Inverse of Order Isomorphism is Order Isomorphism

Theorem
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets.

Let $\phi$ be a bijection from $\left({S, \preceq_1}\right)$ to $\left({T, \preceq_2}\right)$.

Then:
 * $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$

is an order isomorphism
 * $\phi^{-1}: \left({T, \preceq_2}\right) \to \left({S, \preceq_1}\right)$
 * $\phi^{-1}: \left({T, \preceq_2}\right) \to \left({S, \preceq_1}\right)$

is also an order isomorphism.

Proof
Follows directly from the definition of order isomorphism.

Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be an order isomorphism.

Then:
 * $\forall x, y, \in S: x \preceq_1 y \implies \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$

and:


 * $\forall \phi \left({x}\right), \phi \left({y}\right) \in T: \phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies \phi^{-1} \left({\phi \left({x}\right)}\right) \preceq_1 \phi^{-1} \left({\phi \left({y}\right)}\right)$

That is:
 * $\forall \phi \left({x}\right), \phi \left({y}\right) \in T: \phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies x \preceq_1 y$

Hence the result.