this is for holding javascript data
ZoƩ Christoff edited This_example_shows_that_direction__.tex
almost 8 years ago
Commit id: 7e026d155efd971649b06cb8ecd3866f0468cb59
deletions | additions
diff --git a/This_example_shows_that_direction__.tex b/This_example_shows_that_direction__.tex
index 7e53fab..c9b7ecc 100644
--- a/This_example_shows_that_direction__.tex
+++ b/This_example_shows_that_direction__.tex
...
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}