Power Function with Cancellable Element Preserves Strict Ordering in Ordered Semigroup

Theorem
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.

Let $x, y \in S$ be such that:
 * $(1): \quad x \prec y$
 * $(2): \quad$ either $X$ or $y$ (or both) is cancellable for $\circ$.

Let $n \in \N_{>0}$ be a strictly positive integer.

Then:
 * $x^n \prec y^n$

where $x^n$ is the $n$th power of $x$.

Proof
, suppose $x$ is cancellable for $\circ$.

The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
 * $x \prec y \implies x^n \prec y^n$

$\map P 1$ is the case:
 * $x \prec y \implies x \prec y$

which is trivially true.

Thus $\map P 1$ is seen to hold.

Basis for the Induction
We have:

and:

Hence:
 * $x \prec y \implies x \circ x \prec y \circ y$

That is:


 * $x \prec y \implies x^2 \prec y^2$

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $x \preceq y \implies x^k \prec y^k$

from which it is to be shown that:
 * $x \preceq y \implies x^{k + 1} \prec y^{k + 1}$

Induction Step
This is the induction step:

and:

Hence:
 * $x \prec y \implies x \circ x^k \prec y \circ y^k$

That is:


 * $x \prec y \implies x^{k + 1} \prec y^{k + 1}$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: x \prec y \implies x^n \prec y^n$