diff --git a/papers/face_monochromatic_pairs/COMMENTARY.md b/papers/face_monochromatic_pairs/COMMENTARY.md
index de48fcf..4a4395d 100644
--- a/papers/face_monochromatic_pairs/COMMENTARY.md
+++ b/papers/face_monochromatic_pairs/COMMENTARY.md
@@ -107,6 +107,36 @@ or revealed why they don't yield a contradiction on their own:
   `experiments/check_min_flip_structure.py`,
   `experiments/check_minority_location.py`.
 
+## Failed proof route via edge-sharing Kempe constancy
+
+A natural-looking strategy to prove Conjecture 5.1 was:
+
+> **Conjecture (now disproved):** If $K_0$ is an $\{a,b\}$-Kempe cycle
+> of $\varphi$ and $K_1$ is an $\{a,c\}$-Kempe cycle of $\varphi$ that
+> shares an edge with $K_0$, then $h_\varphi$ cannot be constant on
+> both $V(K_0)$ and $V(K_1)$ simultaneously.
+
+Combined with Lemma 5.3 (no clause-3 witness $\Rightarrow$ constancy
+on both $V(K_b)$ and $V(K_c)$), this would have closed Conjecture 5.1.
+It is **false** by a concrete counterexample (Figure in `paper.tex` at
+`\ref{fig:no-two-constant-kempe-counterexample}`). A partial proof
+attempt is preserved in the paper alongside the disproof:
+
+- Step 1 (local CW structure) — unconditional.
+- Step 2 (forced odd-crossing $\Rightarrow |E(K_0) \cap E(K_1)|$ even
+  and $\geq 2$) — unconditional.
+- Step 3 (Heawood face-sum mod 3) — unconditional but does not yield
+  a contradiction on its own.
+- Step 4 (lune-face Case A) — closes the sub-case where two shared
+  a-edges are consecutive on \emph{both} cycles (automatic when
+  $|E(K_0) \cap E(K_1)| = 2$).
+- Step 5 (general case) — open / now known false via the
+  counterexample.
+
+The counterexample shows the general case of the conjecture is
+unsalvageable; the search for a structural proof of Conjecture 5.1
+will need a different angle.
+
 ## Snapshot date
 
 This commentary is current as of commit `e688037`.
diff --git a/papers/face_monochromatic_pairs/paper.pdf b/papers/face_monochromatic_pairs/paper.pdf
index 1ec1cf7..0c6d559 100644
Binary files a/papers/face_monochromatic_pairs/paper.pdf and b/papers/face_monochromatic_pairs/paper.pdf differ
diff --git a/papers/face_monochromatic_pairs/paper.tex b/papers/face_monochromatic_pairs/paper.tex
index fc14793..984e481 100644
--- a/papers/face_monochromatic_pairs/paper.tex
+++ b/papers/face_monochromatic_pairs/paper.tex
@@ -814,8 +814,8 @@ that proof produces a clauses-(1)--(3) witness without ever needing to
 inspect the other Kempe cycle.
 \end{proof}
 
