Inverse of Ordered Semigroup Isomorphism is Isomorphism

Theorem
Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.

Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be a mapping.

Then:
 * $\phi$ is an ordered semigroup isomorphism


 * $\phi^{-1}: \struct {T, *, \preccurlyeq} \to \struct {S, \circ, \preceq}$ is also an ordered semigroup isomorphism.
 * $\phi^{-1}: \struct {T, *, \preccurlyeq} \to \struct {S, \circ, \preceq}$ is also an ordered semigroup isomorphism.

Sufficient Condition
Let $\phi$ be an ordered semigroup isomorphism.

Then by definition $\phi$ is a bijection.

Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.

That is:
 * $\exists \phi^{-1}: \struct {T, *, \preccurlyeq} \to \struct {S, \circ, \preceq}$

From Inverse of Algebraic Structure Isomorphism is Isomorphism:
 * $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is an isomorphism.

Indeed, from Isomorphism Preserves Semigroups, $\phi^{-1}$ is specifically a semigroup isomorphism.

Then from Inverse of Order Isomorphism is Order Isomorphism we have that:
 * $\phi^{-1}: \struct {T, \preccurlyeq} \to \struct {S, \preceq}$ is an order isomorphism.

Sufficient Condition
Let $\phi^{-1}$ be an ordered semigroup isomorphism.

Then $\paren {\phi^{-1} }^{-1}$ is also an ordered semigroup isomorphism.

From Inverse of Inverse of Bijection:
 * $\paren {\phi^{-1} }^{-1} = \phi$

and the result follows.