Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 1

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 strictly increasing mapping.

Then $\phi$ is an order embedding

Proof
If $x \preceq_1 y$, then $x = y$ or $x \prec_1 y$.

If $x = y$, then $\phi(x) = \phi(y)$, so $\phi(x) \preceq_2 \phi(y)$.

If $x \prec_1 y$, then by the definition of strictly increasing:
 * $\phi(x) \prec_2 \phi(y)$

so
 * $\phi(x) \preceq_2 \phi(y)$ by the definition of $\prec_2$.

Thus $x \preceq_1 y \implies \phi(x) \preceq_2 \phi(y)$.

We now need to show that $\phi(x) \preceq_2 \phi(y) \implies x \preceq_1 y$.

Suppose that $x \not\preceq_1 y$.

Since $\preceq_1$ is a total ordering:
 * $y \prec_1 x$.

Thus since $\phi$ is strictly increasing:
 * $\phi(y) \prec_1 \phi(x)$.

Thus $\phi(x) \not\preceq_1 \phi(y)$.

Therefore $x \not\preceq_1 y \implies \phi(x) \not\preceq_2 \phi(y)$.

Taking the contrapositive:
 * $\phi(x) \preceq_2 \phi(y) \implies x \preceq y$