Inversion Mapping on Ordered Group is Dual Order-Isomorphism

Theorem
Let $(G, \circ, \preceq)$ be an ordered group.

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

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

Proof
By Inversion Mapping is Involution and Involution is Bijection, $\iota$ is 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 $\iota(y) \prec \iota(x)$.

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

If $\iota(x) \prec \iota(y)$, then $\iota(\iota(y)) \prec \iota(\iota(x))$ 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.