deletions | additions       diff --git a/appendix_section_Irrational_individuals_Theoretically__.tex b/appendix_section_Irrational_individuals_Theoretically__.tex  index caae86f..1bdb446 100644  --- a/appendix_section_Irrational_individuals_Theoretically__.tex  +++ b/appendix_section_Irrational_individuals_Theoretically__.tex 
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\end{fact}  \begin{proof}  Let $G$ be an influence graph, such that it contains two cycles which are not vertex disjoint. Then there is a vertex $i$ contained in both cycles, such that $i$ has a different successor in each cycle. But since $G$ is functional, $i$ has at most one successor. Contradiction.  \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%BELOW THIS POINT IS ONLY OLDER STUFF%%%%%%%%%%%%%%  There are two stability-related questions, given an initial model:   \begin{enumerate}  \item Is the model stable in the sense that it is a limit point of the dynamics (that is it is a fixpoint of $\Box$, if my thoughts above are correct)? YES!  \item Is the model such that it will eventually reach a limit point, or oscillate instead?  \end{enumerate}  As to 1), if the validity of the fixpoint equation $p \leftrightarrow \Box p $ fails in the model, then we are not in a stable model. Notice that your Proposition 2, if one assumes the frame is functional, states precisely that the model is unstable whenever $p \leftrightarrow \Box p $ is not valid in the model.  As to 2), if neither p nor ~p is a post-fixpoint of $Box$ then the frame is 2-colorable (and vice versa). And that means the model cannot possibly stabilize. @DAVIDE: NO, AS MENTIONED THIS MORNING, THIS DEPENDS ON THE FRAMES BEING SYMMETRIC, SO I NEED TO THINK A BIT BETTER INTO WHAT THE FORMULA MEANS WITHOUT SYMMETRY IMPOSED. I WILL COME BACK TO THIS VERY SOON. (we have seen that for instance a simple triangle can carry non-stabilizing opinions).   The Boolean case of the theorem by Jackson should state conditions under which, no matter what the initial assignment for $p$ is, the model will always stabilize (that is either $p$ or $\lnot p$ is a post-fixpoint of $\Box$). So Jackson’s theorem should state conditions that make a graph non-2-colorable.   Indeed, Jackson's theorem excludes 2-colorable graphs, since they are not aperiodic (the greater divisor of all cycles lengths is 2). Jackson results implies that 2-colorable graphs do not guarantee stabilization for any distribution of opinions (and of influence weights). However, more can be said.   For instance, in the case of \emph{symmetric} frames, all and only models which are not proper two-coloring of the graphs will stabilize. This gives us a full characterization of stabilization. There exists only 2 such colorings (for each connected component) of a graph. And the loop is always of size two.NOTE THAT THIS LAST PARAGRAPH DOES NOT RELY ON THE FRAME BEING FUNCTIONAL, SO THIS IS A MORE GENERAL RESULT ABOUT UNANIMITY RULE IN SYMMETRIC GRAPHS. MAYBE WE SHOULD ALSO INTRODUCE IT AS SUCH THEN.   QUESTIONS:   WHAT PROPERTY DOES THE FORMULA CHARACTERIZING STABILIZATION CORRESPOND TO ON NON-SYMMETRIC FRAMES?   WHAT ARE THE CONDITIONS FOR STABILIZATION IN THE RESTRICTED CASE OF BDG (COMPARED TO THE GENERAL CASE OF JACKSON'S THEOREM