# 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 $T$.

For each $i \in S$ such that $i$ is not maximal in $S$, let:

$f \left({i}\right) < f \left({\operatorname{succ} \left({i}\right)}\right)$

where $\operatorname{succ} \left({i}\right)$ is the immediate successor element of $i$.

Let:

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

Then for each $i, j \in S$ such that $i \prec j$:

$f \left({i}\right) < f \left({j}\right)$

## Proof

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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, $j \in S_i$.

Thus:

$j \not \prec \operatorname{succ} \left({i}\right)$
$\operatorname{succ} \left({i}\right) \preceq j$

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.

$\blacksquare$