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\providecommand{\good}{\operatorname{good}}
\providecommand{\Kc}{\mathfrak{K}}
\providecommand{\Uf}{\mathbb{U}}
\providecommand{\Ereg}{\mathcal{E}}

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\subsection{Payment and reach of the Phase-1 output}\label{ssec:4.9}
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In this subsection we bound the cost of the Phase-1 output by means of the three comparisons of Lemma~\ref{lem:4.density}. Filling everything near a point bounds the cost of a ball, deleting a region without marks bounds its cost by the bonds of a sphere around it, and a good sphere, chosen by averaging over the radii, makes the cost decay geometrically away from the marks. Consequently, no hole lies far from a mark, as Theorem~\ref{thm:4.42} states, and in Proposition~\ref{prop:4.39} we pay for the pieces near the clusters of marks by comparison with the cheapest state that covers the marks and fills the forced blocks of the cluster.

Throughout, $(\XX^1,\Uf^1)$ is the Phase-1 output and $\Ereg:=\Ereg^{\bar S^1}$. By Theorem~\ref{thm:4.30}, $(\XX^1,\Uf^1)$ minimizes $\Ereg$, and each bond of $\bar S^1$ touches $\XX^1$ and has both ends within distance $4L_c$ of a mark. Thus, at $(\XX^1,\Uf^1)$, the frozen bonds count as ordinary bonds, and the penalty vanishes by Lemma~\ref{lem:4.forced}. For a set $A$ of sites we put $\Ereg_A:=\#\{b\in\partial\XX^1:\ b\subseteq A\}+\lambda|(\XX^1\setminus\Uf^1)\cap A|+\lambda_U|\Uf^1\cap A|$, let $B_r(x):=Q_\infty(x,r)\cap\Lamo$, and for a fixed site $x$ let $f(r):=\Ereg_{B_r(x)}$, nondecreasing in $r$. Each competitor $(\XX',\Uf')$ below has $\MM\subseteq\XX'$ and $\Uf'\subseteq\XX'$ a union of blocks, and so satisfies $\Ereg(\XX',\Uf')\ge\Ereg(\XX^1,\Uf^1)$.

\begin{lemma}\label{lem:4.density}\label{lem:4.deletion}\label{lem:4.decay}
(a) For every site $x$ and every $r\ge0$, $f(r)\le g(r):=6(2r+4L_c)^2+\lambda_U(2r+4L_c)^3$. (b) Let $B$ be a set of sites with $\dist(B,\MM)\ge6L_c$, and $\hat B$ the union of the blocks meeting $B$. Then $\Ereg_B\le\mathrm{sl}(B):=f(B,\XX^1\setminus B)+\lambda|\Uf^1\cap(\hat B\setminus B)|$. (c) Put $w:=\lceil120\ell/c_s\rceil$ and $w':=w+2L_c$. If $\dist(B_{r+w'}(x),\MM)\ge6L_c$, then $f(r)\le f(r+w')/5$.
\end{lemma}

\begin{proof}
(a) Let $\hat B$ be the union of the blocks meeting $B_r(x)$, a box of sides less than $2r+4L_c$. We compare with $(\XX^1\cup\hat B,\Uf^1\cup\hat B)$. In this competitor the bonds of $\partial\XX^1$ with both ends in $\hat B$ disappear, at most $|\partial\hat B|\le6(2r+4L_c)^2$ bonds appear, the frozen bonds still touch the hole set, each site of $\hat B$ ends with the weight $\lambda_U$, and the penalty does not grow. It follows that $f(r)\le\Ereg_{\hat B}\le|\partial\hat B|+\lambda_U|\hat B|$.

(b) A forced block has a mark within distance $4L_c$ of each of its sites. Consequently, no block meeting $B$ is forced, and no bond of $\bar S^1$ has an end in $B$. We compare with $(\XX^1\setminus B,\Uf')$, where $\Uf'$ is the union of the blocks of $\Uf^1$ that do not meet $B$. This competitor contains the marks and all forced blocks. In passing to it, the bonds of $\partial\XX^1$ with both ends in $B$ disappear, at most $f(B,\XX^1\setminus B)$ bonds appear, the frozen bonds still touch the hole set, the sites of $B\cap\XX^1$ lose their weights, and those of $\Uf^1\cap(\hat B\setminus B)$ change from $\lambda_U$ to $\lambda$.

