Composite of Order Isomorphisms is Order Isomorphism

Theorem
Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ and $\struct {S_3, \preceq_3}$ be ordered sets.

Let:
 * $\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$

and:
 * $\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$

be order isomorphisms.

Then $\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is also an order isomorphism.

Proof
From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection, as, by definition, an order isomorphism is also a bijection.

From Inverse of Composite Bijection, the inverse of $\psi \circ \phi$ is given by:
 * $\paren {\psi \circ \phi}^{-1} = \phi^{-1} \circ \psi^{-1}$

By definition of composition of mappings:
 * $\map {\psi \circ \phi} x = \map \psi {\map \phi x}$

By definition of order isomorphism, we have:


 * $\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$ is an increasing mapping

and:
 * $\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$ is an increasing mapping.

Hence from Composite of Increasing Mappings is Increasing:
 * $\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is an increasing mapping.

Similarly by definition of order isomorphism:
 * $\phi^{-1}: \struct {S_2, \preceq_2} \to \struct {S_1, \preceq_1}$ is an increasing mapping

and:
 * $\psi^{-1}: \struct {S_3, \preceq_3} \to \struct {S_2, \preceq_2}$ is an increasing mapping.

Hence from Composite of Increasing Mappings is Increasing:
 * $\phi^{-1} \circ \psi^{-1}: \struct {S_3, \preceq_3} \to \struct {S_1, \preceq_1}$ is an increasing mapping.

Hence we have that:
 * $\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is an increasing mapping

and:
 * $\paren {\psi \circ \phi}^{-1}: \struct {S_3, \preceq_3} \to \struct {S_1, \preceq_1}$ is an increasing mapping

and it follows by definition that $\psi \circ \phi$ is an order isomorphism.