Equivalence of Definitions of Order Embedding

Proof
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.

Let $\phi: S \to T$ be a mapping.

Definition 1 implies Definition 2
Let $\phi$ be an order embedding by definition 1.

Then by definition:
 * $\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$

It follows from Order Embedding is Injection that $\phi$ is an injection.

Thus $\phi$ is an order embedding by definition 2.

Definition 2 implies Definition 1
Let $\phi$ be an order embedding by definition 2.

Then condition $(2)$ of that definition directly states:
 * $\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$

Thus $\phi$ is an order embedding by definition 1.

Definition 2 implies Definition 4
Let $\phi$ be an order embedding by definition 2.

Then by definition:
 * $\phi$ is an injection
 * $\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$

Let $T' = \Img S$ be the image of $S$ under $\phi$.

From Injection to Image is Bijection:
 * $\phi {\restriction_{S \times T'} }: S \to T'$ is a bijection.

It follow by definition that $\phi {\restriction_{S \times T'} }: S \to T'$ is an order isomorphism.

Thus $\phi$ is an order embedding by definition 4.

Definition 4 implies Definition 1
Let $\phi$ be an order embedding by definition 4.

Then by definition:
 * the restriction of $\phi$ to $S \times T'$ is an order isomorphism between $\struct {S, \preceq_1}$ and $\struct {T', \preceq_2 \restriction_{T' \times T'} }$.

By definition of order isomorphism:
 * $\forall x, y \in S: x \preceq_1 y \iff \map {\phi {\restriction_{S \times T'} } } x \preceq_2 \map {\phi {\restriction_{S \times T'} } } y$

But by definition of restriction:
 * $\forall z \in S: \map {\phi {\restriction_{S \times T'} } } z = \map \phi z$

Thus:
 * $\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$

and so $\phi$ is an order embedding by definition 1.