Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing

Theorem
Let $\struct {S, \preceq_1}$ be a totally ordered set.

Let $\struct {T, \preceq_2}$ be an ordered set.

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

Then $\phi$ is an order embedding $\phi$ is strictly increasing.

That is:


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


 * $\forall x, y \in S: x \prec_1 y \implies \map \phi x \prec_2 \map \phi y$
 * $\forall x, y \in S: x \prec_1 y \implies \map \phi x \prec_2 \map \phi y$

Forward Implication
Let $\phi$ be an order embedding.

Let $x, y \in S$ with $x \prec_1 y$, where $\prec_1$ is the strict predecessor relation.

Then:

So by definition, $\phi$ is strictly increasing.