ZoĆ© Christoff edited This_example_shows_that_direction__.tex  almost 8 years ago

Commit id: 28d611e9bc974f990afb43d3c63d10ef7221a379

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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 $O^{m}_i(p)=O^m_j(p)$, then \O^{m}_i(p)$ $O^{m}_i(p)$  is stable. \item if $O^{m}_i(p) \neq O^m_j(p)$, then:  %$\O^{m+1}_i(p) = \O^{m}_i(p)$: $i$ does not change his opinion at stage $m+1$ but this does not guarantee that this won't change later on:  \begin{enumerate}  \item If there is a $t\in\mathbb{N}$ such that $\bigwedge_{p \in \Atoms} \O^{m+t}_{R_p(i)}(p) O^{m+t}_{R_p(i)}(p)  \wedge \bigwedge_{\varphi \in \C}$ is consistent, then $\O^{m+t+1}_i(p) $O^{m+t+1}_i(p)  = \O^{m}_j(p)$, O^{m}_j(p)$,  and $\O^{m+t+1}_i(p)$ $O^{m+t+1}_i(p)$  is stable. \item If there is no such $t$, then $i$'s opinion on $p$ will never change: $\O^{m}_i(p)$ $O^{m}_i(p)$  is stable. \end{enumerate}  \end{enumerate}  \end{enumerate}