Power Function Preserves Ordering in Ordered Semigroup/Proof 2

Proof
The proof proceeds by induction.

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

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

which is trivially true.

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

Basis for the Induction
We have:

and:

Hence as $\preceq$ is an ordering and hence transitive:
 * $x \preceq y \implies x \circ x \preceq y \circ y$

That is:


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

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:
 * $x \preceq y \implies x^k \preceq y^k$

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

Induction Step
This is the induction step:

and:

Hence as $\preceq$ is an ordering and hence transitive:
 * $x \preceq y \implies x \circ x^k \preceq y \circ y^k$

That is:


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

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

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