ZoĆ© Christoff edited This_example_shows_that_direction__.tex  almost 8 years ago

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Let $p\in \Atoms$ and $C\subseteq\N$ be a cycle of $G_p$. Consider an arbitrary $i\in \N$. $i$ is at some distance $k$ from some agent $l\in C$. We show that for any such distance $k\in\mathbb{N}$, there is an $n \in \mathbb{N}$, such that $O^{n}_ i(p)$ is stable.  \begin{itemize}  \item If $k=0$: $i$ is in the cycle $C$, hence $\O_{i}(p)$ is stable.   \item if $k=n$: $k=n+1$:  Assume that for agent $j$ at distance $n-1$ $n$  from agent $l$, for some $m \in \mathbb{N}$, $O^{m}_ j(p)$ is stable. We need to consider the following cases: \begin{enumerate}  \item 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 $\bigwedge_{p \in \Atoms} \O^{m}_{R_p(i)}(p) \wedge \bigwedge_{\varphi \in \C}$ is not consistent, then:  \begin{enumerate}  \item if $\O^{m}_i(p)=\O^m_j(p)$, $\O^{m}_i(p)$ is stable.   \item if $\O^{m}_i(p) \neq O^m_j(p)$, $\O^{m+1}_i(p) then:  %$\O^{m+1}_i(p)  = \O^{m}_i(p)$: $i$ does not change his opinion at stage $m+1$ but nothing guarantees this does not guarantee  that this won't change later on: \begin{enumerate}  \item If there is a $t\in\mathbb{N}$ such that $\bigwedge_{p \in \Atoms} \O^{m}_{R_p(i)}(p) \O^{m+t}_{R_p(i)}(p)  \wedge \bigwedge_{\varphi \in \C}$ 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.   \end{enumerate}  \end{enumerate}  \end{enumerate}  \end{enumerate} \end{itemize}    \end{proof}