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

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 3:

Then $\phi: S \to T$ is an order embedding by Definition 1:

Proof
Let $\phi$ be an order embedding by definition 3.

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


 * $(2): \quad \forall x, y, \in S: x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$

Let $x \preceq_1 y$.

Then $x \prec_1 y$ or $x = y$.

If $x \prec_1 y$, then by hypothesis:
 * $\map \phi x \prec_2 \map \phi y$

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

If $x = y$, then:
 * $\map \phi x = \map \phi y$

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

Thus it has been shown that:


 * $x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$

Let $\map \phi x \preceq_2 \map \phi y$.

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

or:
 * $\map \phi x = \map \phi y$

Suppose $\map \phi x \prec_2 \map \phi y$.

Then by hypothesis:
 * $x \prec_1 y$

and so:
 * $x \preceq_1 y$

Suppose $\map \phi x = \map \phi y$.

Then since $\phi$ is injective:
 * $x = y$

and so:
 * $x \preceq_1 y$

Thus in both cases:
 * $x \preceq_1 y$

and so:
 * $\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$

Hence the result:
 * $x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$

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