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\subsection{Section~\ref{sec:5}}\label{S:proofs5}

\paragraph{The dual paths of Lemma~\ref{lem:5.3}.} In fine coordinates with $\Xi=[0,R]\times[0,T]\times\{0\}$, the dual paths are the closed paths of fine cubes in the planes $y=j+\frac12$ around the rectangles of cube centres $[i+\frac12,2R-i-\frac12]\times[i-R+\frac12,R-i-\frac12]$ in $(x,z)$, $0\le j<T$, $0\le i<R$. These paths are disjoint, and in these coordinates the path of $(i,j)$ crosses $\Xi$ once, at $x=i+\frac12$.

\paragraph{Proof of Lemma~\ref{lem:5.12}.} By \cite[Lemma~2.1 and the sentence after it, p.~141]{GK04}, the first number is at most the number $\binom{Dn}{n}/((D-1)n+1)=\binom{Dn}{n-1}/n$ of $n$-vertex subtrees containing the root of the infinite rooted $D$-ary tree, $D:=D_{\max}$, and $\binom{Dn}{n-1}/n\le D^{n-1}n^{n-1}/n!\le(eD)^{n-1}$ as $n!\ge e(n/e)^n$. For the second, we choose the first plaquette at an edge of $C$ of each component among at most $4|C|$ plaquettes ($2^{4|C|}$ choices), the sizes of the components in the order of these roots ($2^a$), and each component by the first count with $D_{\max}=12$.

\paragraph{Proof of Lemma~\ref{lem:5.14}.} We put $\lambda:=\log(2^{k+1}\kpl)$ and $\omega:=|\log\eps|/F_*\ge c_T(k)/\eta\ge c_T(k)=1+\lambda$. By Lemma~\ref{lem:5.12}, $\PD(E)\le2^{4|C|+j}\sum_{a\ge RT}e^{-a(\omega-\lambda)+\omega(1-\eta)RT}\le2^{4|C|+j}\frac{e}{e-1}e^{RT(\lambda-\eta\omega)}\le2^{4|C|+j+2}e^{-RT}$, as $\omega-\lambda\ge1$. Since $j\le3$, $\ED e^{\sigma X}\le e^{\sigma\eta RT}+e^{\sigma RT}\PD(E)\le e^{\sigma\eta RT}(1+2^{4|C|+5})\le e^{\sigma\eta RT+4|C|+4}$.

\paragraph{Proof of Lemma~\ref{lem:5.18}(a), (c) to (e).} We fix the cut $j$ and write $U:=U_j$, $U':=U_j'$, with $U^\alpha$ and $\mathcal W(\alpha)$ as in Definition~\ref{def:5.17}.

(a) Let $(u,v)$ be a bond of $\Sep^*_j$ with $u\in U^\alpha$ and $v\in W$. If $v\in U$, then $v$ is box-adjacent to $u$ and lies in $U^\alpha$, which is impossible. Hence $v\notin U$. If $(u,v)$ is a $\Gp_{t_j}$-bond, then $v\in\KK_j$ because $\KK_j$ is a component, so $v\in U'$ and the bond is a cut bond. Otherwise $v\in\XX_{t_j}$ (a hole face) or the bond is frozen.

(c) Let $W\in\mathcal W(\alpha)$ and $z\in W\cap U'$. We take a $\Gp_{t_j}$-path in $\KK_j$ from $z$ to $U^\alpha$ and let $y$ be the site just before its first entry into $U^\alpha$. The segment from $z$ to $y$ is a box path avoiding $U^\alpha$, so $y\in W$. Also $y\in\KK_j$, and $y\notin U$ because a site of $U$ box-adjacent to $U^\alpha$ lies in $U^\alpha$. Thus $y\in U'$, and the bond from $y$ into $U^\alpha$ is a cut bond in $\partial(U^\alpha,W)$. A bond $(u,v)$ with $u\in U$, $v\notin U$ determines $\alpha$ (the box-component of $u$ in $U$) and $W$ (the box-component of $v$ in $\Lc\setminus U^\alpha$), so distinct pieces are disjoint. Each piece contains one of the $A_j$ cut bonds.

(d) For a cut bond $(u,z)$ with $u\in U^\alpha$, $z\in U'$, the box-component of $z$ in $\Lc\setminus U^\alpha$ belongs to $\mathcal W(\alpha)$. The same argument with $U$ and $U'$ exchanged gives the piece of $\Sep^{**}_j$.

(e) Let $u^*$ be the last site of $U$ on the path, $u^*\in U^{\alpha^*}$. The rest of the path lies in $\Lc\setminus U\subseteq\Lc\setminus U^{\alpha^*}$ and is box-connected. Hence it lies in one box-component $W^*$ of $\Lc\setminus U^{\alpha^*}$, and $W^*\in\mathcal W(\alpha^*)$ because the path ends in $U'$. The bond leaving $u^*$ lies in $\partial(U^{\alpha^*},W^*)$.

