deletions | additions       diff --git a/section_Convergence_label_sec_convergence__.tex b/section_Convergence_label_sec_convergence__.tex  index 36035f4..c31cf33 100644  --- a/section_Convergence_label_sec_convergence__.tex  +++ b/section_Convergence_label_sec_convergence__.tex 
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\fbox{$2) \Rightarrow 1)$}  Assume $S\subseteq\N$ be such that $S$ is a cycle in $G_p$, and for all $i,j\in S$, $O_i(p)=O_j(p)$. Then, for all $j\in S$, and all $x\in\mathbb{N}$, $O^{x}_j(p)=O_i(p)$ and for all $f\in\N\notin S$ with distance $d$ from $f$ to $i$, for all $x\in\mathbb{N}$, such that $x\geq d$, $O^{x}_f(p)=O_i(p)$.   \end{proof}  This trivially implies that the class of opinion profiles which guarantees convergence for {\em any} influence profile, is the one where everybody agrees on everything already. Note that the only stable distributions of opinions are the ones where, in each connected component in $G$, all members have the same opinion, i.e, on BDPs, converging and reaching a consensus (within each connected component) are equivalent, unlike equivalent.   %,unlike  in the stochastic case. Moreover, for an influence profile where influence graphs have at most diameter $d$ and the smallest cycle in components with diameter $d$ is of length $c$, it is easy to see that if a consensus is reached, it will be reached in at most $d-c$ steps, which is at most $n-1$. \medskip