deletions | additions       diff --git a/appendix_section_Irrational_individuals_Theoretically__.tex b/appendix_section_Irrational_individuals_Theoretically__.tex  index 7f55354..e41ac38 100644  --- a/appendix_section_Irrational_individuals_Theoretically__.tex  +++ b/appendix_section_Irrational_individuals_Theoretically__.tex 
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\item[] FOR ALL $j\in\{1,\ldots,m\}$, $G_pj$ IS NOT TWO-COLORABLE, OR  \item[] THERE IS A $j\in\{1,\ldots,m\}$, SUCH THAT $\O$ PROPERLY COLORS $G_pj$, THAT IS SUCH THAT FOR ALL $i,j\in\N$ WITH $iRj$, $O{p_i}=1$ if and only if $O{p_j}=1$.  \end{itemize}  \end{proposition} \subsection{Backdrop: extra graph theoretical remark (WE DON'T NEED THIS ONE)}  First, for functional graphs, no node in a cycle can have any successor outside of the cycle. Therefore, all cycles have to be vertex-disjoint:  \begin{fact} \label{fact:vertex}  Let $G$ be an influence graph. All cycles in $G$ are vertex-disjoint.  \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}