Inversion Mapping Reverses Ordering in Ordered Group

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

Let $x, y \in G$.

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

Then the following equivalences hold: $(\text{OG} 3)\quad x \le y \iff y^{-1} \le x^{-1}$

$(\text{OG}3')\quad x < y \iff y^{-1} < x^{-1}$

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

Thus by User:Dfeuer/CRG3, we obtain the first result:


 * $x \le y \iff y^{-1} \le x^{-1}$

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

Thus by again User:Dfeuer/CRG3, we obtain the second result:


 * $x < y \iff y^{-1} < x^{-1}$