Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 5

Theorem
Let $p$ be a prime.

Let $b \in Z_{> 0}$ such that $b, p$ are coprime.

Let $\sequence{d_n}$ be a sequence of $p$-adic digits.

Let $\sequence{r_n}$ be a sequence of integers:
 * $(\text a) \quad \forall n \in \N: r_n = d_{n+1} b + p r_{n+1}$
 * $(\text b) \quad \exists n_0 \in \N : \forall n \ge n_0 : -b \le r_n \le 0$

Then:
 * $\exists \mathop m, l \in \N : \forall n \ge m: r_n = r_{n + l}$ and $d_n = d_{n + l}$

Proof
By hypothesis the set $\set{r_n : n \in \N}$ of values of the sequence $\sequence{r_n}$ is a subset of:
 * $\set{r_0, r_1, \ldots, r_{n_0}} \cup \set{-b, -b + 1, -b + 2, \ldots, 2, 1, 0}$

It follows that the set $\set{r_n : n \in \N}$ takes only finitely many values.

Hence:
 * $\exists m_0, l \in \N : l > 0 : r_{m_0} = r_{ {m_0} + l}$

Lemma 10
Let $m = m_0 + 1$

The proof now proceeds by induction.

For all $n \ge m$, let $\map P {n}$ be the proposition:
 * $r_n = r_{n+l}$ and $d_n = d_{n + l}$

Basis for the Induction
$\map P {m}$ is the proposition:
 * $r_m = r_{m+l}$ and $d_m = d_{m + l}$

It has already been shown that:
 * $r_{m_0} = r_{ {m_0}+l}$

From lemma 10:
 * $d_m = d_{m+l}$
 * $r_m = r_{m+l}$

This proves proposition $\map P {m}$.

Induction Hypothesis
Let $n \ge m$.

The induction hypothesis is the proposition $\map P {n}$:
 * $r_n = r_{n+l}$ and $d_n = d_{n + l}$

It has to be shown that the proposition $\map P {n+1}$ is true:
 * $r_{n + 1} = r_{n + l + 1}$ and $d_{n + 1} = d_{n + l + 1}$

Induction Step
From the induction hypothesis:
 * $r_n = r_{n+l}$

From lemma 10:
 * $d_{n + 1} = d_{n + l + 1}$
 * $r_{n + 1} = r_{n + l + 1}$

Hence:
 * $\forall n \ge m: r_n = r_{n + l}$ and $d_n = d_{n + l}$

The result follows.