Proposition \[prop\_sym3\] implies that $Q(j)$ is non-decreasing; since $Q(j)<M$ it means that $Q(j)$ must converge to a finite limit.

Let $A_j$ be the event that at time $t_j$ there are $3$ or more zeroes in a row in $v(t_j)$. On $A_j$ there is a $k\in\M$ such that $v_k(t_j)=v_{k+1}(t_j)=v_{k+2}(t_j)=0$ but $v_{k-1}(t_j)>0$, (unless all $v_k=0$ but then the argument is similar). Then, with a probability exceeding $1/M$, at time $t_j+1$ new particle gets adsorbed at $k+2$, yielding by Proposition \[prop\_sym3\] that for all $j'>j$ we have $v_{k-1}(t_{j'})>0$, $v_{k}(t_{j'})=0$, $v_{k}(t_{j'})>0$, hence the event $B_j:=\{Q(j+1)\ge Q(j)+1\}$ occurs as well. Therefore, $$\P(B_j\| \F_{t_j})\ge \frac 1M\, \P(A_j\| \F_{t_j}),$$ where $\F_{t_j}$ denotes the sigma-algebra generated by $v(t)$ by time $t_j$. Combining this with the second Borel–Cantelli lemma, we obtain {A\_j } &=&{\_j ¶(A\_j \_[t\_j]{})=} {\_j ¶(B\_j \_[t\_j]{})=}\
&=&{B\_j } ={Q(j)} leading to a contradiction.

\[prop\_sym5\] Let $$W(j)=\left|\left\{k:\ v_{k-1}(t_j)=0, v_{k}(t_j)>0,
v_{k+1}(t_j)>0, v_{k-1}(t_j)=0, \right\}\right|$$ be the number of ‘doubles’. Then $W(j)$ is non-increasing.

Let us investigate how we can obtain a subsequence $(0,\ast,\ast,0)$ starting at position $k-1$ at time $t_{j+1}$, where $\ast$ stands for a positive element. One possibility is that at time $t_j$ we already have such a subsequence there; this does not increase $W(j)$. The other possibilities 