\paragraph{Proof of Lemma~\ref{lem:5.22}.} \emph{Shells.} We partition $\Lc$ into \emph{shell boxes} as a tensor product, cutting each coordinate axis of $\Lc$ into intervals of lengths in $[6L_c,12L_c)$ (possible since every coarse side of $\Lc$ is at least $\ell\ge308L_c$, Lemma~\ref{lem:3.3}(c)). We index the boxes by $\iota\in\prod_\mu[0,n^{\mathrm{sh}}_\mu)$ and put $\delta(b,b'):=|\iota-\iota'|_\infty$ and $n_{\min}:=\min_\mu n^{\mathrm{sh}}_\mu$. For a box $b_0$ and $i\ge1$ the \emph{shell} $\mathrm{Sh}_i(b_0)$ is the union of the boxes $b$ with $\delta(b,b_0)\in\{3i,3i+1,3i+2\}$.

\emph{Shell claim: if $\mathrm{Sh}=\mathrm{Sh}_i(b_0)$ with $3i+2\le(n_{\min}-1)/2$ contains a whole cell of $U_j$ and a whole cell of $U_j'$ (cells at time $t_j$), then each of $\Sep^*_j$ and $\Sep^{**}_j$ has at least $(c_1/24)L_c^2$ bonds with both ends in $\mathrm{Sh}$.} We write $U:=U_j$. Let $A'$ be the set of sites joined to $U$ by a box path with no bond of $\Sep^*_j$, and $B':=\Lc\setminus A'$. Then $U_j'\subseteq B'$ by Lemma~\ref{lem:5.18}(e), and every box bond between $A'$ and $B'$ lies in $\Sep^*_j$. We put $W:=A'\cap\mathrm{Sh}$. The sets $W$ and $\mathrm{Sh}\setminus W$ contain a cell each, hence at least $c_1L_c^3$ sites each, and it suffices to find $(c_1/24)L_c^2$ box bonds inside $\mathrm{Sh}$ between them. The boxes of $\mathrm{Sh}$ form a face-connected family: since $3i+2\le(n_{\min}-1)/2$, each index sphere of radius $3i$, $3i+1$, $3i+2$ around $b_0$ keeps, in each direction, at least one of its two faces inside the index box, so its clipped faces form a connected surface, and consecutive spheres are face-adjacent. Put $f_b:=|W\cap b|/|b|$.

\emph{Case 1: $f_b\le\frac12$ for all boxes $b$ of $\mathrm{Sh}$.} The box Cheeger inequality of the proof of Lemma~\ref{lem:5.22} in each box (sides less than $12L_c$) gives at least $|W|/(24L_c)\ge c_1L_c^2/24$ bonds.

\emph{Case 2: $f_b\ge\frac12$ for all $b$.} The same argument applies to $\mathrm{Sh}\setminus W$.

\emph{Case 3: otherwise.} There are face-adjacent boxes $b_1,b_2$ with $f_{b_1}\ge\frac12>f_{b_2}$. In the box $b_1\cup b_2$ (sides less than $24L_c$) both $W$ and its complement have at least $\frac12(6L_c)^3=108L_c^3$ sites, and the same inequality gives at least $108L_c^3/(48L_c)>2L_c^2>(c_1/24)L_c^2$ bonds.

The same argument with $U$ and $U_j'$ exchanged gives the bound for $\Sep^{**}_j$.

\emph{Counting shells.} Let $b_0$ be the shell box of $x$. The $U_j$-end of the cut bond of Lemma~\ref{lem:5.22}(i) lies within $\rho/2+2$ of $x$, hence at index distance less than $\rho/(12L_c)+2$ from $b_0$, and the far site of (ii) lies at index distance more than $3\rho/(4L_c)-1$. A c-path in $U_j$ from the former to the latter changes the index distance by at most $1$ per step, and cells lie in one block, of side less than the box sides. Hence each shell $\mathrm{Sh}_i(b_0)$, $i\in I(\rho):=\{i\in\ZZ:\rho/(36L_c)+1\le i\le\rho/(4L_c)-1\}$, contains a whole cell of $U_j$, and likewise of $U_j'$. These shells are disjoint, lie within $9\rho$ of $x$, satisfy $3i+2\le(n_{\min}-1)/2$, and number $|I(\rho)|\ge\rho/(9L_c)$ for $\rho\ge27L_c$. In this range, the shell claim applied in each shell gives $(\rho/(9L_c))(c_1/24)L_c^2=c_{\mathrm{sh}}\rho L_c$ bonds of $\Sep^*_j$, and as many of $\Sep^{**}_j$.

\paragraph{Proof of Lemma~\ref{lem:5.38}(b).} We take the breadth-first spanning tree from $o$ of the graph defining anchoring, with fixed orders. Along this tree, the block set is determined by the following data:
\begin{itemize}
\item the sizes of the components ($2^b$ choices);
\item the anchors of the children of $o$ in $Z(\Xi')$, $|Z(\Xi')|\le2a$, and of the children of each component $C$ among its blocks ($2^{2a+k_o}$ and $2^{|C|+k_C}$ choices, $k_o+\sum_Ck_C=m$);
\item the positions of the root blocks relative to their anchors ($(2d_0+1)^3$ choices each);
\item the components, $*$-connected sets of given sizes containing their root blocks ($(26e)^{|C|}$ choices by Lemma~\ref{lem:5.12} with $D_{\max}=26$).
\end{itemize}
With $\sum_C|C|=b$ the product is $2^b\cdot2^{2a+b+m}(2d_0+1)^{3m}(26e)^b$.