dual_decomposition: trim to the minimal-counterexample section

Remove the Introduction and Strategy sections and everything after the
separating-cycle definition (no-separating-triangle lemma, 5-connectivity
proposition, and the Step 2-6 stubs). Rename the section heading from
"Step 1: The minimal counterexample" to "The minimal counterexample", drop
the now-unused separating-cycle definition, and adjust the lead-in to mention
only the degree reduction. Remaining: reduction-to-triangulations lemma,
minimal-counterexample definition, |V|>=12 remark, and minimum-degree-5 lemma.

Co-Authored-By: Claude Opus 4.7 (1M context) <noreply@anthropic.com>

papers/dual_decomposition_minimal_counterexamples/paper.tex

@@ -48,66 +48,12 @@ graph, edge connectivity, cyclic edge cut, Tait colouring, $3$-edge-colouring}
 
 \maketitle
 
-\section{Introduction}
-
-% TODO: introduction.
-
-By Tait's theorem, a plane triangulation $G$ is properly $4$-vertex-colourable if
-and only if its dual cubic graph $G'$ is properly $3$-edge-colourable. We propose
-to obstruct a minimal counterexample to the Four Colour Theorem by decomposing its
-dual along a minimum edge cut, $3$-edge-colouring the two resulting pieces, and
-recombining the colourings.
-
-\section{Strategy}
-
-% Outline of the intended argument. Each step below is to be made precise and
-% proved in its own section; nothing here is yet established.
-
-The argument proceeds in six steps.
-
-\begin{enumerate}
-  \item \textbf{Minimal counterexample.} Assume a smallest maximal planar graph
-        $G$ admitting no proper $4$-colouring. Standard reductions let us take $G$
-        to be $5$-connected, so its dual $G'$ is a cyclically $5$-edge-connected
-        cubic plane graph.
-
-  \item \textbf{Dualise.} Pass to the dual cubic plane graph $G'$, where, by
-        Tait, $4$-colourability of $G$ is equivalent to $3$-edge-colourability of
-        $G'$.
-
-  \item \textbf{Minimum edge cut.} Take a smallest edge cut of $G'$ separating it
-        into two components $A$ and $B$. (The size of the smallest cyclic edge cut
-        equals the cyclic edge connectivity of $G'$; record which cut is used and
-        why it is minimal.)
-
-  \item \textbf{Cap to cubic.} Add the minimum number of edges (and, if needed,
-        vertices) to each of $A$ and $B$ to absorb the severed cut edges, yielding
-        two cubic plane graphs $A^{+}$ and $B^{+}$, each smaller than $G'$.
-
-  \item \textbf{Colour the pieces.} Since $A^{+}$ and $B^{+}$ are smaller than the
-        minimal counterexample's dual, each admits a proper $3$-edge-colouring.
-
-  \item \textbf{Reconnect.} Prove that the two $3$-edge-colourings can be made
-        compatible across the cut --- after a colour permutation and Kempe-type
-        adjustments --- so that they glue to a proper $3$-edge-colouring of $G'$,
-        contradicting the assumption that $G$ is a counterexample.
-\end{enumerate}
-
-\begin{remark}
-Steps (1)--(5) are assembled from standard machinery (minimality, Tait duality,
-edge-cut/girth duality, and the inductive existence of colourings on smaller cubic
-graphs). The load-bearing claim is step (6): the compatibility of the two
-boundary colourings across the cut. The parity of colours along an edge cut in a
-$3$-edge-colouring is constrained, and the goal is to show these constraints can
-always be satisfied simultaneously on both sides.
-\end{remark}
-
-\section{Step 1: The minimal counterexample}
+\section{The minimal counterexample}
 
 Throughout, a \emph{triangulation} is a simple plane graph, with a fixed
 embedding, in which every face --- including the outer face --- is bounded by a
-triangle. We first reduce to triangulations, then record the degree and
-connectivity properties a smallest counterexample must have.
+triangle. We first reduce to triangulations, then record the degree properties a
+smallest counterexample must have.
 
 \begin{lemma}[Reduction to triangulations]
 \label{lem:triangulate}
@@ -169,78 +115,4 @@ $G$, a contradiction.
 Hence $\delta(G) \ge 5$.
 \end{proof}
 
-For the connectivity reduction we recall that, in a triangulation, the relevant
-small separators are short cycles.
-
-\begin{definition}[Separating cycle]
-A cycle $C$ in a triangulation $G$ is \emph{separating} if it is not a face
-boundary; equivalently, both the open interior and the open exterior of $C$ (with
-respect to the embedding) contain at least one vertex of $G$.
-\end{definition}
-
-\begin{lemma}[No separating triangle]
-\label{lem:no-sep-tri}
-A minimal counterexample $G$ has no separating triangle; consequently $G$ is
-$4$-connected.
-\end{lemma}
-
-\begin{proof}
-Suppose $T = xyz$ is a separating triangle. It splits $G$ into the triangulation
-$G_{\mathrm{in}}$ induced by $T$ and the vertices interior to $T$, and the
-triangulation $G_{\mathrm{out}}$ induced by $T$ and the vertices exterior to $T$;
-in each, $T$ bounds a face. Both are triangulations on fewer vertices than $G$,
-so by minimality each has a proper $4$-colouring. Permuting colours in
-$G_{\mathrm{out}}$, we may assume the two colourings agree on $\{x,y,z\}$ (the
-three corners receive distinct colours in each, so a colour permutation aligns
-them). The two colourings then glue to a proper $4$-colouring of $G$, a
-contradiction. A triangulation with no separating triangle and $\delta \ge 5$ is
-$4$-connected, since a minimal vertex cut of size $\le 3$ in a triangulation is
-the vertex set of a separating cycle of that length, and a $3$-cut would give a
-separating triangle.
-\end{proof}
-
-\begin{proposition}[$5$-connectivity]
-\label{prop:5conn}
-A minimal counterexample $G$ may be taken to be $5$-connected; equivalently, $G$
-has no separating triangle and no separating $4$-cycle.
-\end{proposition}
-
-\begin{proof}[Discussion]
-By Lemma~\ref{lem:no-sep-tri} it remains to eliminate separating $4$-cycles. This
-is the classical reducibility step of Birkhoff: a shortest separating $4$-cycle
-$C = w x y z$ bounds inner and outer triangulations; one $4$-colours both by
-minimality and reconciles their colourings on $C$ using Kempe-chain exchanges,
-the only obstruction being the two ways the four corners can be coloured with
-three or four colours. Carrying out the exchanges shows the colourings can always
-be aligned, so $G$ has no separating $4$-cycle. As with
-Lemma~\ref{lem:no-sep-tri}, the absence of separating $3$- and $4$-cycles in a
-triangulation with $\delta \ge 5$ is equivalent to $5$-connectivity. We take this
-reduction as standard and refer to the literature on the Four Colour Theorem for
-the detailed Kempe-chain bookkeeping.
-\end{proof}
-
-The output of this step --- a $5$-connected triangulation $G$ with $\delta(G) \ge
-5$ and no separating triangle or $4$-cycle --- is exactly the hypothesis the dual
-construction of Step~2 consumes.
-
-\section{Step 2: The dual cubic graph}
-
-% TODO.
-
-\section{Step 3: The minimum edge cut}
-
-% TODO.
-
-\section{Step 4: Capping to cubic planar graphs}
-
-% TODO.
-
-\section{Step 5: Colouring the pieces}
-
-% TODO.
-
-\section{Step 6: Reconnecting the colourings}
-
-% TODO. (The crux.)
-
 \end{document}