Strictly Increasing Mapping on Well-Ordered Class

Theorem
Let $\struct {S, \prec}$ be a strictly well-ordered class.

Let $\struct {T, <}$ be a strictly ordered class.

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

For each $i \in S$ such that $i$ is not maximal in $S$, let:
 * $\map f i < \map f {\map \Succ i}$

where $\map \Succ i$ is the immediate successor element of $i$.

Let:
 * $\forall i, j \in S: i \preceq j \implies \map f i \le \map f j$

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

Proof
By Non-Greatest Element of Well-Ordered Class has Immediate Successor, $\map \Succ i$ is guaranteed to exist.

Let $i \prec j$.

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

Then $\map \Succ i$ is the minimal element of $S_i$.

By supposition, $j \in S_i$.

Thus:
 * $j \nprec \map \Succ i$

Since Well-Ordering is Total Ordering:
 * $\map \Succ i \preceq j$

Thus by supposition:
 * $\map f {\map \Succ i} \le \map f j$

Since $\map f i < \map f {\map \Succ i}$:
 * $\map f i < \map f j$

by transitivity.