at position $i$ or (possibly) the $0$ at position $i+2$. On this event, the configuration in (B) at time $t_{j+1}$ becomes $$\dots  a\ \underbrace{\underbrace{0\ 1}\ \underbrace{0 \ 1}\ \dots\  \underbrace{0 \ 1}}_{2l-2\mbox{ elements}}\ 0 \ 0\ 0 \ ? \dots  \tag{$\mathrm{B}'$}$$ By iterating this argument until $l=0$, we conclude that eventually there will be a chunk ‘$\dots a\ 0\ 0\dots$’ on some step $t_{j'}$ which in turn at time $t_{j'+1}$ will become ‘$
\dots a-1\ 0\ ?\dots $’with probability at least $\frac 13$, resulting in $S_{j'+1}\le S_{j'}-1$. This yields the statement of Proposition \[prop\_a\_men\] with $K=M$ and $\eps=3^{-M}$.

\[prop\_a\_conv\] With probability $1$, there is a $j_0=j_0(\omega)$ such that $$S(j)=0 \mbox{ for all } j\ge j_0.$$ Additionally, the state ${\bf 0}=(0,0,\dots,0)$ is recurrent for the Markov chain $v(t)$.

The first statement trivially follows from Proposition \[prop\_a\_men\]. Next observe that at times $t_j\ge
t_{j_0}$ the sequence $(v_1(t_j),\dots,v_{M}(t_j))$ consists only of $0$s and $1$s locally looking either like $$\dots  1\ \underbrace{\underbrace{0\ 0}\ \underbrace{0 \ 0}\ \dots\  \underbrace{0 \ 0}}_{2l\mbox{ elements}}\ 1 \dots  \tag{C}$$ or $$\dots  1\ \underbrace{\underbrace{0\ 0}\ \underbrace{0 \ 0}\ \dots\  \underbrace{0 \ 0}}_{2l\mbox{ elements}}\ 0\ 1 \dots  \tag{D}$$ With positive probability even-located $0$s are picked before odd-located $0$s, hence at time 