\section{The fold, and the proof of Theorem~\ref*{thm:smooth}}\label{sec:fold}

Throughout this section $c>1$, and $\mu_1,\mu_2$ are as in Theorem~\ref{thm:smooth}: ergodic
translation-invariant gradient Gibbs measures of slope $0$; by Lemma~\ref{lem:SFE}(iii) they are equilibrium
states. By
Corollary~\ref{cor:cases} it remains to treat the case
\[
\mub(E_{\Z\Z})=1,
\]
which we assume until the end of Subsection~\ref{subsec:BK}. Fix the band width $M:=2$ (any integer
$M\ge2$ would do; $M\ge2$ is used in Lemma~\ref{lem:band}(iv)). Only $I^-=\Z$ is used in the
construction.

The idea is Sheffield's repeated swap \cite[Section~8.6.3]{Sheffield}. When every level carries infinite
clusters of both signs, no single swap is available, but one can exchange the two copies on infinitely
many bands of levels at once, so that the difference is folded into a strip (Figure~\ref{fig:fold}).
The steps are:
\begin{enumerate}
\item[(1)] a \emph{band index} $k(x)$, defined from the infinite minus clusters at the levels $jM$, which
changes by at most one between neighbouring faces (Lemma~\ref{lem:band});
\item[(2)] the \emph{fold} $R$, which applies to each face the composition of swaps prescribed by its band
index; it is admissible and preserves the joint weight (Lemma~\ref{lem:fold}), and in the image every
infinite cluster at the levels $0$ and $M$ is confined to one \emph{double band}, a pair of adjacent
bands (Lemma~\ref{lem:image});
\item[(3)] a \emph{local reconstruction}: on a box, the image and boundary data of size $O(\text{perimeter})$
determine the original, so the fold does not change the free energy per face
(Lemmas~\ref{lem:reconstruct}--\ref{lem:FEsource});
\item[(4)] hence the image is again an equilibrium state with independent residual energies, and the
Lanford--Ruelle theorem makes it a Gibbs measure (Proposition~\ref{prop:imageFE},
Lemmas~\ref{lem:residual}--\ref{lem:LR}). Sheffield obtains this step from his variational principle
\cite[Lemma~8.3.2 and Theorem~2.5.2]{Sheffield}; the image here carries residual energies and a hard
constraint between the copies, and the subshift formulation handles both;
\item[(5)] a Burton--Keane argument limits the number of infinite clusters of the image
(Lemma~\ref{lem:BK}); then either some double band carries an infinite cluster, and the lowest such
double band gives a canonical height, so that both measures are smooth, or none does, and a symmetry of
the image gives a contradiction.
\end{enumerate}

\subsection{The band index}

For a triple $\om$ with $I^-(\om)=\Z$ and $j\in\Z$, let $T_j(\om)$ be the union of the infinite minus FK
clusters at level $jM$.

\begin{lemma}[The band index]\label{lem:band}\leavevmode
\begin{enumerate}
\item[(i)] $T_j\subset T_{j+1}$; $x\in T_j$ implies $D(x)<jM$; and every face lies in $T_j$ for all large
$j$. Hence $k(x):=\min\{j:x\in T_{j+1}\}$ is a well-defined integer, $x\in T_{k(x)+1}\setminus T_{k(x)}$,
and $D(x)<(k(x)+1)M$.
\item[(ii)] If $x\in T_j$, $y$ is NN-adjacent to $x$ and $D(y)<jM$, then $y\in T_j$.
\item[(iii)] For NN-adjacent faces $|k(x)-k(y)|\le1$, and if $k(y)=k(x)+1=:j$ then $D(x)=jM-1$ and
$D(y)=jM$.
\item[(iv)] For the two faces $x,y$ of a diagonal $d$, $|k(x)-k(y)|\le1$, and if $k(y)=k(x)+1=:j$ then
exactly one of the following holds: ($\alpha$) $D(y)=jM$; ($\beta$) $D(y)=jM+1$, $D(x)=jM-1$ and $D=jM$
on $d^\perp$, so that $d$ is $jM$-monotone; ($\gamma$) $D(x)=D(y)=jM-1$, $D=jM$ on $d^\perp$ and
$n_d\ge1$, so that $d$ is a closed $jM$-saddle of sign $-$.
\item[(v)] $k(V_{2M}\om)=k(\om)+2$, and $k$ is covariant under translations.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) $T_j\subset T_{j+1}$ by Lemma~\ref{lem:levels}. Let $C$ be an infinite minus cluster at some level
$j_0M$, $y\in C$, and $y=y_0,\dots,y_m=x$ an NN path. For $jM>\max(j_0M,\max_iD(y_i))$ all $y_i$ are minus
faces at level $jM$, $C$ lies in an infinite minus cluster at level $jM$, and so does $x$.
(ii) An NN edge between two minus faces is an edge of $G_{jM}$.
(iii) Let $k(x)<k(y)$ and $j:=k(x)+1$. Then $x\in T_j$ and $y\notin T_j$, so $D(y)\ge jM$ by~(ii),
$D(x)\le jM-1$, and Lemma~\ref{lem:diff}(i) gives equality in both. Since $x\in T_{j+1}$ and
$D(y)=jM<(j+1)M$, (ii) gives $y\in T_{j+1}$, that is $k(y)=j$.
(iv) Let $z\in d^\perp$. By (iii), $|k(x)-k(y)|\le2$, and if $k(y)=k(x)+2$ then $k(z)=k(x)+1=:j$, and
(iii) for the pairs $\{x,z\}$ and $\{z,y\}$ gives $D(z)=jM$ and $D(z)=(j+1)M-1$, which is impossible
for $M\ge2$. Let $k(y)=k(x)+1=j$; then $x\in T_j$, $y\notin T_j$ and $D(x)\le jM-1$. If $D(y)\ge jM$,
then $D(y)\le D(x)+2$ gives ($\alpha$), or $D(y)=jM+1$ and $D(x)=jM-1$, and then $D=jM$ on $d^\perp$ by
Lemma~\ref{lem:diff}(i), which is ($\beta$). If $D(y)<jM$: if $D(z)<jM$ for some $z\in d^\perp$, then
$z\in T_j$ by (ii) applied to $x$, and $y\in T_j$ by (ii) applied to $z$, a contradiction; so $D=jM$ on
$d^\perp$, $D(x)=D(y)=jM-1$, and the bridge $d$ is closed because $y\notin T_j$, which is ($\gamma$).
(v) is clear.
\end{proof}

\subsection{The folding map}

For $k\in\Z$ let
\[
g_k(p_1,p_2):=\begin{cases}(p_1+kM/2,\ p_2-kM/2)&k\text{ even},\\ (p_2-(k+1)M/2,\ p_1+(k+1)M/2)&k\text{ odd,}\end{cases}
\qquad s_\ell(p_1,p_2):=(p_2-\ell,\ p_1+\ell).
\]
Then $g_j=g_{j-1}\circ s_{jM}$ for every $j$: crossing from band $j-1$ into band $j$ adds one swap at level
$jM$. Define $R(\om):=(\psi_1',\psi_2',n')$ by
\[
(\psi_1',\psi_2')(x):=g_{k(x)}(\psi_1(x),\psi_2(x)),\qquad
n'_d:=n_d+\tfrac12\bigl([\mathsf a(\psi_1';d)+\mathsf a(\psi_2';d)]-[\mathsf a(\psi_1;d)+\mathsf a(\psi_2;d)]\bigr),
\]
and write $D':=\psi_2'-\psi_1'$; thus $D'=D-kM$ where $k$ is even and $D'=(k+1)M-D$ where $k$ is odd.

\input{fig-fold}

\begin{lemma}[The fold is admissible and weight-preserving]\label{lem:fold}\leavevmode
\begin{enumerate}
\item[(i)] $R(\om)$ is a triple: $\psi_1',\psi_2'\in\Omega$ and $n'\ge0$. In the notation of
Lemma~\ref{lem:band}(iv), $n'_d=n_d$ if $k$ is constant on $d$ or in case ($\alpha$), $n'_d=n_d+1$ in
case ($\beta$), and $n'_d=n_d-1$ in case ($\gamma$). Moreover $\tau_d(R\om)=\tau_d(\om)$ for every $d$.
\item[(ii)] $R(V_{2M}\om)=R(\om)$ in $\Om$, and $R$ commutes with translations, so $R$ is defined on
$\Om/\langle V_{2M}\rangle$.
\item[(iii)] $\psi_1'+\psi_2'=\psi_1+\psi_2$.
\item[(iv)] $D'\le M-1$ where $k$ is even and $D'\ge1$ where $k$ is odd. For NN-adjacent faces with
$k(y)=k(x)+1=j$, $(D'(x),D'(y))=(1,0)$ if $j$ is even and $(M-1,M)$ if $j$ is odd.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) If $k(x)=k(y)$, both faces are transformed by the same $g_k$, which shifts each copy by its own
constant and, for odd $k$, exchanges them; NN increments and the $\mathsf a$-sum on a diagonal are preserved.
If $k(y)=k(x)+1=j$, then $(\psi'(x),\psi'(y))=(g_{j-1}\psi(x),g_{j-1}s_{jM}\psi(y))$. For an NN pair,
$D(y)=jM$ by Lemma~\ref{lem:band}(iii), so $s_{jM}$ acts trivially at $y$. For a diagonal, the $\mathsf a$-sum of
the image equals that of the pair in which only $y$ is swapped at level $jM$: in case ($\alpha$) nothing
changes, in case ($\beta$) a $jM$-monotone diagonal becomes a saddle ($+2$), and in case ($\gamma$) a
saddle becomes monotone ($-2$), by Lemma~\ref{lem:diff}(ii); in case ($\gamma$), $n_d\ge1$. (ii)
$g_{k+2}(p_1,p_2+2M)=g_k(p_1,p_2)+(M,M)$. (iii) and (iv) follow from the formulas and
Lemma~\ref{lem:band}(i),(iii).
\end{proof}

\subsection{Structure of the image}

Put $\Pi_m:=\{x:k(x)\in\{2m,2m+1\}\}$ and $\Pi'_m:=\{x:k(x)\in\{2m-1,2m\}\}$, the \emph{double bands}. FK clusters of the image are
those of the triple $R(\om)$, defined with $n'$.

\begin{lemma}[Clusters of the image]\label{lem:image}\leavevmode
\begin{enumerate}
\item[(a)] Every minus FK cluster at level $0$ and every plus FK cluster at level $M$ of $R(\om)$ is
finite.
\item[(b)] Every plus FK cluster at level $0$ of $R(\om)$ is contained in one $\Pi_m$, and every minus FK
cluster at level $M$ of $R(\om)$ is contained in one $\Pi'_m$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) Let $C'$ be a minus cluster at level $0$ of the image. Its faces have $D'<0$, hence even $k$ and
$D<kM$. NN-adjacent faces of $C'$ have the same $k$ (both even, differing by at most $1$), and they are
NN-adjacent minus faces at level $kM$ of $\om$. Let $d=\{x,y\}$ be an open bridge of $C'$: $D'(x)=D'(y)=-1$,
$D'=0$ on $d^\perp$, $n'_d=0$. Then $k(x)=k(y)=:k$ (both even, Lemma~\ref{lem:band}(iv)) and
$D(x)=D(y)=kM-1$. For $z\in d^\perp$, $k(z)\in\{k-1,k,k+1\}$: $k(z)=k+1$ would give $D(z)=(k+2)M$, too
large; $k(z)=k-1$ would give $D(z)=kM$ with $z\in T_k$, impossible; so $k(z)=k$, $D(z)=kM$, $d$ is a
$kM$-saddle of sign $-$ of $\om$, and $n_d=n'_d=0$. So $C'$ is contained in one minus FK cluster at level
$kM$ of $\om$, which is finite because its faces are not in $T_k$. Plus clusters at level $M$ are
treated in the same way with $k$ odd: $D'>M$ if and only if $D<kM$, and a bridge has $D'=M+1$ on $d$ and
$M$ on $d^\perp$, while $k(z)=k\pm1$ would give $D(z)=kM$ with $z\in T_k$, or $D(z)=(k+2)M$.
(b) By Lemma~\ref{lem:fold}(iv), an NN pair in different $\Pi$'s ($k=2m-1$ and $2m$) has $D'=(1,0)$,
so NN edges of the plus graph at level $0$ stay in one $\Pi_m$. Let $d=\{x,y\}$ be a $0$-saddle of sign
$+$ of the image with $k(x)=2m-1$ and $k(y)=2m=:j$. Then $D(x)=jM-1$ and $D(y)=jM+1$, which is case
($\beta$) of Lemma~\ref{lem:band}(iv), and $n'_d=n_d+1\ge1$: the bridge is closed. At level $M$, an NN
pair with $k=2m$ and $2m+1$ has $D'=(M-1,M)$, and a saddle of sign $-$ at level $M$ joining $k(x)=j-1$
(even) and $k(y)=j$ (odd) has $D(x)=jM-1$, $D(y)=jM+1$, case ($\beta$) again.
\end{proof}

\subsection{Local reconstruction}

Let $\Lambda$ be a box, $\partial^*\Lambda$ the set of faces of $\Lambda$ that are $*$-adjacent to its
complement, $b_0$ a fixed corner of $\Lambda$, $\operatorname{diam}$ the NN-diameter of $\Lambda$, and
$L_\Lambda:=M+2+2\operatorname{diam}$. In the following, ``gradients on $\Lambda$'' are the increments
between pairs of faces that both lie in $\Lambda$. For $\om$ with $I^-=\Z$ put
\begin{itemize}
\item $X_\Lambda(\om):=(\nabla\psi_1,\nabla\psi_2$ on $\Lambda$; $D\bmod2M$ on $\Lambda$; $n_d$ for the
diagonals $d\subset\Lambda)$,
\item $Y_\Lambda(\om):=(\nabla\psi_1',\nabla\psi_2'$ on $\Lambda$; $D'\bmod2M$ on $\Lambda$;
$D'^{\mathrm{tr}}:=\max(-1,\min(M+1,D'))$ on $\Lambda$; $n'_d$ for $d\subset\Lambda)$,
\item $B_\Lambda(\om):=\bigl((k(b)-k(b_0))_{b\in\partial^*\Lambda};\ k(b_0)\bmod2;\
\max(-L_\Lambda,\min(L_\Lambda,D'(b_0)))\bigr)$.
\end{itemize}
All three are functions on $\Om/\langle V_{2M}\rangle$. Since $k$ is $1$-Lipschitz along the NN-connected
ring $\partial^*\Lambda$, $B_\Lambda$ takes at most $3^{|\partial^*\Lambda|}\cdot2\cdot(2L_\Lambda+1)$
values.

\begin{lemma}[Reconstruction]\label{lem:reconstruct}
$(X_\Lambda,B_\Lambda)$ and $(Y_\Lambda,B_\Lambda)$ determine each other, and
$\prod_{d\subset\Lambda}q^{\tau_d}$ takes the same value at $X_\Lambda$ and at $Y_\Lambda$.
\end{lemma}

\begin{proof}
The statement about $\tau$ is Lemma~\ref{lem:fold}(i).

\emph{Fact 1.} If $k$ is not constant on $\Lambda$, then $|D'(b_0)|<L_\Lambda$: some NN pair of $\Lambda$ has
different values of $k$, so $D'\in\{0,1,M-1,M\}$ on it (Lemma~\ref{lem:fold}(iv)), and $D'$ is
$1$-Lipschitz along NN steps.

\emph{Fact 2.} $T_j\cap\Lambda$ is the set of faces of $\Lambda$ joined, by a path of minus faces and edges of
$G_{jM}(\om)$ inside $\Lambda$, to a face $b\in\partial^*\Lambda$ with $k(b)<j$. Indeed a face joined to such
a $b$ lies in its infinite cluster; conversely, a path in an infinite cluster from $x\in\Lambda$ to the
complement of $\Lambda$ has a last face in $\Lambda$ before its first exit, and since edges of $G$ join
$*$-adjacent faces, that face is in $\partial^*\Lambda$ and in $T_j$.

\emph{From $(X_\Lambda,B_\Lambda)$ to $Y_\Lambda$.} If the third entry of $B_\Lambda$ is $\pm L_\Lambda$, then
$k\equiv k(b_0)$ on $\Lambda$ by Fact~1; the image on $\Lambda$ is the original ($k(b_0)$ even) or the
original with the copies exchanged ($k(b_0)$ odd), $D'\equiv\pm D\pmod{2M}$, $D'^{\mathrm{tr}}$ is the
constant $-1$ or $M+1$ according to the sign, and $n'=n$. Otherwise $D'(b_0)$ is known; choose the
representative modulo $V_{2M}$ with $k(b_0)\in\{0,1\}$. Then $D(b_0)$, which equals $D'(b_0)+k(b_0)M$ or
$(k(b_0)+1)M-D'(b_0)$, is known, hence the honest $D$ on $\Lambda$, hence all $T_j\cap\Lambda$ by Fact~2,
hence $k$ on $\Lambda$ and $Y_\Lambda$.

\emph{From $(Y_\Lambda,B_\Lambda)$ to $X_\Lambda$.} In the first case invert as above. Otherwise the honest $D'$
on $\Lambda$ is known, and it suffices to recover $k$ on $\Lambda$. For $\kappa\in\Z$ let
$\mathcal T_\kappa:=\{x\in\Lambda:k(x)<\kappa\}$; by Fact~2, $\mathcal T_\kappa=\emptyset$ for
$\kappa\le\min_{\partial^*\Lambda}k$. We claim that $\{x\in\Lambda:k(x)=\kappa\}$ is the set of faces of
$\Lambda\setminus\mathcal T_\kappa$ joined to a \emph{seed} by a path in the graph $\mathfrak G_\kappa$, where
$\mathfrak G_\kappa$ is the plus graph at level $0$ of the image if $\kappa$ is odd, and the minus graph at
level $M$ of the image if $\kappa$ is even, restricted to the faces of $\Lambda\setminus\mathcal T_\kappa$ and
to bridges $d\subset\Lambda\setminus\mathcal T_\kappa$; the seeds are the faces $b\in\partial^*\Lambda$ with
$k(b)=\kappa$ and the faces of $\Lambda\setminus\mathcal T_\kappa$ that are NN-adjacent to $\mathcal T_\kappa$.
This determines $\mathcal T_{\kappa+1}$ from $\mathcal T_\kappa$, the image and $B_\Lambda$, hence $k$ on
$\Lambda$ by induction on $\kappa$.

We prove the claim for odd $\kappa$; the even case is the same with level $M$. Put $j:=\kappa+1$.
(1) Faces with $k=\kappa$ are plus faces at level $0$ of the image ($D'\ge1$). For two such faces, an NN
edge is an edge of both $G_{jM}(\om)$ (both have $D<jM$) and $\mathfrak G_\kappa$. A bridge of $G_{jM}(\om)$
between them ($D=jM-1$ on $d$, $jM$ on $d^\perp$, $n_d=0$) has $k=j$ on $d^\perp$, because faces with
$D=jM$ are not in $T_j$ and are NN-adjacent to faces with $k=\kappa$; so $D'=1$ on $d$, $D'=0$ on
$d^\perp$ and $n'_d=n_d=0$, a bridge of $\mathfrak G_\kappa$. Conversely, a bridge of $\mathfrak G_\kappa$
between two faces with $k=\kappa$ has $D=jM-1$ on $d$, and for $z\in d^\perp$, where $D'(z)=0$: $k(z)=\kappa$
is impossible (odd band, $D'\ge1$), $k(z)=\kappa-1$ would give $D(z)=(\kappa-1)M$, too small for $M\ge2$,
so $k(z)=j$ and $D(z)=jM$, a bridge of $G_{jM}(\om)$.
(2) Seeds have $k=\kappa$: a face of $\Lambda\setminus\mathcal T_\kappa$ has $k\ge\kappa$, and if it is
NN-adjacent to a face with $k<\kappa$, then $k=\kappa$.
(3) No edge of $\mathfrak G_\kappa$ joins a face with $k=\kappa$ to one with $k=\kappa+1$
(Lemma~\ref{lem:image}(b); its proof is local). Hence every face joined to a seed in $\mathfrak G_\kappa$
has $k=\kappa$.
(4) Conversely, let $k(x)=\kappa$ with $x\in\Lambda$. Then $x$ lies in an infinite minus cluster of
$G_{jM}(\om)$. Follow a path in it from $x$ until it is about to leave $\Lambda$ or to enter
$\mathcal T_\kappa$. Up to that point its faces are in $T_j\setminus\mathcal T_\kappa$, that is, have
$k=\kappa$, and its edges are edges of $\mathfrak G_\kappa$ by~(1). If the next step leaves $\Lambda$, the
current face is a seed of the first kind. If the next step enters $\mathcal T_\kappa$, it is an NN step
(a bridge at level $jM$ has $D=jM-1$ on both faces, while faces with $k<\kappa$ have
$D<\kappa M\le jM-2$), so the current face is a seed of the second kind.
\end{proof}

\subsection{Free energy of the image}

We now compare free energies per face. The original pair, seen through the pattern $X_\Lambda$, has the
free energy of two independent equilibrium states (Lemma~\ref{lem:FEsource}(i)); by Lemma~\ref{lem:reconstruct}
the image, seen through $Y_\Lambda$, has at most that much plus the entropy of the boundary data
$B_\Lambda$, which is of the order of the perimeter (Lemma~\ref{lem:FEcompare}). Since no pair process can
do better than two equilibrium states (Lemma~\ref{lem:FEsource}(ii)), the image is itself optimal, and this
forces its residual energies to be independent and geometric (Lemma~\ref{lem:residual}) and its pair
marginal to be a Gibbs measure (Lemma~\ref{lem:LR}).


Let $\Psh$ be the law on $\Om/\langle V_{2M}\rangle$ of the class of $\om^{(\vartheta)}$, where the gradient
triple has law $\mub$ and $\vartheta$ is uniform on $\{0,\dots,2M-1\}$ and independent. Then $\Psh$ is
translation-invariant and $\Psh(I^-=\Z)=1$. Let
\[
\rhob:=\Psh\circ R^{-1},
\]
a translation-invariant probability measure on $\Om$ (Lemma~\ref{lem:fold}(ii)).

For a box $\Lambda$ put $W_\Lambda:=\prod_{d\subset\Lambda}q^{\tau_d}=\prod_{d\subset\Lambda}c^{\mathsf a(\psi_1;d)+\mathsf a(\psi_2;d)}q^{n_d}$,
$N_\Lambda:=\#\{d\subset\Lambda\}$, and for the law $p$ of a pattern $Z_\Lambda$ ($Z=X$ or $Y$) put
$\mathrm{FE}(Z_\Lambda):=\sum_zp(z)\log(p(z)/W_\Lambda(z))$. For a translation-invariant law $Q$ of a pair
process (a gradient pair, possibly with decorations) put
\[
P^{(2)}(Q):=\mathsf h(Q)+\log c\cdot Q\Bigl(\sum_{d\in\{d^+_0,d^-_0\}}\bigl[\mathsf a(\psi_1;d)+\mathsf a(\psi_2;d)\bigr]\Bigr).
\]

\begin{lemma}[The fold costs no free energy]\label{lem:FEcompare}
$\mathrm{FE}(Y_\Lambda\text{ under }\Psh)\le\mathrm{FE}(X_\Lambda\text{ under }\Psh)
+\log\bigl(3^{|\partial^*\Lambda|}\cdot2\cdot(2L_\Lambda+1)\bigr)$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:reconstruct}, $(X_\Lambda,B_\Lambda)\mapsto(Y_\Lambda,B_\Lambda)$ is a bijection between the
supports that preserves $W_\Lambda$. So $\sum p(x,b)\log(p(x,b)/W_\Lambda(x))=\sum p'(y,b)\log(p'(y,b)/W_\Lambda(y))=:J$.
Now $\mathrm{FE}(X_\Lambda)=J+H(B_\Lambda\mid X_\Lambda)$ and $\mathrm{FE}(Y_\Lambda)=J+H(B_\Lambda\mid Y_\Lambda)$,
with $H$ the conditional Shannon entropy, and $0\le H\le\log\#\{\text{values of }B_\Lambda\}$.
\end{proof}

\begin{lemma}[Free energy of the original pair]\label{lem:FEsource}\leavevmode
\begin{enumerate}
\item[(i)] $\lim_\Lambda|\Lambda|^{-1}\mathrm{FE}(X_\Lambda\text{ under }\Psh)=-(2P^*+2\log c)+2\log(1-q)$.
\item[(ii)] For every translation-invariant law $Q$ of a pair process on admissible pairs, with
decorations each of which takes, on a box $\Lambda$, at most $e^{o(|\Lambda|)}$ values given the gradient
pair on $\Lambda$: $P^{(2)}(Q)\le2P^*+2\log c$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) Under $\Psh$ the pattern $X_\Lambda$ consists of independent pieces with laws $\mu_1|_\Lambda$,
$\mu_2|_\Lambda$, the uniform law on $\Z/2M$ (the value of $D\bmod2M$ at one face), and $\pi$ on the
diagonals. So $\mathrm{FE}(X_\Lambda)=-H(\mu_1|_\Lambda)-H(\mu_2|_\Lambda)-\log2M-\log c\sum_{d\subset\Lambda}
(\mu_1(\mathsf a_d)+\mu_2(\mathsf a_d))+N_\Lambda\log(1-q)$. Divide by $|\Lambda|$: $N_\Lambda/|\Lambda|\to2$, and
$\mathsf h(\mu_i)+\log c\cdot\mu_i(\mathsf a(d^+_0)+\mathsf a(d^-_0))=P(\mu_i)+\log c=P^*+\log c$ because $\mu_i$ is an
equilibrium state.
(ii) The Shannon entropy of the decorated pattern on $\Lambda$ is at most that of the two gradient
patterns plus $o(|\Lambda|)$, so $\mathsf h(Q)\le\mathsf h(Q^1)+\mathsf h(Q^2)$ for the two gradient
marginals, and $P(Q^i)\le P^*$.
\end{proof}

The decorations used below are $D'\bmod2M$ and $D'^{\mathrm{tr}}$. On a box $\Lambda$, given the gradient
pair on $\Lambda$, the first is determined by its value at one face, and the second by
$\max(-1-\operatorname{diam},\min(M+1+\operatorname{diam},D'(x_0)))$ for a face $x_0\in\Lambda$; so they take
at most $2M(2M+4+2\operatorname{diam})$ values, which is $e^{o(|\Lambda|)}$.

\begin{proposition}[The image is an equilibrium state]\label{prop:imageFE}
Let $Y^{\mathrm{pr}}$ be the pair part of $Y$ (the gradients of $\psi_1',\psi_2'$, $D'\bmod2M$ and
$D'^{\mathrm{tr}}$) and $\rho^{\mathrm{pr}}$ its law under $\rhob$. Then
\begin{enumerate}
\item[(i)] $P^{(2)}(\rho^{\mathrm{pr}})=2P^*+2\log c$;
\item[(ii)] $\lim_\Lambda|\Lambda|^{-1}\mathcal H\bigl(\operatorname{law}Y_\Lambda\,\big\|\,\operatorname{law}Y^{\mathrm{pr}}_\Lambda\otimes\pi_\Lambda\bigr)=0$,
where $\mathcal H(\cdot\,\|\,\cdot)$ is relative entropy and $\pi_\Lambda$ the product of $\pi_0$ over the diagonals in
$\Lambda$.
\end{enumerate}
\end{proposition}

\begin{proof}
Since $W_\Lambda(y,n)=c^{\sum\mathsf a}\prod q^{n_d}$ and $\pi_\Lambda(n)=(1-q)^{N_\Lambda}\prod q^{n_d}$, the chain rule
gives
\[
\mathrm{FE}(Y_\Lambda)=\Bigl[-H(Y^{\mathrm{pr}}_\Lambda)-\log c\sum_{d\subset\Lambda}\rhob\bigl(\mathsf a(\psi_1';d)+\mathsf a(\psi_2';d)\bigr)\Bigr]
+\mathcal H\bigl(\operatorname{law}Y_\Lambda\,\big\|\,\operatorname{law}Y^{\mathrm{pr}}_\Lambda\otimes\pi_\Lambda\bigr)+N_\Lambda\log(1-q).
\]
The bracket divided by $|\Lambda|$ tends to $-P^{(2)}(\rho^{\mathrm{pr}})$, and the relative entropy is
nonnegative. By Lemmas~\ref{lem:FEcompare} and~\ref{lem:FEsource}(i),
$-P^{(2)}(\rho^{\mathrm{pr}})+\limsup|\Lambda|^{-1}\mathcal H(\cdots)\le-(2P^*+2\log c)$, and by
Lemma~\ref{lem:FEsource}(ii), $P^{(2)}(\rho^{\mathrm{pr}})\le2P^*+2\log c$. Both claims follow.
\end{proof}

\begin{lemma}[Residual energies of the image]\label{lem:residual}
Under $\rhob$, $n'$ is independent and identically distributed with law $\pi_0$ and independent of the pair
$(\psi_1',\psi_2')\in\Om$.
\end{lemma}

\begin{proof}
Fix a box $\Lambda_0$ and let $H_0:=\mathcal H(\operatorname{law}Y_{\Lambda_0}\,\|\,\operatorname{law}Y^{\mathrm{pr}}_{\Lambda_0}\otimes\pi_{\Lambda_0})$.
For a large box $\Lambda$ containing $m$ disjoint translates $\Lambda^{(1)},\dots,\Lambda^{(m)}$ of $\Lambda_0$, with
$n^{(i)}$ the restriction of $n'$ to the diagonals in $\Lambda^{(i)}$,
\[
\begin{aligned}
\mathcal H(\operatorname{law}Y_\Lambda\,\|\,\operatorname{law}Y^{\mathrm{pr}}_\Lambda\otimes\pi_\Lambda)
&\ge\sum_i\E\Bigl[\mathcal H\bigl(\operatorname{law}(n^{(i)}\mid Y^{\mathrm{pr}}_\Lambda,n^{(1)},\dots,n^{(i-1)})\,\big\|\,\pi\bigr)\Bigr]\\
&\ge\sum_i\E\Bigl[\mathcal H\bigl(\operatorname{law}(n^{(i)}\mid Y^{\mathrm{pr}}_{\Lambda^{(i)}})\,\big\|\,\pi\bigr)\Bigr]=mH_0,
\end{aligned}
\]
by monotonicity of relative entropy under marginalisation, the chain rule, convexity (conditioning on
less decreases the expected relative entropy) and translation invariance. Taking $m|\Lambda_0|/|\Lambda|\to1$
and using Proposition~\ref{prop:imageFE}(ii) gives $H_0=0$. So for every box $\Lambda_0$, $n'$ on the
diagonals of $\Lambda_0$ has law $\pi$ and is independent of $Y^{\mathrm{pr}}_{\Lambda_0}$; applying this to boxes
$\Lambda_1\supset\Lambda_0$ and letting $\Lambda_1\uparrow F$, $n'$ on $\Lambda_0$ is independent of the whole
process $Y^{\mathrm{pr}}$. Finally the pair $(\psi_1',\psi_2')\in\Om$ is a measurable function of
$Y^{\mathrm{pr}}$: the honest $D'$ is determined by the gradients and by $D'^{\mathrm{tr}}$ at any face where
$0\le D'\le M$, and such faces exist because $k$ is not constant (Lemma~\ref{lem:fold}(iv)).
\end{proof}

\begin{lemma}[Equilibrium implies Gibbs]\label{lem:LR}
The pair marginal of $\rhob$ satisfies the DLR equations with honest difference: for every finite
$A\subset F$ and every bounded measurable function $f$ of pairs $(\psi_1',\psi_2')$ modulo common shifts,
$\rhob(f)=\rhob((\gamma_A\otimes\gamma_A)f)$.
\end{lemma}

\begin{proof}
Fix an integer $N\ge3$. Let $\mathbb Y_N\subset((\Z/N)^2)^F$ be the set of pairs $(y_1,y_2)$ with
$y_i(x+e_j)-y_i(x)\in\{0,1\}\bmod N$ for all $x,i,j$; it is a nearest-neighbour subshift of finite type,
hence has the topological Markov property \cite[p.~11]{BGMT}. Every element of $\mathbb Y_N$ is the
reduction mod $N$ of a pair of admissible functions, unique up to adding multiples of $N$ to each: the
four increments around a block lie in $\{0,1\}$ and sum to $0\bmod N$, hence to $0$, so the increments
integrate on $\Z^2$. Let $\Phi$ be the interaction with
$\Phi_{B(x)}(y):=-\log w_x(y_1)-\log w_x(y_2)$ and $\Phi_A:=0$ for every other $A$; it is bounded, of
finite range and translation-invariant, hence absolutely summable.

Let $\rho_N$ be the law of $((\psi_1'+U)\bmod N,(\psi_2'+U)\bmod N)$, with $(\psi_1',\psi_2')$ a
representative of the pair under $\rhob$ and $U$ uniform on $\Z/N$ and independent. It is
translation-invariant and its entropy is $\mathsf h(\rho^{\mathrm{pr}})$, since both processes are the
gradient pair with decorations of subexponential complexity. By Proposition~\ref{prop:imageFE}(i),
$\mathsf h(\rho_N)-\rho_N(f_\Phi)=P^{(2)}(\rho^{\mathrm{pr}})-2\log c=2P^*$, and by
Lemma~\ref{lem:FEsource}(ii) no invariant measure on $\mathbb Y_N$ does better, so $\rho_N$ is an
equilibrium measure for $f_\Phi$. By the Lanford--Ruelle theorem for subshifts of finite type
(Theorem~\ref{thm:BGMT}), $\rho_N$ is a Gibbs measure for $\Phi$: for every finite $A$, the conditional law of
the pattern on $A$ given the pattern off $A$ is the Boltzmann distribution on the patterns $u$ with
$u\vee y_{A^c}\in\mathbb Y_N$.

Let $A$ be a box. Its complement is connected, so the patterns $u$ with $u\vee y_{A^c}\in\mathbb Y_N$
correspond, by the lifting property applied to the whole configuration, to the admissible fillings of
$A$ for each copy, and the Boltzmann weights are those of $\gamma_A\otimes\gamma_A$. Hence
$\rhob(f)=\rhob((\gamma_A\otimes\gamma_A)f)$ for every $f$ that is a bounded function of the gradient pair on
a finite window and of $D'(x_0)\bmod N$. For a bounded $f$ of the gradient pair on a window and of the
honest $D'(x_0)$, let $f_N$ be the same function evaluated at the representative of $D'(x_0)\bmod N$ in
$(-N/2,N/2]$; then $f_N\to f$ and $(\gamma_A\otimes\gamma_A)f_N\to(\gamma_A\otimes\gamma_A)f$ pointwise and
boundedly as $N\to\infty$, since the kernel changes $D'(x_0)$ by at most $|A|$. Such $f$ generate the
$\sigma$-field, which proves the claim for boxes. For a general finite $A$ choose a box $B\supset A$; by
consistency of the specification, $(\gamma_B\otimes\gamma_B)(\gamma_A\otimes\gamma_A)=\gamma_B\otimes\gamma_B$, so
$\rhob(\gamma_A\otimes\gamma_A)=\rhob(\gamma_B\otimes\gamma_B)(\gamma_A\otimes\gamma_A)=\rhob(\gamma_B\otimes\gamma_B)=\rhob$.
\end{proof}

For a finite $A$ whose complement is not connected the correspondence used for boxes fails: the
mod-$N$ kernel allows the heights inside a hole to shift by a multiple of $N$. This is why boxes are
treated first.

\begin{proposition}[The image is a Gibbs measure]\label{prop:imageGibbs}
$\rhob$ is a translation-invariant probability measure on $\Om$ with $\rhob\gb_\Lambda=\rhob$ for every finite
$\Lambda$.
\end{proposition}

\begin{proof}
Lemmas~\ref{lem:residual} and~\ref{lem:LR}.
\end{proof}

\subsection{Burton--Keane, and the proof of Theorem~\ref*{thm:smooth}}\label{subsec:BK}

Statements about ``the image'' concern $R(\om)$ with $\om$ distributed according to $\Psh$. Its law $\rhob$
satisfies Proposition~\ref{prop:imageGibbs}, so Lemmas~\ref{lem:finiteswap} and~\ref{lem:MM}--\ref{lem:exch}
apply to $\rhob$ directly; no lift is needed, since $\rhob$ is a probability measure on triples with honest
difference.

\begin{lemma}[Sublinear growth along the axes]\label{lem:sublinear}
$\Psh$-almost surely, for $i=1,2$ and both signs, $h_1(\pm me_i)/m\to0$, $h_2(\pm me_i)/m\to0$ and
$D'(\pm me_i)/m\to0$ as $m\to\infty$, where $h_i:=2\psi_i-x_1-x_2$ is normalised by $h_i(0)=0$.
Consequently $h_1':=2\psi_1'-x_1-x_2=(h_1+h_2)/2-D'$ and $h_2':=(h_1+h_2)/2+D'$ satisfy the same.
\end{lemma}

\begin{proof}
For $\mu_i$: $\xi:=h(e_1)-h(0)$ is bounded, and by Birkhoff's theorem $(h(me_1)-h(0))/m$ converges almost
surely to $f:=\E[\xi\mid\text{invariant }\sigma\text{-field of the shift by }e_1]$. Since
$|h(me_1+e_2)-h(me_1)|\le1$, $f$ is also invariant under the shift by $e_2$, hence constant by
ergodicity, and equal to the slope $0$. For $D'$: it is an integer field with translation-invariant law
and NN increments bounded by $1$; Birkhoff's theorem gives almost sure convergence of $D'(me_1)/m$, and
the limit is $0$ because $D'(me_1)$ has the law of $D'(0)$. The formulas for $h_i'$ follow from
Lemma~\ref{lem:fold}(iii).
\end{proof}

\begin{lemma}[Finitely many infinite clusters]\label{lem:BK}
$\rhob$-almost surely the plus graph at level $0$ has finitely many infinite clusters, and so has the
minus graph at level $M$.
\end{lemma}

\begin{proof}
We treat the plus graph $\mathfrak G$ at level $0$; the other case is symmetric (exchange the copies and
shift by $M$). Suppose that with positive probability there are infinitely many infinite clusters, and
choose $r$ such that with positive probability $\Lambda_r:=[-r,r]^2$ meets at least three of them.

\emph{Modification.} Let $A_0$ be the smallest even integer at least $\max_{\Lambda_r}h_1'+2r+2$, let
$G(x):=A_0-|x_1|-|x_2|$, an admissible height in $h$-coordinates, and $S_G:=\{x:G(x)\ge h_1'(x)+2\}$; then
$\Lambda_r\subset S_G$. By Lemma~\ref{lem:sublinear} the set $\{x:h_2'(x)+|x|_1<A_0\}$ is finite almost
surely: if $h_2'(x)-h_2'(0)\le-|x|_1+t$, then comparing along $0\to(x_1,0)\to x$ and $0\to(0,x_2)\to x$ and
using $|h_2'(y)-h_2'(y')|\le|y-y'|_1$ gives $h_2'(x_1,0)-h_2'(0)\le-|x_1|+t$ and
$h_2'(0,x_2)-h_2'(0)\le-|x_2|+t$, which by Lemma~\ref{lem:sublinear} bounds $|x_1|$ and $|x_2|$. So we can fix
$n$ such that with positive probability $\Lambda_r$ meets at least three infinite clusters and $G\le h_2'$
off $\Lambda_n$; call this event $\mathcal A$. On $\mathcal A$ define the modified triple $m(\om')$ by
replacing $h_2'$ with $h_2'':=h_2'\vee(G\wedge(h_1'+2))$ and keeping $\psi_1'$ and $n'$. By
Lemma~\ref{lem:lattice} it is admissible. It differs from $\om'$ only on faces of $\Lambda_n$; its
difference $D''=\max(D',\min((G-h_1')/2,1))\ge D'$; $D''\ge1$ on $S_G\cap\Lambda_n$; and faces with
$D''\ge1>D'$ lie in $S_G\cap\Lambda_n$. Since $\gb_{\Lambda_n}(\{m(\om')\}\mid\om')>0$ and $\rhob\gb_{\Lambda_n}=\rhob$,
we have $\rhob(\mathcal T)\ge\int_{\mathcal A}\gb_{\Lambda_n}(\{m(\om')\}\mid\om')\,\rhob(\dd\om')>0$ for every event
$\mathcal T$ that contains all $m(\om')$ with $\om'\in\mathcal A$.

\emph{Trifurcation.} Let $\mathcal T$ be the event that there is a finite connected set $\hat Q$ of
vertices of $\mathfrak G$, contained in $\Lambda_n$, such that $\mathfrak G\setminus\hat Q$ has at least three
infinite components adjacent to $\hat Q$. We claim that $m(\om')\in\mathcal T$ for $\om'\in\mathcal A$. Write
$\mathfrak G'$ and $\mathfrak G''$ for the graphs of $\om'$ and $m(\om')$. The set $S_G\cap\Lambda_n$ is
NN-connected, since $G-h_1'$ does not decrease along an NN step that decreases $|x|_1$, so every monotone
path to the origin stays in $S_G$; and it consists of plus faces of $\mathfrak G''$. Let $\hat Q$ be the
component of $\mathfrak G''$ restricted to $\Lambda_n$ that contains it.
(1) Every vertex of $\mathfrak G''$ outside $\hat Q$ is a plus face of $\om'$ with $D''=D'$, and every edge
of $\mathfrak G''$ between two such vertices is an edge of $\mathfrak G'$: for an NN edge this is clear; for a
bridge $d$, $D''=1$ on $d$ and $D''=0$ on $d^\perp$ force $D'=1$ on $d$ and then $D'=0$ on $d^\perp$ (since
$D'\le D''$ and by Lemma~\ref{lem:diff}(i)), and $n'$ is unchanged. Hence every component of
$\mathfrak G''\setminus\hat Q$ is contained in one cluster of $\mathfrak G'$.
(2) If $C$ is an infinite cluster of $\mathfrak G'$ meeting $\Lambda_r$, then $C\cup\hat Q$ is connected in
$\mathfrak G''$: faces of $C$ stay plus faces, NN edges stay, and for a bridge $d$ of $C$ either $D''=0$
on $d^\perp$ and it stays an open bridge, or some $z\in d^\perp$ has $D''(z)\ge1$ and the two faces of $d$
are joined through $z$. Removing the finite set $\hat Q$ from the connected, infinite, locally finite
graph on $C\cup\hat Q$ leaves finitely many components, one of which is infinite, adjacent to $\hat Q$, and
contained in $C$. Three clusters $C$ give three distinct such components by~(1).

\emph{Counting.} By translation invariance each translate $\Lambda_n+(2n+1)z$ carries the translated event
with probability $\delta:=\rhob(\mathcal T)>0$, so the expected number of such tiles inside $\Lambda_N$ grows like
$\delta N^2/(2n+1)^2$. The sets $\hat Q$ of distinct tiles are disjoint and connected, and each other $\hat Q$
lies in one component of $\mathfrak G\setminus\hat Q$; so after contracting all of them to points, each point
still separates at least three infinite components. The counting argument of Burton and
Keane~\cite{BurtonKeane} (count the leaves of the resulting forest on the boundary) bounds their number by
the number $O(N)$ of faces $*$-adjacent to $\Lambda_N$, since bridges join diagonal faces. This is a contradiction for large $N$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:smooth}]
Let $\mu_1\ne\mu_2$ be as in the theorem. By Corollary~\ref{cor:cases} either both are smooth or
$\mub(E_{\Z\Z})=1$; assume the latter. Let $\mathcal M$ be the set of $m\in\Z$ such that $\Pi_m$ contains an
infinite plus FK cluster at level $0$ of the image, and $\mathcal M'$ the set of $m$ such that $\Pi'_m$
contains an infinite minus FK cluster at level $M$ of the image, both computed for a representative of
$\om$ modulo $V_{2M}$, which shifts them by $1$. By Lemmas~\ref{lem:image}(b) and~\ref{lem:BK},
$\mathcal M$ and $\mathcal M'$ are finite almost surely.

\emph{Case 1: $\Psh(\mathcal M\ne\emptyset)>0$ or $\Psh(\mathcal M'\ne\emptyset)>0$.} Say the first. A gradient
triple $(\nabla\psi_1,\nabla\psi_2,n)$ has $2M$ classes of honest lifts modulo $V_{2M}$, of which $\Psh$
picks one uniformly. Let $\mathcal E_0$ be the event, of the gradient triple, that $\mathcal M$ is finite
for every one of the $2M$ classes and nonempty for at least one; $\mub(\mathcal E_0)>0$. On $\mathcal E_0$,
each class with $\mathcal M\ne\emptyset$ has exactly one honest lift $D$ for which $\min\mathcal M=0$; let
$\mathfrak d$ be the smallest of these at most $2M$ lifts, which differ by constants. We apply
Lemma~\ref{lem:canonical} with $\mathcal U$ the space of $n$, $\varpi=\pi$ and $\mathcal E=\mathcal E_0$.
Condition~(a) holds since $\mathcal E_0$ is translation-invariant. For~(b), let $\tilde\psi_2$ differ from
$\psi_2$ on a finite set and fix an honest lift $D$; the lift $\tilde D:=D+(\tilde\psi_2-\psi_2)$ of the new
triple still has $I^-=\Z$ (Lemma~\ref{lem:robust}). Let $S$ be a box containing all blocks that meet the
changed set, and $S^+$ its $*$-neighbourhood.
\begin{enumerate}
\item[(1)] If $jM$ exceeds $D$ and $\tilde D$ on $S^+$, the minus graphs at level $jM$ of the two triples
coincide: all faces of $S^+$ are minus faces for both, and a $jM$-saddle cannot meet $S$, since it would
lie in $S^+$ and contain faces with $D=jM$; so the bridges are the same.
\item[(2)] If $jM$ is at most $D$ and $\tilde D$ on $S^+$, there is no minus face at level $jM$ in $S^+$ for
either triple, no $jM$-saddle of sign $-$ meets $S$, and again the minus graphs coincide.
In both cases $T_j$ is unchanged.
\item[(3)] For the finitely many remaining $j$, $T_j$ changes by finitely
many faces: a face of the old $T_j$ is in the new one unless it lies in $S$ or in a finite component of
its old cluster minus $S$, and symmetrically, as in Lemma~\ref{lem:robust}.
\end{enumerate}
So $k$, and hence $R(\om)$, changes on finitely many faces and diagonals only. As in the proof of
Lemma~\ref{lem:robust}, an infinite plus cluster at level $0$ of the old image contains an infinite
connected piece that is unchanged, is contained in a cluster of the new image, and lies in the same
$\Pi_m$, since old and new $k$ agree there and the new cluster lies in one $\Pi_m$ by
Lemma~\ref{lem:image}(b); and symmetrically. So $\mathcal M$, computed in the lifts $D$ and $\tilde D$, is
the same set, the canonical lifts of corresponding classes differ by $\tilde\psi_2-\psi_2$, and so do their
minima: $\mathfrak d$ changes by $\tilde\psi_2-\psi_2$. The same argument applies to finite changes of
$\psi_1$, for which $\mathfrak d$ changes by $-(\tilde\psi_1-\psi_1)$. Lemma~\ref{lem:canonical}, in both
versions, shows that $\mu_1$ and $\mu_2$ are smooth.

\emph{Case 2: $\mathcal M=\emptyset=\mathcal M'$ almost surely.} Then, with Lemma~\ref{lem:image}(a),
$\rhob$-almost surely the image has no infinite FK cluster at level $0$ and none at level $M$. By
Proposition~\ref{prop:imageGibbs} and the proof of Lemma~\ref{lem:exch} applied to the probability
measure $\rhob$ at level $0$, $\rhob$ is invariant under the exchange $(\psi_1',\psi_2',n')\mapsto(\psi_2',\psi_1',n')$,
so $D'(0)$ and $-D'(0)$ have the same law. The same argument at level $M$ (swap all clusters at level
$M$ contained in $\Lambda_N$; in the limit $(\psi_1',\psi_2')\mapsto(\psi_2'-M,\psi_1'+M)$) shows that $D'(0)$
and $2M-D'(0)$ have the same law. Hence $D'(0)$ and $D'(0)+2M$ have the same law, which is impossible
for an integer-valued random variable. So Case~2 does not occur.
\end{proof}