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.

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

As $\phi$ is an order isomorphism, we have:
 * $\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \map \phi {x_1} \preceq_2 \map \phi {y_1}$

As $\psi$ is an order isomorphism, we have:
 * $\forall x_2, y_2 \in S_2: x_2 \preceq_2 y_2 \implies \map \psi {x_2} \preceq_3 \map \psi {y_2}$

By setting $x_2 = \map \phi {x_1}, y_2 = \map \phi {y_1}$, it follows that:
 * $\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \map \psi {\map \phi {x_1} } \preceq_3 \map \psi {\map \phi {y_1} }$

Similarly we can show that:
 * $\forall x_3, y_3 \in S_3: x_3 \preceq_3 y_3 \implies \map {\phi^{-1} } {\map {\psi^{-1} } {x_3} } \preceq_1 \map {\phi^{-1} } {\map {\psi^{-1} } {y_3} }$