If every $a_i\le 1$ and $\parallel a\parallel \ge 1$, the strict right minima form a two-sided increasing list on which $S_a$ increases by exactly $1$ at each successive index

The integers $S_a(0),S_a(1),\dots ,S_a(m-1)$ have a least element $\mu$: by induction on $t$, every list $S_a(0),\dots ,S_a(t)$ has a least element, since the order on $\mathbb{Z}$ is total, so adjoining one further integer to a list with a least element leaves it with one.

For every $M\in \mathbb{Z}$ the set $T_M:=\{i\in \mathbb{Z}:S_a(i)\le M\}$ is nonempty. If $M\ge 0$ then $S_a(0)=0\le M$. If $M<0$ then put $t:=-M$, a positive integer; induction on $t$ with the quasiperiodicity clause of [F1] gives $S_a(-tm)=-tk$, and $tk\ge t$ because $k\ge 1$, so $S_a(-tm)\le -t=M$.

Each $T_M$ has an upper bound. Let $i\in T_M$ and write $i=qm+r$ with $0\le r<m$ by [L4], so $S_a(i)=qk+S_a(r)\ge qk+\mu$ by [F1] and step 1.1, whence $qk\le M-\mu$. If $q\ge 1$ then $q\le qk$ because $k\ge 1$, so $q\le M-\mu$; and if $q\le 0$ then $q\le 0$. So in either case $q\le B$, where $B$ is the greater of $0$ and $M-\mu$, and therefore $i\le Bm+m-1$.

By [L3] the set $T_M$ has a greatest element $j_M$. Every $i>j_M$ lies outside $T_M$, so $S_a(i)>M\ge S_a(j_M)$, and $j_M$ is a strict right minimum. Its value is exactly $M$: the one-step difference identity of [F1] gives $S_a(j_M+1)-S_a(j_M)=a_{j_M<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>m}\le 1$ by hypothesis, while $S_a(j_M+1)>M$ since $j_M+1$ is outside $T_M$, so $M<S_a(j_M+1)\le S_a(j_M)+1\le M+1$ and hence $S_a(j_M)=M$.

At most one strict right minimum has a given value: if $j<j^{\mathrm{\prime }}$ are both strict right minima then $S_a(j^{\mathrm{\prime }})>S_a(j)$, so their values differ. With step 3.1 this gives, for each $M$, exactly one $j\in R$ with $S_a(j)=M$; write $\rho (M)$ for it. Every $j\in R$ satisfies $j=\rho (S_a(j))$ by that uniqueness, so $\rho$ is onto $R$, and it is injective because $S_a(\rho (M))=M$; by [L5] it is a bijection $\mathbb{Z}\to R$. This is clause 1.

$\rho$ is strictly increasing: if $M<M^{\mathrm{\prime }}$ and $\rho (M^{\mathrm{\prime }})\le \rho (M)$, then either $\rho (M^{\mathrm{\prime }})=\rho (M)$, forcing $M=M^{\mathrm{\prime }}$, or $\rho (M^{\mathrm{\prime }})<\rho (M)$, and then the strict right minimum property of $\rho (M^{\mathrm{\prime }})$ gives $M=S_a(\rho (M))>S_a(\rho (M^{\mathrm{\prime }}))=M^{\mathrm{\prime }}$; both contradict $M<M^{\mathrm{\prime }}$. And $\rho (M)+m$ is a strict right minimum of value $M+k$: for $i>\rho (M)+m$ we have $i-m>\rho (M)$, so the quasiperiodicity clause of [F1] gives $S_a(i)=S_a(i-m)+k>S_a(\rho (M))+k=S_a(\rho (M)+m)$, and $S_a(\rho (M)+m)=M+k$; hence $\rho (M+k)=\rho (M)+m$ by step 4.1. This is clause 2.

Fix $j_0\in \mathbb{Z}$. Iterating clause 2 by induction gives $\rho (M+tk)=\rho (M)+tm$ for every $t\in \mathbb{N}$, so the set $\{M:\rho (M)\ge j_0\}$ is nonempty, taking $t$ with $tm\ge j_0-\rho (0)$, and bounded below, since for $t$ with $\rho (0)-tm<j_0$ every $M\le -tk$ has $\rho (M)\le \rho (-tk)=\rho (0)-tm<j_0$; let $M_0$ be its least element by [L3]. Then $\rho (M_0-1)<j_0\le \rho (M_0)$, so $\rho (M_0+k)=\rho (M_0)+m\ge j_0+m$ and $\rho (M_0+k-1)=\rho (M_0-1)+m<j_0+m$. Since $\rho$ is strictly increasing and every member of $R$ is some $\rho (M)$, the members of $R$ in $\{j_0,\dots ,j_0+m-1\}$ are exactly $\rho (M_0),\dots ,\rho (M_0+k-1)$, and $i\mapsto \rho (M_0+i)$ is a bijection from the natural number $k$ onto that set; so by [L6] the set is finite with exactly $k$ elements, which is clause 3.