Strictly Increasing Mapping on Well-Ordered Class

Theorem
Let $ \left({S, \prec}\right)$ be a strictly well-ordered class.

Let $\left({T, <}\right)$ be a strictly ordered class.

Let $f$ be a mapping from $S$ to $T$.

Suppose that for each $i \in S$ such that $i$ is not maximal in $S$:
 * $f \left({i}\right) < f \left({\operatorname{succ} \left({i}\right)}\right)$

where $\operatorname{succ} \left({i}\right)$ is the Definition:Immediate Successor Element of $i$.

Suppose also that:
 * $\forall i, j \in S: i \preceq j \implies f \left({i}\right) \le f \left({j}\right)$

Then for each $i, j \in S$ such that $i \prec j$:
 * $f \left({i}\right) < f \left({j}\right)$

Proof
By Non-Maximal Element of Well-Ordered Class has Immediate Successor, $\operatorname{succ} \left({i}\right)$ is guaranteed to exist.

Let $i \prec j$.

Let $S_i := \left\{{q : i \prec q}\right\}$.

Then $\operatorname{succ} \left({i}\right)$ is the minimal element of $S_i$.

By supposition, $j \in S_i$, so $j \not \prec \operatorname{succ} \left({i}\right)$.

Since Well-Ordering is Total Ordering, $ \operatorname{succ} \left({i}\right) \preceq j$.

Thus by supposition:
 * $f \left({\operatorname{succ} \left({i}\right)}\right) \le f \left({j}\right)$

Since $f \left({i}\right) < \left({\operatorname{succ} \left({i}\right)}\right)$:
 * $f \left({i}\right) < f \left({j}\right)$

by transitivity.