Theorem: Strictly Increasing Mapping Between Wosets Implies Order Isomorphism

Let $J$ and $E$ be well-ordered sets.
Let there exist a mapping $k: J \to E$ which is strictly increasing.
Then $J$ is order isomorphic to $E$ or an initial segment of $E$.

Proof:
If the sets considered are empty or singletons, the theorem holds vacuously or trivially.
Suppose $J, E$ both have at least two elements.
Let $e_0 = \min E$, the smallest element of $E$.
Define the mapping:
:$h: J \to E$:
:$\map h \alpha = \begin {cases} \map \min {E \setminus h \sqbrk {S_\alpha} } & : h \sqbrk {S_\alpha} \ne E \\ e_0 & : h \sqbrk {S_\alpha} = E \end {cases}$
where $S_\alpha$ is the initial segment determined by $\alpha$ and $h \sqbrk {S_\alpha}$ is the image of $S_\alpha$ under $h$.
By the Principle of Recursive Definition for Well-Ordered Sets, this construction is well-defined and uniquely determined.
Observe that:
{{begin-eqn}}
{{eqn | l = h \sqbrk {S_\alpha}
      | r = \set {\map h x \in E: \exists x \in J: \map h x = \map \min {E \setminus h \sqbrk {S_\alpha} } }
      | c = {{Defof|Image of Subset under Mapping|image of a subset}}
}}
{{eqn | r = \set {\map h x \in E: \exists x \in J: \map h x \prec \map h \alpha}
}}
{{eqn | r = S_{\map h \alpha}
      | c = {{Defof|Initial Segment|initial segment}}
}}
{{end-eqn}}
This equality also holds if $\map h \alpha = e_0$, by Initial Segment Determined by Smallest Element is Empty.
We claim that $\map h \alpha \preceq \map k \alpha$ for all $\alpha \in J$.
{{AimForCont}} there is some $a \in J$ such that $\map h a \not \preceq \map k a$.
Then $\map k a \prec \map h a$ by the trichotomy law.
Because $\map h a$ has an element preceding it, $\map h a \ne e_0$.
Thus $\map k a \prec \map \min {E \setminus h \sqbrk {S_a} }$ by the construction of $k$.
Then $\map k a \in h \sqbrk {S_a}$, because it precedes the smallest element that isn't in $h \sqbrk {S_a}$.
Recall that $h \sqbrk {S_a} = S_{\map h a}$.
Then $\map k a \in S_{\map h a}$.
This implies that $\map h a \prec \map k a$, contradicting the assumption that $\map h a \not \preceq \map k a$
From this contradiction we can conclude:
:$\map h \alpha \preceq \map k \alpha$ for all $\alpha \in J$.
{{AimForCont}} there is some $\alpha \in J$ such that $h \sqbrk {S_\alpha} = E$.
Recall that $h \sqbrk {S_\alpha} = S_{\map h \alpha}$.
Thus, were such an $\alpha$ to exist, then it would succeed all elements in $E$.
It particular, it would also succeed $\map k \alpha$.
But we showed above that $\map h \alpha \preceq \map k \alpha$.
From this contradiction we see that there cannot be any $\alpha \in J$ with $h \sqbrk {S_\alpha} = E$.
Thus the definition of $h$ can be simplified:
:$h: J \to E$:
:$\map h \alpha = \map \min {E \setminus h \sqbrk {S_\alpha} }$
Then the hypotheses of Characterization of Strictly Increasing Mapping on Woset are satisfied.
Thus $h$ is a strictly increasing mapping and its image is $E$ or an initial segment of $E$.
From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $h$ is also injective.
From Injection to Image is Bijection, $h$ is also bijective to its image.
We conclude that there is an order isomorphism from $J$ to $E$, or from $J$ to an initial segment of $E$.
{{qed}}