Inversion Mapping on Ordered Group is Dual Order-Isomorphism

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

Let $\iota: G \to G$ be the inversion mapping, defined by $\phi \left({x}\right) = x^{-1}$.

Then $\iota$ is a dual order-isomorphism.

Proof
By Inversion Mapping is Involution and Involution is Permutation, $\iota$ is a permutation and so by definition bijective.

Let $x, y \in G$ such that $x \prec y$.

Then $y^{-1} \prec x^{-1}$ by Group Inverse Reverses Ordering in Ordered Group.

Thus $\iota \left({y}\right) \prec \iota \left({x}\right)$.

Since this holds for all $x$ and $y$ with $x \prec y$, $\iota$ is strictly decreasing.

If $\iota \left({x}\right) \prec \iota \left({y}\right)$, then $\iota \left({\iota \left({y}\right)}\right) \prec \iota \left({\iota \left({x}\right)}\right)$ by the above.

Thus by Inverse of Group Inverse: $y \prec x$.

Therefore, $\iota$ reverses ordering in both directions, and is thus a dual isomorphism.