Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing

Theorem
Let $\left({S, \preceq_1}\right)$ be a totally ordered set and let $\left({T, \preceq_2}\right)$ be an ordered set.

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

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

That is:


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

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

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.