-\begin{theorem}[Constant Heawood on two edge-sharing Kempe cycles is impossible]
-\label{thm:no-two-constant-kempe-cycles}
+\begin{conjecture}[Constant Heawood on two edge-sharing Kempe cycles --- \textbf{FALSE}]
+\label{conj:no-two-constant-kempe-cycles}
 Let $H$ be a cubic plane graph with a proper $3$-edge-colouring
 $\varphi$, fix a colour $a \in \{1, 2, 3\}$, and let $\{b, c\} =
 \{1, 2, 3\} \setminus \{a\}$. Let $K_0$ be an $\{a, b\}$-Kempe cycle
@@ -823,9 +823,56 @@ of $\varphi$ and $K_1$ an $\{a, c\}$-Kempe cycle of $\varphi$ such that
 $E(K_0) \cap E(K_1) \neq \emptyset$ (equivalently, $K_0$ and $K_1$
 share at least one colour-$a$ edge). If $h_\varphi$ is constant on
 $V(K_0)$, then $h_\varphi$ is \emph{not} constant on $V(K_1)$.
-\end{theorem}
+\end{conjecture}
+
+\begin{remark}[Disproof of
+Conjecture~\ref{conj:no-two-constant-kempe-cycles}]
+\label{rem:no-two-constant-kempe-cycles-counterexample}
+Conjecture~\ref{conj:no-two-constant-kempe-cycles} is \emph{false}:
+there exists a cubic plane graph $H$ with a proper $3$-edge-colouring
+$\varphi$ admitting an $\{a, b\}$-Kempe cycle $K_0$ and an
+$\{a, c\}$-Kempe cycle $K_1$ which share a colour-$a$ edge and on which
+$h_\varphi$ is simultaneously constant on $V(K_0)$ and on $V(K_1)$. A
+concrete counterexample is recorded in
+Figure~\ref{fig:no-two-constant-kempe-counterexample}.
+
+The (partial) proof attempt below establishes the constraint
+$|E(K_0) \cap E(K_1)| \geq 2$ (Step~2) and closes the sub-case where
+two shared a-edges are consecutive on \emph{both} cycles
+(``Case~A''/Step~4), but the general claim fails because in the
+counterexample no pair of shared a-edges is consecutive on both
+cycles --- the $K_1$-arc between two $K_0$-consecutive shared a-edges
+itself passes through other shared a-edges, breaking the lune-face
+assumption.
+\end{remark}
+
+\begin{figure}[h]
+\centering
+% TODO: replace placeholder with the actual counterexample drawing
+% (e.g., the whiteboard photo or a TikZ rendering).
+\fbox{\parbox{0.7\textwidth}{\centering
+\emph{Placeholder for the counterexample figure.}\\[2pt]
+A cubic plane graph $H$ with three edge colours
+(red, blue, teal) showing two Kempe cycles sharing a colour edge,
+on which $h_\varphi$ is simultaneously constant on both cycles.\\
+See \texttt{figures/no-two-constant-kempe-counterexample.\{png,pdf\}}
+once the image is committed to the repo.}}
+\caption{Counterexample to
+Conjecture~\ref{conj:no-two-constant-kempe-cycles}.}
+\label{fig:no-two-constant-kempe-counterexample}
+\end{figure}
+
+\begin{proof}[Partial proof attempt (now superseded by
+Remark~\ref{rem:no-two-constant-kempe-cycles-counterexample})]
+\textbf{Note.}
+Conjecture~\ref{conj:no-two-constant-kempe-cycles} is false (see the
+counterexample in
+Remark~\ref{rem:no-two-constant-kempe-cycles-counterexample} /
+Figure~\ref{fig:no-two-constant-kempe-counterexample}). The argument
+below is preserved as partial progress: Steps~1--2 are unconditional
+and Step~4 closes the sub-case where some pair of shared a-edges is
+consecutive on both $K_0$ and $K_1$.
 
-\begin{proof}[Proof attempt]
 Suppose for contradiction that $h_\varphi$ is constant on both
 $V(K_0)$ and $V(K_1)$, and that $K_0, K_1$ share a colour-$a$ edge
 $e = (u, w)$. Then $u, w \in V(K_0) \cap V(K_1)$ are consecutive on
@@ -935,14 +982,16 @@ contribution is a multiple of $3$ and drops out.) Hence
 \quad\text{for every face } \Phi \text{ of } K_0 \cup K_1.
 \end{equation}
 
-\textbf{Step 4 (lune-face contradiction, Case A).} Pick any two
-\emph{consecutive} shared a-edges on the $K_0$-walk: $e_1 = (p, p')$
-and $e_2 = (q, q')$, separated by a $K_0$-arc $A_1$ from $p'$ to $q$
-of length $m$. The arc begins with the colour-$b$ edge at $p'$ and
-ends with the colour-$b$ edge at $q$, so $m$ is odd. Both $K_0$ and
-$K_1$ traverse $e_1$ and $e_2$, so the four vertices
-$\{p, p', q, q'\}$ are shared. Two cases for the cyclic order in which
-$K_1$ visits these four:
+\textbf{Step 4 (lune-face contradiction, ``Case A'' sub-case).}
+Suppose there exist two shared a-edges $e_1 = (p, p')$ and
+$e_2 = (q, q')$ which are consecutive on \emph{both} the $K_0$-walk
+\emph{and} the $K_1$-walk (so the $K_0$-arc from $p'$ to $q$ and the
+$K_1$-arc from $p'$ to $q$ both have all-non-shared interior). In
+particular, when $|E(K_0) \cap E(K_1)| = 2$ this is automatic, but in
+general it is an extra hypothesis. Let $A_1$ be the $K_0$-arc from
+$p'$ to $q$ of length $m$. The arc begins with the colour-$b$ edge at
+$p'$ and ends with the colour-$b$ edge at $q$, so $m$ is odd. Two
+cases for the cyclic order in which $K_1$ visits $\{p, p', q, q'\}$:
 
 \begin{description}
 \item[Case A.] $K_1$ visits them in the same cyclic order