Add contraction-lift proof-strategy sketch toward Conjecture 5.7

Section 5.6 sketches an inductive route to the simple-resolution md4
surjectivity conjecture:

- Lemma 5.8 (good contraction): every md4 triangulation on n >= 7
  vertices has a degree-4 vertex with an md4-preserving diagonal
  contraction. Empirically true at n=7..11; proof obligation called
  out.
- Lemma 5.9 (lift): given a labelled preimage of the contracted
  triangulation, reinserting the contracted vertex at the
  diagonal-bounded quadrilateral yields a preimage of the original
  triangulation. Proof obligation called out.
- Inductive scheme paragraph chains the two lemmas with the octahedron
  at n=6 as the base case, citing the n=7 hand-verification (already
  scripted in experiments/inductive_lift_check.py).

Lemmas are stated without proof; the three remaining proof
obligations are explicit.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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@@ -539,6 +539,102 @@ vertices is not reachable by the algorithm; at $n = 10$, two further
iso-classes with four degree-$3$ vertices and high-degree hubs fail to
appear among algorithm outputs.
\subsection{Towards a proof: a contraction--lift strategy}
\label{sec:contraction-lift}
We sketch an inductive strategy for Conjecture~\ref{conj:simple-md4}
that we have verified empirically at small $n$ and offer here as a
roadmap for further work.
Let $T$ be a plane triangulation with minimum degree at least $4$, and
let $v \in V(T)$ be a degree-$4$ vertex with cyclic neighbors
$a, b, c, d$ (in the cyclic order inherited from $T$'s planar
embedding). Removing $v$ from $T$ exposes the $4$-cycle $abcd$, which we
retriangulate by adding one of the two diagonals $(a, c)$ or $(b, d)$.
We call this operation \emph{contraction at $v$ along diagonal
$(a, c)$}, denoted $T_{v, (a, c)}$. The contraction is \emph{valid} when
the chosen diagonal is not already an edge of $T$.
\begin{lemma}[Good contraction]
\label{lem:good-contraction}
Let $T$ be a plane triangulation on $n \geq 7$ vertices with minimum
degree at least $4$. Then there exist a degree-$4$ vertex $v \in V(T)$,
with cyclic neighbors $a, b, c, d$, and an unordered pair
$\{a, c\}$ such that:
\begin{enumerate}
\item $(a, c) \not\in E(T)$;
\item $\deg_T(b) \geq 5$ and $\deg_T(d) \geq 5$.
\end{enumerate}
Under these conditions $T_{v, (a, c)}$ is a plane triangulation on
$n - 1$ vertices with minimum degree at least $4$.
\end{lemma}
The conditions of Lemma~\ref{lem:good-contraction} ensure that the
contraction is valid (1) and md$_4$-preserving (2): the only vertices
whose degree changes under $T \to T_{v, (a, c)}$ are $a, b, c, d$, with
$\deg(a)$ and $\deg(c)$ unchanged (each loses the edge to $v$ but gains
the edge from the diagonal), while $\deg(b)$ and $\deg(d)$ each decrease
by $1$.
The lemma is empirically true at $n = 7, \ldots, 11$ for every md$_4$
iso-class; we conjecture it holds for all $n \geq 7$. The $n = 6$ case
is excluded: the unique md$_4$ iso-class is the octahedron, in which
every vertex has all four cyclic neighbors at degree $4$ and so no
contraction preserves md$_4$. The octahedron is therefore the base case
of the proposed induction.
\begin{lemma}[Lift]
\label{lem:lift}
Let $T$ be a plane triangulation with minimum degree at least $4$, and
suppose Lemma~\ref{lem:good-contraction} applies via vertex $v$ and
diagonal $(a, c)$ with $T_{v, (a, c)}$ the resulting contraction. Let
$H$ be a plane triangulation on $V(T_{v, (a, c)}) = V(T) \setminus \{v\}$
and $S$ a level source of $H$ such that the algorithm of
Section~\ref{sec:flip-algorithm} applied to $(H, S)$ produces
$T_{v, (a, c)}$ as a labelled simple graph. Define the \emph{lift}
$G \;:=\; H[a, b, c, d, v]$ by:
\begin{itemize}
\item adding vertex $v$ to $V(H)$;
\item removing the edge $(a, c)$ from $E(H)$;
\item adding the four edges $(v, a), (v, b), (v, c), (v, d)$.
\end{itemize}
Then $G$ is a plane triangulation on $|V(T)|$ vertices, and the
algorithm of Section~\ref{sec:flip-algorithm} applied to $(G, S)$
produces $T$.
\end{lemma}
Lemma~\ref{lem:lift} requires that $(a, c) \in E(H)$ and that the two
triangles of $H$ bordering $(a, c)$ have boundary
$\{a, b, c, d\}$. When these conditions hold, the lift restores a
degree-$4$ vertex $v$ inserted into the quadrilateral $abcd$; when they
fail, the lift is undefined and a different labelled preimage $H$ must
be chosen.
\paragraph{Inductive scheme.}
Conjecture~\ref{conj:simple-md4} would follow from
Lemmas~\ref{lem:good-contraction} and~\ref{lem:lift} together with the
existence at each step of a labelled preimage $H$ satisfying the lift's
side conditions. The base case is the octahedron at $n = 6$, which is
empirically a simple level resolution
(Observation~\ref{obs:md4-simple-resolution}). The inductive step takes
an md$_4$ target $T$ on $n$ vertices, applies
Lemma~\ref{lem:good-contraction} to obtain an md$_4$ contraction
$T_{v, (a, c)}$ on $n - 1$ vertices, invokes the inductive hypothesis to
produce a labelled preimage $H$, and applies Lemma~\ref{lem:lift} to
lift $H$ to $G$ with $\mathrm{alg}(G, S) = T$.
We have verified the entire scheme by hand for the unique md$_4$
iso-class at $n = 7$: contraction at $v = 2$ along diagonal $(4, 3)$
yields the octahedron on six vertices labelled $\{0, 1, 3, 4, 5, 6\}$;
a labelled preimage $H$ exists with source $S = \{0, 1, 6\}$; lifting
along $(4, 3, v = 2)$ produces a triangulation $G$ on seven vertices on
which the algorithm with source $S$ recovers $T$ exactly. The principal
remaining work is a proof of Lemma~\ref{lem:good-contraction} for all
$n \geq 7$, a proof of Lemma~\ref{lem:lift} (which involves analysing
how the algorithm's depth-guided flips interact with the added vertex
$v$), and a guarantee that a label-faithful preimage $H$ always
exists.
\begin{question}
\label{q:terminate-all-n}
Does Phase~1 terminate for all $(G, S)$? Equivalently, is there an