(c) We put $A:=B_{r+w'}(x)\setminus B_r(x)$ and apply (b) to $B_s:=B_s(x)$ for $r\le s<r+w$. Since a site of $\XX^1$ at distance exactly $s+1$ from $x$ has at most six bonds into $B_s$, we have $\sum_sf(B_s,\XX^1\setminus B_s)\le6|\XX^1\cap A|$. A site of $\Uf^1$ outside $B_s$ lies in a block meeting $B_s$ for fewer than $2L_c$ values of $s$, and therefore $\sum_s\lambda|\Uf^1\cap(\hat B_s\setminus B_s)|\le2\lambda L_c|\Uf^1\cap A|=24|\Uf^1\cap A|$. Combining the two bounds, we obtain $\sum_s\mathrm{sl}(B_s)\le30|\XX^1\cap A|\le30(\lambda^{-1}+\lambda_U^{-1})\Ereg_A\le(30\ell/c_s)\Ereg_A$, using $\lambda^{-1}+\lambda_U^{-1}=L_c/12+\ell/(2c_s)\le\ell/c_s$. Thus some $s$ has $\mathrm{sl}(B_s)\le\Ereg_A/4$. The bonds and sites counted in $\Ereg_A$ are not counted in $f(r)$. Therefore $\Ereg_A\le f(r+w')-f(r)$, and $f(r)\le f(s)\le\mathrm{sl}(B_s)\le(f(r+w')-f(r))/4$.
\qed
\end{proof}

\begin{theorem}\label{thm:4.42}
Let $J$ be the least integer with $5^J>g((J+1)w')$, and $R_*:=(J+1)w'+6L_c$. Every site of $\XX^1$ lies within distance $R_*$ of a mark, and $R_*=O((\ell/c_s)\log(\ell/c_s))$, uniformly in $|\MM|$, $\Lamo$, $C$ and $D$.
\end{theorem}

\begin{proof}
Let $\dist(x,\MM)>R_*$. Then $\dist(B_{(J+1)w'}(x),\MM)>6L_c$, and applying Lemma~\ref{lem:4.density}(c) $J$ times from $r=w'$, together with Lemma~\ref{lem:4.density}(a), we obtain $f(w')\le5^{-J}f((J+1)w')\le5^{-J}g((J+1)w')<1$. So no bond of $\partial\XX^1$ has both ends in the box $B_{w'}(x)$, which therefore lies in $\XX^1$ or misses it. In the first case $f(w')\ge\lambda_U|B_{w'}(x)|>1$. Therefore $x\notin\XX^1$. With the arithmetic of Section~\ref{S:C1}, this proves Theorem~\ref{thm:3.21}(b).
\qed
\end{proof}

\begin{proposition}\label{prop:4.39}
Theorem~\ref{thm:3.21}(a) holds with $\Cst:=C_1+2C_{\rm seal}$, $C_1:=7+27\cdot24.6/\vartheta$, and $\Cfree:=\Cst(2+1/(2c_s)+1/3696)$.
\end{proposition}

