surrounded by $0$s. Similarly, if $v_k(t_j)>0$ and $v_{k+1}(t_j)>0$ but $v_{k-1}(t_j)=v_{k+2}(t_j)=0$, then points $\{k,k+1\}$ remain surrounded by $0$s at time $t_{j+1}$.

Finally, if point $k$ is surrounded by non-zeros on both sides, that is $v_{k-1}(t_j)$, $v_{k}(t_j)$ and $v_{k+1}(t_j)$ are all positive, we have $v_k(t_{j+1})=v_k(t_j)-1$.

Consequently, all sequences of non-zero elements of length $\ge 3$ are bound to disappear, and no such new sequence can arise as $j$ increases.

\[prop\_sym3\] For any $k\in\M$, if for some $s$ $$v_{k-1}(s)>0, \ v_{k}(s)=0, \ v_{k+1}(s)>0$$ then for all $j$ such that $t_j\ge s$ $$v_{k-1}(t_{j})>0, \ v_{k}(t_{j})=0, \ v_{k+1}(t_{j})>0.$$

This immediately follows from the fact that there must be a particle adsorbed at point $k$ during the time interval $(t_{j_0},t_{j_0+1}]$ where $j_0=\max\{j:\ t_j\le s\}$, and that would imply that $v_{k\pm 1}(t_{j_0+1})\ge v_{k\pm 1}(t_{j_0})$ while $v_{k}(t_{j_0+1})=0$. Now an induction on $j$ finishes the proof.

\[prop\_sym4\] For $j$ large enough, in the sequence of $v_k(t_j)$, $k\in\M$, there are no more than two $0$s in a row, that is $$\mbox{if }v_k(t_j)=0\mbox{ then either }v_{k-1}(t_j)>0\mbox{ or
 }v_{k+1}(t_j)>0, \mbox{ or both}.$$

Suppose $j$ is so large that already there are no consecutive subsequences of positive elements of length $\ge 2$ in $(v_1,\dots,v_M)$ (see Proposition \[prop\_sym2\]). Let $$Q(j)=\left|\left\{k:\ v_{k-1}(t_j)>0, v_{k}(t_j)=0,
v_{k-1}(t_j)>0, \right\}\right|.$$ 