Extension Theorem for Total Orderings

Theorem
Let the following conditions be fulfilled:


 * $(1):\quad$ Let $\left({S, \circ, \preceq}\right)$ be a totally ordered commutative semigroup
 * $(2):\quad$ Let all the elements of $\left({S, \circ, \preceq}\right)$ be cancellable
 * $(3):\quad$ Let $\left({T, \circ}\right)$ be an inverse completion of $\left({S, \circ}\right)$.

Then:


 * $(1):\quad$ The relation $\preceq'$ on $T$ satisfying $\forall x_1, x_2, y_1, y_2 \in S: x_1 \circ \left({y_1}\right)^{-1} \preceq' x_2 \circ \left({y_2}\right)^{-1} \iff x_1 \circ y_2 \preceq x_2 \circ y_1$ is a well-defined relation
 * $(2):\quad$ $\preceq'$ is the only total ordering on $T$ compatible with $\circ$
 * $(3):\quad$ $\preceq'$ is the only total ordering on $T$ that induces the given ordering $\preceq$ on $S$.

Proof
By Inverse Completion is Commutative Semigroup:
 * every element of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$.

Proof that Relation is Well-Defined
First we need to show that $\preceq'$ is well-defined.

So we need to show that if $x_1, x_2, y_1, y_2, z_1, z_2, w_1, w_2 \in S$ satisfy:

... then $x_2 \circ \left({w_2}\right)^{-1} = y_2 \circ \left({z_2}\right)^{-1}$.

We have:

and:

So:

Thus $\preceq'$ is a well-defined relation on $T$.

Proof that Relation is Transitive
To show that $\preceq'$ is transitive:

Thus $\preceq'$ is transitive.

Proof that Relation is Total Ordering
To show that $\preceq'$ is a total ordering on $T$ compatible with $\circ$:

Let $z_1, z_2 \in T$.

Then $\exists x_1, x_2, y_1, y_2 \in S: z_1 = x_1 \circ y_1^{-1}, z_2 = x_2 \circ y_2^{-1}$.

Then $z_1 \circ y_1 = x_1, z_2 \circ y_2 = x_2$.

Suppose that WLOG that $z_1 \circ y_1 \circ y_2 \preceq z_2 \circ y_2 \circ y_1$.

As $\preceq$ is a total ordering on $S$, it's either that or $z_2 \circ y_2 \circ y_1 \preceq z_1 \circ y_1 \circ y_2 $.

So:

and we see that $\preceq'$ is a total ordering on $T$.

Proof that Relation is Compatible with Operation
Let $x_1, x_2, y_1, y_2 \in T$ such that $x_1 \preceq' x_2, y_1 \preceq' y_2$.

We need to show that $x_1 \circ y_1 \preceq' x_2 \circ y_2$.

Let:
 * $x_1 = r_1 \circ s_1^{-1}, x_2 = r_2 \circ s_2^{-1}$
 * $y_1 = u_1 \circ v_1^{-1}, y_2 = u_2 \circ v_2^{-1}$

We have:

and

Because of the compatibility of $\preceq$ with $\circ$ on $S$, we have:

Thus compatibility is proved.

Proof about Restriction of Relation
To show that the restriction of $\preceq'$ to $S$ is $\preceq$:

Conversely:

Proof that Relation is Unique
To show that $\preceq'$ is unique:

Let $\preceq^*$ be any ordering on $T$ compatible with $\circ$ that induces $\preceq$ on $S$.

Then:

Then:

So:

Hence as every element of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$, the orderings $\preceq^*$ and $\preceq'$ are identical.