Equivalence of Definitions of Order Isomorphism

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

Definition 1 implies Definition 2
Let $\phi:S \to T$ be an order isomorphism by Definition 1.

Then $\phi$ is bijective, and thus trivially surjective.

Let $x, y \in S$.

Then by Definition 1:


 * $x \preceq_1 y \implies \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$

Suppose that $\phi \left({x}\right) \preceq_2 \phi \left({y}\right)$.

Then by Definition 1:


 * $\phi^{-1}\left({\phi \left({x}\right)}\right) \preceq_1 \phi^{-1} \left({\phi \left({y}\right)}\right)$

By the definition of inverse:


 * $x \preceq_1 y$.

Thus by Rule of Implication:


 * $\phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies x \preceq_1 y$.

So:


 * $x \preceq_1 y \iff \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$

As this holds for all $x, y \in S$ and $\phi$ is surjective, $\phi$ is an order isomorphism by Definition 2.

Definition 2 implies Definition 1
Let $\phi:S \to T$ be an order isomorphism by Definition 2.

Then $\phi$ is a surjective order embedding.

By Order Embedding is Injection, $\phi$ is injective.

As it is also surjective, $\phi$ is bijective, and thus satisfies the first condition of Definition 1.

Since $\phi$ is an order embedding:
 * $\forall x, y \in S: \left({x \preceq_1 y \iff \phi\left({x}\right) \preceq_2 \phi\left({y}\right)}\right)$

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

This satisfies the second condition of Definition 1.

Furthermore:


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

Let $p, q \in T$ and let $p \preceq_2 q$.

Then since $\phi$ is bijective, it has an inverse $\phi^{-1}$ such that


 * $\phi\left({\phi^{-1}\left({p}\right)}\right) = p$
 * $\phi\left({\phi^{-1}\left({q}\right)}\right) = q$

Thus $\phi\left({\phi^{-1}\left({p}\right)}\right) \preceq_2 \phi\left({\phi^{-1}\left({q}\right)}\right)$.

Therefore:


 * $\phi^{-1}\left({p}\right) \preceq_1 \phi^{-1}\left({q}\right)$.

Thus we see that:


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

Thus we have satisfied the final condition of Definition 1.