this is for holding javascript data
ZoƩ 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.