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

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

Let $\phi: S \to T$ be a mapping such that for all $x, y \in S$:


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

Then:
 * $(1): \quad \phi$ is injective


 * $(2): \quad \forall x, y, \in S: x \prec_1 y \iff \phi \left({x}\right) \prec_2 \phi \left({y}\right)$

That is, if $\phi$ is an order embedding by Definition 1 then it is also an order embedding by Definition 3.

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

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

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

It remains to be shown that:
 * $x \prec_1 y \iff \phi \left({x}\right) \prec_2 \phi \left({y}\right)$

Suppose first that $x \prec_1 y$.

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

Thus by the premise:
 * $\phi \left({x}\right) \preceq_2 \phi\left({y}\right)$

Since $\phi$ is injective:
 * $\phi \left({x}\right) \ne \phi\left({y}\right)$

Therefore:
 * $\phi \left({x}\right) \prec_2 \phi\left({y}\right)$

Suppose instead that $\phi \left({x}\right) \prec_2 \phi\left({y}\right)$

Then:
 * $\phi \left({x}\right) \preceq_2 \phi\left({y}\right)$

and:
 * $\phi \left({x}\right) \ne \phi\left({y}\right)$

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.