first part of the Theorem.

Finally, note that if the initial configuration is empty, the conditions of Proposition \[prop\_sym5\] are fulfilled with no ‘doubles’ at all, i.e. $W(0)=0$. Consequently, for all $j\ge 0$ we have that there are no consecutive non-zero elements in $v_k(t_j)$, yielding the final statement of the Theorem.

Appendix {#binf}
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In this section we briefly describe the long-time behaviour of the growth process generated by the dynamics, where a particle is allocated at random to a site with maximum potential. The process is trivial in both the symmetric and asymmetric cases. Consider the symmetric case, i.e. $U_i=\{i-1,i,i+1\}$, $i\in\M$. It is easy to see that with probability $1$, there exists $k$ such that either $$\label{sym_inf}
\lim_{t\to\infty} \frac{\xi_k(t)}t=1 \ \mbox{ and } \ \sup_{i\ne k}
 \xi_i(t)<\infty$$ or $$\label{sym_inf1}
 \lim_{t\to\infty} \frac{\xi_k(t)}t=\lim_{t\to\infty} \frac{\xi_{k+1}(t)}t=\frac 12
  \ \mbox{ and } \ \sup_{i\notin \{k,k+1\}}  \xi_i(t)<\infty.$$ Indeed, recall the formula for the potential given by (\[eq:sympot\]). Then $u(t)$ is a Markov chain with transition probabilities given by ¶(u\_[i]{}(t+1)=u\_[i]{}(t)+1\_[i{k-1,k,k+1}]{})= for $k\in\M$, where $$S_{\max}=\left\{i:\ u_i(t)=\max_{i\in\M} u_i(t)\right\}\subseteq
\M$$ is the set of those $i$ for which $u_i(t)$ equals the maximum value.

Observe that if at time $s$ the adsorption/allocation occurs at point $i$, then $S_{\max}(s+1)\subseteq \{i-1,i,i+1\}$. In 