Consider four nodes of $H(\Gamma)$, indexed by $J_0$, $J_1$, $J_2$ and $J_3$.
Suppose $J_0$ and $J_1$ are adjacent, as are $J_2$ and $J_3$.
These intersections may be configured as seen in Figure \ref{fig:config four}, where there are $n$ edges between nodes $J_1$ and $J_2$ and $m$ edges between $J_0$ and $J_3$.
Without loss of generality suppose $n+2\leq m$.
These intersections may be configured as seen in Figure \ref{fig:config four}, where there are at least $n$ edges between nodes $J_1$ and $J_2$ and at least $m$ edges between $J_0$ and $J_3$.
Without loss of generality $n\leq m$.
\begin{figure}
\centering
\begin{tikzpicture}
...
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@@ -64,7 +64,91 @@ Suppose that the edge between $J_0$ and $J_1$ belongs to $A_b(\Gamma)$, as does
Then nodes of $H(\Gamma\cup\set{\eta})$ will be calculated for $J_0\cup J_1$ and $J_2\cup J_3$.
These two nodes will be adjacent if $|(J_0\cup J_1)\cap(J_2\cup J_3)| = |J_0\cup J_1|-1$.
Note that if $0 < n = |J_1|$ then $J_2$ has no indices in common with $J_1$.
Likewise, $J_0$ has no indices in common with $J_3$.
The two nodes of the next generation therefore cannot be adjacent.
Since $J_1$ and $J_2$ are separated by $n$ edges $|J_1\cap J_2| = |J_1| - n$.
The nodes $J_0$ and $J_2$ are separated by $n+1$ edges, and so $|J_0\cap J_2| = |J_1| -(n+1)$.
Therefore, $J_2$ contains $J_1\setminus J_0$ but does not contain any of $J_0\setminus J_1$ and $(J_0\cup J_1)\cap J_2= J_1\cap J_2$.
Likewise, $|J_1\cap J_3| = |J_1| -(n+1)$ and $J_1$ contains none of $J_3\setminus J_2$.
The nodes $J_0$ and $J_3$ are separated by the smaller of $n+2$ and $m$ edges.
Since $m \geq n$, this means that $|J_1| -(n+2)\leq |J_0\cap J_3| \leq |J_1| - n$.
We consider each of the three possibilities in turn.
\newcommand{\PolyConfig}[1]{
\foreach\x in {0,1,...,#1}{
\coordinate (P\x) at (90-\x*360/#1+180/#1:1);}
\foreach\x in {0,1}{\filldraw[black] (P\x) circle (1pt) node[above] {$J_\x$};}
\foreach\x in {2,3}{\filldraw[black] (P\x) circle (1pt) node[below] {$J_\x$};}
\foreach\x in {4,...,#1}{\filldraw[black] (P\x) circle (1pt);}
