Inverse of Infimum in Ordered Group is Supremum of Inverses

Theorem
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group.

Let $x, y \in G$.

Then:
 * $\set {x, y}$ admits an infimum in $G$


 * $\set {x^{-1}, y^{-1} }$ admits a supremum in $G$
 * $\set {x^{-1}, y^{-1} }$ admits a supremum in $G$

in which case:
 * $\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$

Sufficient Condition
Let $\set {x, y}$ admits an infimum $c$ in $G$.

Then:

and so $c^{-1}$ is an upper bound of $\set {x^{-1}, y^{-1} }$.

Suppose $d$ is an upper bound of $\set {x^{-1}, y^{-1} }$.

Then:

That is, $d^{-1}$ is a lower bound of $\set {x, y}$.

But because $c$ is an infimum of $\set {x, y}$:

That is, for an arbitrary upper bound $d$ of $\set {x^{-1}, y^{-1} }$:
 * $c^{-1} \preccurlyeq d$

and so $c^{-1}$ is a supremum of $\set {x^{-1}, y^{-1} }$ by definition.

That is:
 * $\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$

Necessary Condition
Let $\set {x^{-1}, y^{-1} }$ admit a supremum $c$ in $G$.

Then:

and so $c^{-1}$ is a lower bound of $\set {x, y}$.

Suppose $d$ is a lower bound of $\set {x, y}$.

Then:

That is, $d^{-1}$ is an upper bound of $\set {x, y}$.

But because $c$ is a supremum of $\set {x^{-1}, y^{-1} }$:

That is, for an arbitrary lower bound $d$ of $\set {x, y}$:
 * $d \preccurlyeq c^{-1}$

and so $c^{-1}$ is an infimum of $\set {x, y}$ by definition.

That is:
 * $\paren {\inf \set {x, y} } = \sup \set {x^{-1}, y^{-1} }^{-1}$

from which it follows from that:
 * $\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$