le p$.

Since this holds for all $p \in S$, $\le$ is reflexive.

$\le$ is transitive
Suppose that $p \le q$ and $q \le r$.

Then there are three possibilities:

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

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

Thus $p \le r$

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

Because $\preceq$ is transitive, $p \preceq a$.

Since $p \preceq a$ and $b \preceq r$, $p \le r$.

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

Because $\preceq$ is transitive, $b \preceq r$.

Since $p \preceq a$ and $b \preceq r$, $p \le r$.

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 $b \preceq a$, contradicting the premise.

Since this holds for all $p$, $q$, and $r$, $\le$ is transitive.

$\le$ is antisymmetric
Suppose that $p \le q$ and $q \le p$.

There are three possibilities to consider:

Proof 2
The reflexive property of $\preceq'^-$ follows trivially from the reflexive property of $\preceq$.

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

$\preceq'^-$ is antisymmetric:

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

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

$x_0 \preceq' x_1 \preceq' \cdots \preceq' x_n$ and $y_0 \preceq' y_1 \preceq' \cdots \preceq' y_m$.

In the statement $x_0 \preceq' x_1 \preceq' \cdots \preceq' x_n$, only one of the relations can usefully be $a\preceq' b$, because any segment of the form $a \preceq' b \preceq' \cdots \preceq' a \preceq' b$ can be collapsed to $a \preceq' b$.

Thus the statement can be put in the form
 * $x_0 \preceq x_1 \preceq \cdots \preceq x_n$ or the form
 * $x_0 \preceq \cdots \preceq x_j \preceq a \preceq' b \preceq x_{j+3} \preceq \cdots \preceq x_n$.

In the first case, transitivity of $\preceq$ implies $x_0 \preceq x_n$, or $x \preceq y$.

In the second case, transitivity of $\preceq$ implies
 * $x_0 \preceq a \preceq' b \preceq x_n$, that is, $x \preceq a$ and $b \preceq y$.

Similarly, we see that the statement $y_0 \preceq' y_1 \preceq' \cdots \preceq' y_m$ implies that
 * $y \preceq x$ or both $y \preceq a$ and $b \preceq x$.

If $x \preceq y$ and $y \preceq x$, then $x=y$ by antisymmetry of $\preceq$, contradicting the assumption that $x \ne y$.

If, however, $x \preceq a$ and $b \preceq y$, we must have $x \preceq'^- a$ and $b \preceq'^- y$.

Since $y \preceq'^- x$ and $\preceq'^-$ is transitive, $b \preceq'^- x$, so $b \preceq'^- a$. But $a \preceq'^- b$, so $a=b$.

This contradicts the assumption that $a$ and $b$ are not $\preceq$-comparable.

The same contradiction occurs if we suppose that $y \preceq a$ and $b \preceq x$,

Thus assuming $\preceq'^-$ is not anti-symmetric leads to a contradiction, so
 * $\preceq'^-$ is anti-symmetric.