Inversion Mapping Reverses Ordering in Ordered Group/Corollary

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

Let $x \in G$.

Then the following equivalences hold:

Proof 1
By User:Dfeuer/OG3:

Since $e^{-1} = e$, the theorem holds.

Proof 2
By the definition of an ordered group, $\le$ is a relation compatible with $\circ$.

Thus by User:Dfeuer/CRG4, we obtain the first two results:


 * $(\operatorname{OG}4.1):\quad x \le e \iff e \le x^{-1}$
 * $(\operatorname{OG}4.2):\quad e \le x \iff x^{-1} \le e$

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

Thus by again User:Dfeuer/CRG4, we obtain the remaining results:


 * $(\operatorname{OG}4.1'):\quad x < e \iff e < x^{-1}$
 * $(\operatorname{OG}4.2'):\quad e < x \iff x^{-1} < e$