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 $p, q \in S$:


 * $p \preceq_1 q \iff \phi \left({p}\right) \preceq_2 \phi \left({q}\right)$

Then $\phi$ is injective and for all $p, q \in S$:


 * $p \prec_1 q \iff \phi(p) \prec_2 \phi(q)$

That is, if $\phi$ is an order embedding by Definition 1 then it is also an order embedding by Definition 2.

Proof
Suppose that for all $p, q \in S$:


 * $p \preceq_1 q \iff \phi(p) \preceq_2 \phi(q)$

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

We must show that:


 * $p \prec_1 q \iff \phi(p) \prec_2 \phi(q)$

Suppose first that $p \prec q$.

Then $p \preceq q$ and $p ≠ q$.

Thus by the premise, $\phi \left({p}\right) \preceq \phi\left({q}\right)$.

Since $\phi$ is injective, $\phi \left({p}\right) ≠ \phi\left({q}\right)$.

Therefore $\phi \left({p}\right) \prec \phi\left({q}\right)$.

Suppose instead that $\phi \left({p}\right) \prec \phi\left({q}\right)$.

Then $\phi \left({p}\right) \preceq \phi\left({q}\right)$ and $\phi \left({p}\right) \ne \phi\left({q}\right)$.

By the premise, $p \preceq q$.

By the substitutive property of equality, $p \ne q$.

Thus $p \prec q$.