Inverse of Ordered Ring Isomorphism is Ordered Ring Isomorphism

Theorem
Let $\left({S, +, \circ, \preceq}\right)$ and $\left({T, \oplus, *, \preccurlyeq}\right)$ be ordered rings.

Let $\phi: S \to T$ be an ordered ring isomorphism.

Then $\phi^{-1}: T \to S$ is also an ordered ring isomorphism.

Proof
By definition, $\phi$ is a bijection.

By Bijection iff Inverse is Bijection, $\phi^{-1}$ is also a bijection.

By definition, an ordered ring isomorphism from $\phi: \left({S, +, \circ, \preceq}\right) \to \left({T, \oplus, *, \preccurlyeq}\right)$ is:


 * an order isomorphism from the poset $\left({S, \preceq}\right)$ to the poset $\left({T, \preccurlyeq}\right)$
 * a group isomorphism from the group $\left({S, +}\right)$ to the group $\left({T, \oplus}\right)$
 * a semigroup isomorphism from the semigroup $\left({S, \circ}\right)$ to the semigroup $\left({T, *}\right)$.

From Inverse of Order Isomorphism is Order Isomorphism, $\phi^{-1}: \left({T, \preccurlyeq}\right) \to \left({S, \preceq}\right)$ is an order isomorphism.

From Inverse of Algebraic Structure Isomorphism is Isomorphism:
 * $\phi^{-1}: \left({T, \oplus}\right) \to \left({S, +}\right)$ is a group isomorphism
 * $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is a semigroup isomorphism.

From Isomorphism of Abelian Groups, $\phi^{-1}: \left({T, \oplus}\right) \to \left({S, +}\right)$ preserves the commutativity of $S$ and $T$.

Hence the result.