Ordering Compatible with Group Operation is Strongly Compatible

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

Let $x, y, z \in \left({G, \circ, \le}\right)$.

Then the following equivalences hold:


 * $x \le y \iff x \circ z \le y \circ z$
 * $x \le y \iff z \circ x \le z \circ y$
 * $x < y \iff x \circ z < y \circ z$
 * $x < y \iff z \circ x < z \circ y$

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

By Relation Compatible with Group is Strongly Compatible, $\le$ is strongly compatible with $\circ$.

Thus by the definition of strong compatibility, we obtain the first two results:


 * $x \le y \iff x \circ z \le y \circ z$
 * $x \le y \iff z \circ x \le z \circ y$

By Transitive Antisymmetric Relation Compatible with Group Induces Compatible Strict Ordering, $< \mathop = \left({\le \cap \Delta_S^c}\right)$ is compatible with $\circ$.

Thus by Relation Compatible with Group is Strongly Compatible, $<$ is strongly compatible with $\circ$, whence:


 * $x < y \iff x \circ z < y \circ z$
 * $x < y \iff z \circ x < z \circ y$

and so the result is established.