Strictly Increasing Mapping on Well-Ordered Class

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

Let $(T,<)$ 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 $S$, $f(i) < f(\operatorname{succ}(i))$, where $\operatorname{succ}(i)$ is the Definition:Immediate Successor Element of $i$, guaranteed to exist by Non-Maximal Element of Well-Ordered Class has Immediate Successor.

Suppose also that for each $i, j\in S$ such that $i \preceq j$, $f(i) \le f(j)$.

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

Proof
Let $i \prec j$.

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

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

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

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

Thus by supposition, $f( \operatorname{succ} (i)) \le f(j)$. Since $f(i) < f( \operatorname{succ} (i))$,
 * $f(i) < f(j)$

by transitivity.