Extension Theorem for Total Orderings

Theorem
Let the following conditions be fulfilled:


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

Then:


 * 1) 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) $$\preceq'$$ is the only total ordering on $$T$$ compatible with $\circ$;
 * 3) $$\preceq'$$ is the only total ordering on $$T$$ that induces the given ordering $$\preceq$$ on $$S$$.

Proof
By Inverse Completion 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.