\begin{proof}
By Theorem~\ref{thm:4.30}(i), $S^1=\emptyset$. We compare with the competitor that replaces $Y_P$ by the marks and the forced blocks of $P$. The sets $Y_P$ cover $\XX^1$ by Theorem~\ref{thm:4.42}. Marks of distinct parts are more than $2R_*+2$ apart, and hence no bond joins $P^{+R_*}$ to $P'^{+R_*}$. Moreover, a block of $\Uf^1$ is connected and lies in $\XX^1$. The marks of a set $3Q$ are within distance $6L_c\le r'$ of each other, and hence those of a forced block lie in one part. The union $U_F(P)$ of the forced blocks with marks in $P$ lies in $Y_P$, by Lemma~\ref{lem:4.forced} and because a forced block lies within $4L_c$ of its marks. It has at most $27N(P)/(\vartheta L_c^2)$ blocks, since a mark lies in at most $27$ sets $3Q$. The competitor $\XX':=(\XX^1\setminus Y_P)\cup P\cup U_F(P)$, $\Uf':=(\Uf^1\setminus Y_P)\cup U_F(P)$ contains the marks, $\Uf'$ is a union of blocks containing all forced blocks, and the change is confined to $Y_P$, which no bond joins to the rest of $\XX^1$. A bond of $\bar S^1$ that loses contact with the hole set touches $Y_P$, so the marks of its two blocks lie within $R_*+4L_c<r'$ of $P$ and therefore in $P$. There are at most $C_{\rm seal}N(P)$ such bonds by Proposition~\ref{prop:4.sealcount}, each costing at most $2$. Since $|\partial Q|+\lambda_U|Q|<24.6L_c^2$ and $\lambda=12/L_c\le1$ ($L_c\ge35$ by item~(e) of Section~\ref{ssec:D.5}), minimality implies $|\partial Y_P|+\lambda|Y_P\setminus\Uf^1|+\lambda_U|Y_P\cap\Uf^1|\le(6+\lambda)N(P)+24.6\cdot27N(P)/\vartheta+2C_{\rm seal}N(P)\le\Cst N(P)$.
The second inequality follows from $|Y_P|/\ell\le(\lambda\ell)^{-1}\lambda|Y_P\setminus\Uf^1|+(\lambda_U\ell)^{-1}\lambda_U|Y_P\cap\Uf^1|$, with $(\lambda\ell)^{-1}\le1/3696$ and $(\lambda_U\ell)^{-1}=1/(2c_s)$.
\qed
\end{proof}

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\subsection{Phase 2: insulation, density bound and cuts}\label{ssec:4.4}\label{ssec:4.7}
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In this subsection we first bound the insulation in Phase~2, in Lemma~\ref{lem:4.21}, and then prove the density bound (R9), Lemma~\ref{lem:3.18} and Proposition~\ref{prop:3.19}. Each hole site added in Phase~2 outside the removed cells is hull material, whose removal does not lower the hull functional, while the Phase-1 insulation $\mathrm{ins}^1:=\partial\XX^1$ is a fixed budget, bounded in Theorem~\ref{thm:3.21}(a). Throughout, $R_t$ is the union of the cells removed in Phase~2 before time $t$. A cell $\Delta$ removed in Phase~2 is removed at a hull-closed time $t_\Delta$, at which $w(\Delta)$ is evaluated. We denote by $s_\Delta$ the number of bonds of $E(\Delta)$ frozen in Phase~2 before $t_\Delta$, converted ones included, with no end in $\XX^1$.

\begin{lemma}\label{lem:4.21}
(a) Let $t$ be a hull-closed time of Phase~2, $N'$ a box, $W':=(\XX_t\setminus(\XX^1\cup R_t))\cap N'$, $\mathrm{Ins}_{N'}(Y):=\#\{b\in E(N'):\text{exactly one end in }Y\}+2\#\{b\in\bar S_t\cap E(N'):\text{no end in }Y\}$, and $\mathrm{cut}_{N'}$ the number of bonds across $\partial N'$ with inner end in $W'$ whose outer end lies in $\XX_t$ or which are frozen. Then
\begin{equation}\label{eq:4.I5}
\mathcal A_2(E(N');t)\le\mathrm{Ins}_{N'}(\XX^1\cup R_t)+\mathrm{cut}_{N'}.
\end{equation} (b) Every cell $\Delta$ removed in Phase~2 satisfies $w(\Delta)\le|\mathrm{ins}^1\cap E(\Delta)|+2s_\Delta+|\partial\Delta|$, hence $c_s|\partial\Delta|<|\mathrm{ins}^1\cap E(\Delta)|+2s_\Delta$.
\end{lemma}

