Inverse of Supremum in Ordered Group is Infimum of Inverses

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

Let $x, y \in G$.

Then:
 * $\set {x, y}$ admits a supremum in $G$


 * $\set {x^{-1}, y^{-1} }$ admits an infimum in $G$
 * $\set {x^{-1}, y^{-1} }$ admits an infimum in $G$

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

Proof
Let:

Recall from Inverse of Group Inverse:

From Inverse of Infimum in Ordered Group is Supremum of Inverses:

Then:
 * $\set {a, b}$ admits an infimum in $G$


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

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

Substituting back for $a$ and $b$:


 * $\set {x^{-1}, y^{-1} }$ admits an infimum in $G$


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

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

Hence from :


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