at time $t_j$ are (0,0,0,0),  (0,0,0,),  (0,0,,0 ), (0,0,,),\
(0,,0,0),  (0,,0,), (,,0,0),  (,,0,). By careful examination of all of the configurations above, we conclude that the subsequence $(0,\ast,\ast,0)$ cannot arise at time $t_{j+1}$. Consequently, $W(j)$ cannot increase.

\[prop\_sym6\] For $j$ large enough, in the sequence of $v_k(t_j)$, $k\in\M$, there are no consecutive subsequences of the form $(\ast,\ast,0,0)$ or $(0,0,\ast,\ast)$ where each $\ast$ stands for any positive number; that is there is no $k$ such that v\_[k]{}(t\_j)=v\_[k+1]{}(t\_j)=0 & &\
&v\_[k+2]{}(t\_j)&gt;0 v\_[k+3]{}(t\_j)&gt;0&\
&v\_[k-1]{}(t\_j)&gt;0 v\_[k-2]{}(t\_j)&gt;0.&

Omitting $(t_j)$, without loss of generality suppose $v_{k}>0$, $v_{k+1}>0$, $v_{k+2}=v_{k+3}=0$. Then either at some time $j_1>j$ we will have $v_{k+3}(t_{j_1})>0$ (hence the configuration $(\ast,\ast,0,0)$ gets destroyed), or with probability at least $\frac 13$ for each $j'\ge j$ we have adsorption at position $k+3$ at some time during the time interval $(t_{j'},t_{j'+1}]$. This would imply that $v_{k+1}(t_{j'+1})=v_{k+1}(t_{j'})-1$. Hence, in a geometrically distributed number of times, we obtain $0$ at position $k+1$, and thus the configuration $(\ast,\ast,0,0)$ gets destroyed. On the other hand, by Proposition \[prop\_sym5\], the number of doubles is non-increasing, so no new configurations of this type can arise. Consequently, eventually all configurations $(\ast,\ast,0,0)$ and $(0,0,\ast,\ast)$ will disappear.

\[prop\_sym7\] For $j$ large enough, in the sequence of $v_k(t_j)$, 