Composite of Ordered Ring Isomorphisms is Ordered Ring Isomorphism

Theorem
Let $\left({S_1, +_1, \circ_1, \preccurlyeq_1}\right), \left({S_2, +_2, \circ_2, \preccurlyeq_2}\right), \left({S_3, +_3, \circ_3, \preccurlyeq_3}\right)$ be ordered rings.

Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered ring isomorphisms.

Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered ring isomorphism.

Proof
From Composite of Order Isomorphisms, $\psi \circ \phi: \left({S_1, \preccurlyeq_1}\right) \to \left({S_3, \preccurlyeq_3}\right)$ is an order isomorphism.

From Composite of Isomorphisms on Algebraic Structures, $\psi \circ \phi$ is an algebraic structure isomorphism.

From Isomorphism Preserves Groups, it follows that $\psi \circ \phi$ is a group isomorphism from $\left({S_1, +_1}\right)$ to $\left({S_3, +_3}\right)$.

From Isomorphism Preserves Semigroups, it follows that $\psi \circ \phi$ is a semigroup isomorphism from $\left({S_1, \circ_1}\right)$ to $\left({S_3, \circ_3}\right)$.

Hence the result.