Composite of Order Isomorphisms is Order Isomorphism

Theorem
Let $$\left({S_1, \le_1}\right)$$, $$\left({S_2, \le_2}\right)$$ and $$\left({S_3, \le_3}\right)$$ be posets.

Let $$\phi: \left({S_1, \le_1}\right) \to \left({S_2, \le_2}\right)$$ and $$\psi: \left({S_2, \le_2}\right) \to \left({S_3, \le_3}\right)$$ be isomorphisms.

Then $$\psi \circ \phi: \left({S_1, \le_1}\right) \to \left({S_3, \le_3}\right)$$ is also an isomorphism.

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

By definition of composition of mappings, $$\psi \circ \phi \left({x}\right) = \psi \left({\phi \left({x}\right)}\right)$$.

As $$\phi$$ is an isomorphism, we have:

$$\forall x_1, y_1 \in S_1: x_1 \le_1 y_1 \Longrightarrow \phi \left({x_1}\right) \le_2 \phi \left({y_1}\right)$$

As $$\psi$$ is an isomorphism, we have:

$$\forall x_2, y_2 \in S_2: x_2 \le_2 y_2 \Longrightarrow \psi \left({x_2}\right) \le_3 \psi \left({y_2}\right)$$

By setting $$x_2 = \phi \left({x_1}\right), y_2 = \phi \left({y_1}\right)$$, it follows that:

$$\forall x_1, y_1 \in S_1: x_1 \le_1 y_1 \Longrightarrow \psi \left({\phi \left({x_1}\right)}\right) \le_3 \psi \left({\phi \left({y_1}\right)}\right)$$.

Similarly we can show that:

$$\forall x_3, y_3 \in S_3: x_3 \le_3 y_3 \Longrightarrow \phi^{-1} \left({\psi^{-1} \left({x_3}\right)}\right) \le_1 \phi^{-1} \left({\psi^{-1} \left({y_3}\right)}\right)$$.