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 $$x, y \in S$$.


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

So:

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


 * To show that $$\preceq'$$ is transitive:

Thus $$\preceq'$$$ is transitive.


 * To show that $$\preceq'$$ is a total ordering on $$T$$ compatible with $\circ$:


 * To show that the restriction of $$\preceq'$$ to $$S$$ is $$\preceq$$:

Conversely:


 * To show that $$\preceq'$$ is unique:

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

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.