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 function from $S$ to $T$.

Suppose that for each $i \in S$, $f(i) < f(i^+)$.

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

By Well-Ordering Determines Minimal Elements, $S_i$ has a minimal element, which is, by definition, $i^+$.

By supposition, $j \in S_i$, so $j \not\prec i^+$.

Since Well-Ordering is Total Ordering, $i^+ \preceq j$.

Thus by supposition, $f(i^+) \le f(j)$. Since $f(i) < f(i^+)$,
 * $f(i) < f(j)$

by transitivity.