ZoĆ© Christoff edited This_example_shows_that_direction__.tex  almost 8 years ago

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\begin{theorem}\label{theorem:resistantsufficient}  Let $\G$ be an influence profile, $\O$ be an opinion profile, and $\C$ a set of constraints. Then the following holds:  \begin{enumerate}  \item[] If all $p\in \Atoms$, for all $C\subseteq\N$ $S\subseteq\N$  such that $C$ $S$  is a cycle in $G_{p}$, and all $i,j\in C$: $\O_i(p)=\O_j(p)$, S$: $O_i(p)=O_j(p)$,  then \item[] the resistant BDP for $\O$, $\G$ and $\C$ converges.  \end{enumerate}  \end{theorem}  \begin{proof}  Assume that for all $p\in \Atoms$, for all $C\subseteq\N$ $S\subseteq\N$  such that $C$ $S$  is a cycle in $G_{p}$, for all $i,j\in C$: $\O_i(p)=\O_j(p)$. Consider an arbitrary $p\in \Atoms$, and an arbitrary $i\in \N$. Let $k$ be the distance from $i$ to $l$, where $l$ is the closest agent in a cycle $C\subseteq\N$ of $G_p$. We show that for any such $k\in\mathbb{N}$, there exists an $n \in \mathbb{N}$, such that $\O^{n}_ i(p)$ is stable.  \begin{itemize}  \item If $k=0$: $i\in C$, hence $\O_{i}(p)$ is stable.