deletions | additions       diff --git a/This_example_shows_that_direction__.tex b/This_example_shows_that_direction__.tex  index eec3d33..5c6f72b 100644  --- a/This_example_shows_that_direction__.tex  +++ b/This_example_shows_that_direction__.tex 
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Consider an arbitrary issue $p\in \Atoms$, and an arbitrary agent $i\in \N$. $i$ is at some distance $k$ from some agent $l$ in a cycle $C$ of $G_p$. We show that for any $k\in\mathbb{N}$, $O_i(p)$ stabilizes.  \begin{itemize}  \item If $k=0$: $i$ is in the cycle $C$, hence $\O_{i}(p)$ is already stable.   \item if $k=n$: Assume that for agent $j$ at distance $n-1$ from agent $l$, $j$'s opinion on $p$ stabilizes, that is, stabilizes:  for some $m \in \mathbb{N}$, $O^{m}_ j(p)$ is stable.We need to consider the following cases:  \begin{itemize}  \item if If  $\bigwedge_{p \in \Atoms} \O^{m}_{R_p(i)}(p) \wedge \bigwedge_{\varphi \in \C}$ is consistent, then $\O^{m+1}_i(p)=\O^m_j(p)$, and $\O^{m+1}_i(p)$ is stable.   \item if If  $\bigwedge_{p \in \Atoms} \O^{m}_{R_p(i)}(p) \wedge \bigwedge_{\varphi \in \C}$ is not consistent, then then:  \begin{itemize}  \item if $\O^{m}_i(p)=\O^m_j(p)$, $\O^{m}_i(p)$ is stable.   \item if $\O^{m}_i(p) \neq O^m_j(p)$, $\O^{m+1}_j(p) $\O^{m+1}_i(p)  = \O^{m}_j(p)$. \O^{m}_i(p)$:  $i$ does not change his opinion at stage $m+1$ but nothing guarantees that this won't change later on. Consider the two possible subcases: on:  \begin{itemize}  \item If there is a $t\in\mathbb{N}$ such that $\bigwedge_{p \in \Atoms} \O^{m}_{R_p(i)}(p) \wedge \bigwedge_{\varphi \in \C}$ is consistent, then $\O^{m+t+1}_i(p) = \O^{m}_j(p)$, and $\O^{m+t+1}_i(p)$ is stable.  \item If there is no such $t$, then $i$'s opinion on $p$ will never change: $\O^{m}_i(p)$ is stable.