Ordering Compatible with Group Operation is Strongly Compatible

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

Let $x, y, z \in G$.

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

Then the following equivalences hold:

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

Thus by User:Dfeuer/CRG1, we obtain the first two results:


 * $(\operatorname{OG}1.1):\quad x \le y \iff x \circ z \le y \circ z$
 * $(\operatorname{OG}1.2):\quad x \le y \iff z \circ x \le z \circ y$

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

Thus by again User:Dfeuer/CRG1, we obtain the remaining results:


 * $(\operatorname{OG}1.1'):\quad x < y \iff x \circ z < y \circ z$
 * $(\operatorname{OG}1.2'):\quad x < y \iff z \circ x < z \circ y$

and so the theorem is established.