Power Function Preserves Ordering in Ordered Group

Theorem
Let $\left({S,\circ, \le}\right)$ be an ordered Group.

Let $x,y \in S$. Let $n \in \N_{>0}$ be a strictly positive integer. Let $<$ be the reflexive reduction of $\le$.

Then the following hold:
 * If $x \le y$ then $x^n \le y^n$
 * If $x < y$ then $x^n < y^n$

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

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

By the definition of an ordering, $\le$ is transitive.

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

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

By the definition of an ordered group, $\left({S,\circ}\right)$ is a group, and therefore a semigroup.

Thus the theorem holds by Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements.