\section{Two tools}\label{app:tools}

\subsection{The Lanford--Ruelle theorem for subshifts of finite type}

Let $\mathcal A$ be a finite alphabet and $\mathcal X\subset\mathcal A^{\Z^2}$ a subshift of finite type. An interaction
$\Phi=(\Phi_A)_{A\subset\Z^2\text{ finite}}$ assigns to each finite $A$ a function of the configuration on $A$;
it is translation-invariant and absolutely summable if $\Phi_{A+x}=\Phi_A\circ\theta_x$ and
$\sum_{A\ni0}\|\Phi_A\|_\infty<\infty$. Put $f_\Phi:=\sum_{A\ni0}\Phi_A/|A|$. A translation-invariant
probability measure $\mu$ on $\mathcal X$ is an \emph{equilibrium measure} for $f_\Phi$ if it maximises
$\mathsf h(\mu)-\mu(f_\Phi)$ over translation-invariant probability measures on $\mathcal X$ (for the six-vertex
potential on $\mathbb X$ this is an equilibrium state in the sense of Section~\ref{subsec:FE}); it is a \emph{Gibbs
measure} for $\Phi$ if, for every finite $A$, the conditional law of the configuration on $A$ given the
configuration $y$ off $A$ is proportional to $\exp(-\sum_{B\cap A\ne\emptyset}\Phi_B(u\vee y))$ on the
patterns $u$ with $u\vee y\in\mathcal X$ \cite[(7)]{BGMT}.

\begin{theorem}[{\cite[Theorem~1.1(b)]{BGMT}}]\label{thm:BGMT}
Let $\mathcal X$ be a subshift of finite type over a finite alphabet on $\Z^2$ and $\Phi$ a translation-invariant
absolutely summable interaction. Then every equilibrium measure for $f_\Phi$ is a Gibbs measure for $\Phi$.
\end{theorem}

The theorem is proved in \cite{BGMT} for subshifts with the topological Markov property on countable
amenable groups, and every subshift of finite type has that property \cite[p.~11]{BGMT}. It is used in
Lemma~\ref{lem:LR} for the nearest-neighbour subshift $\mathbb Y_N$ and a bounded interaction of finite
range.

\subsection{Anti-concentration}

\begin{lemma}[Erd\H{o}s~\cite{Erdos}]\label{lem:LO}
Let $m_1,\dots,m_T$ be nonzero integers and $\varepsilon_1,\dots,\varepsilon_T$ independent uniform signs. Then
$\sup_x\Pp(\sum_i\varepsilon_im_i=x)\le\binom{T}{\lfloor T/2\rfloor}2^{-T}$, which tends to $0$ as $T\to\infty$.
\end{lemma}

\begin{proof}
Replacing $\varepsilon_i$ by $-\varepsilon_i$ we may assume $m_i>0$. For fixed $x$, the sets $\{i:\varepsilon_i=+1\}$ of the
solutions form an antichain, since enlarging the set strictly increases the sum, so there are at most
$\binom{T}{\lfloor T/2\rfloor}$ of them by Sperner's theorem.
\end{proof}