Power Function Preserves Ordering in Ordered Group/Corollary/Proof 2

Theorem
Let $\left({G, \circ, \le}\right)$ be an ordered group with identity $e$.

Let $<$ be the reflexive reduction of $\le$.

Let $x \in G$.

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

Then the following hold:


 * $x \le e \implies x^n \le e$
 * $e \le x \implies e \le x^n$
 * $x < e \implies x^n < e$
 * $e < x \implies e < x^n$

Proof
By the definition of an ordered group, $\le$ is a transitive relation compatible with $\circ$.

By Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements:
 * $x \le e \implies x^n \le e^n$
 * $e \le x \implies e^n \le x^n$

By Identities are Idempotent, $e$ is idempotent with respect to $\circ$.

Thus we obtain the first two results:
 * $x \le e \implies x^n \le e$
 * $e \le x \implies e \le x^n$

By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $<$ is compatible with $\circ$.

By Reflexive Reduction of Transitive Antisymmetric Relation is Transitive, $<$ is transitive.

Thus by the same method as above, we obtain the remaining results:


 * $x < e \implies x^n < e$
 * $e < x \implies e < x^n$