Equivalence of Definitions of Order Embedding/Definition 1 implies Definition 3

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

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

Let $\phi: S \to T$ be an order embedding by Definition 1:

Then $\phi: S \to T$ is an order embedding by Definition 3:

Proof
Let $\phi$ be an order embedding by definition 1.

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

$\phi$ is injective by Order Embedding is Injection.

It remains to be shown that:
 * $x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$

Suppose first that $x \prec_1 y$.

Then $x \preceq_1 y$ and $x \ne y$.

Thus by the premise:
 * $\map \phi x \preceq_2 \map \phi y$

Since $\phi$ is injective:
 * $\map \phi x \ne \map \phi y$

Therefore:
 * $\map \phi x \prec_2 \map \phi y$

Suppose instead that $\map \phi x \prec_2 \map \phi y$

Then:
 * $\map \phi x \preceq_2 \map \phi y$

and:
 * $\map \phi x \ne \map \phi y$

By the premise:
 * $x \preceq_1 y$

By the substitutive property of equality:
 * $x \ne y$

Thus:
 * $x \prec_1 y$

Thus $\phi$ is an order embedding by definition 3.