\begin{proof}
(a) We put $\XX:=\XX_t$ and $V:=\Lamo\setminus\XX$. Since $B_t=\MM\cup\Uf^1\cup R_t\subseteq\XX^1\cup R_t\subseteq\XX$ by Proposition~\ref{prop:4.4}(a), $W'$ misses $B_t$ and $\mathrm{Ins}_{N'}(\XX\setminus W')=\mathrm{Ins}_{N'}(\XX^1\cup R_t)$, while $\mathrm{Ins}_{N'}(\XX)=\mathcal A_2(E(N');t)$, since by Lemma~\ref{lem:4.2} the bonds of $\bar S_t$ with no end in $\XX$ are those of $S_t$. This proves (a) if $W'=\emptyset$. Otherwise $F^{\bar S_t}_\lambda(\XX\setminus W')>F^{\bar S_t}_\lambda(\XX)$, because $\XX=H^{\bar S_t}_\lambda(B_t)$. Written bond by bond as in \eqref{eq:4.locrem}, this inequality reads $f_{E\setminus\bar S_t}(W',V)+\lambda|W'|<f(W',\XX\setminus W')+\sum_{b\in\bar S_t\text{ between }W',V}(w_b-1)+\sum_{b\in\bar S_t,\,b\subseteq W'}w_b$, with $f_{E\setminus\bar S_t}$ counting unfrozen bonds. When we pass from $\mathrm{Ins}_{N'}(\XX\setminus W')$ to $\mathrm{Ins}_{N'}(\XX)$, the contribution of a bond of $E(N')$ between $W'$ and $\XX\setminus W'$ changes by $-1$, that of a bond of $\bar S_t$ inside $W'$ by $-2$, that of a bond between $W'$ and $V$ by $+1$ if unfrozen and by $-1$ if frozen, and that of every other bond by $0$. Inserting the inequality, we see that the bonds between $W'$ and $\XX\setminus W'$ inside $N'$ cancel, frozen bonds inside $N'$ with an end in $W'$ get the coefficient $w_b-2\le0$, and the bonds counted in $\mathrm{cut}_{N'}$ remain, each with coefficient at most $1$. We conclude that $\mathcal A_2(E(N');t)<\mathrm{Ins}_{N'}(\XX^1\cup R_t)+\mathrm{cut}_{N'}-\lambda|W'|$.

(b) We take $N':=\Delta$ and $t:=t_\Delta$. Since $\Delta\cap R_t=\emptyset$ and, by Theorem~\ref{thm:4.30}, each bond of $\bar S^1$ touches $\XX^1$, $\mathrm{Ins}_\Delta(\XX^1\cup R_t)=|\partial\XX^1\cap E(\Delta)|+2\#\{b\in\bar S_t\cap E(\Delta):\text{no end in }\XX^1\}\le|\mathrm{ins}^1\cap E(\Delta)|+2s_\Delta$, and $\mathrm{cut}_\Delta\le|\partial\Delta|$. The last claim follows from the trigger.
\qed
\end{proof}

\begin{proposition}[Density bound]\label{prop:4.22}
At every state satisfying (S-rem), in particular at the output, (R9) holds, and $(3+c_s)|\partial\Delta|<54(3+c_s)\ell^2$.
\end{proposition}

\begin{proof}
Since a frozen bond is a missing pair of neighbours counted at both ends, the density count of Section~\ref{sec:stage1} is two-sided, namely the hull's own count $\mathcal A_2$. If $\Delta\not\subseteq B$, then $\mathcal A_2(E(\Delta))=w(\Delta)\le w_*$ by (S-rem). If $\Delta\subseteq B\subseteq\XX$, this count is $0$. A bond of $\partial\Delta\cap E$ weighs at most $2$. This yields the bound in which these bonds are counted as well, and the last assertion holds as $|\partial\Delta|<54\ell^2$ by (G3). For the constant $c_0$ of the conversion into fine plaquettes we refer to Section~\ref{S:B1.c0}.
\qed
\end{proof}


We recall that Phase~2 fills only by removal and freezes by the seal and the cut rule, and that the cut rule fires only at hull-closed states without auxiliary material, where Theorem~\ref{thm:4.rest} holds.

\begin{lemma}\label{lem:3.18}
Let the firings $j=1,2,\dots$ of the cut rule have times $t_j$, components $\Kc_j$ and sides $U_j$, $\Kc_j\setminus U_j$. (a) Every bond of $\partial_{\Gp}U$ joins two different cells, so it crosses a block face; $\partial_{\Gp}U$ is a union of whole interfaces. (b) After the cut $j$ no $\Gp$-bond joins $U_j$ to $\Kc_j\setminus U_j$; every later component of $\Gp$ lies in $U_j$, in $\Kc_j\setminus U_j$, or outside $\Kc_j$. The sides of the cuts form a laminar family. (c) $|\partial_{\Gp}U_j|\ge c_gL_c^2$ for every cut. (d) Cut bonds support no hull material, alone or with seal bonds, filled blocks and box faces, and cut bonds block no tube of another interface. (e) If $J(U)>0$, then $\partial_{\Gp}U$ contains two interfaces at distance $>R_{\rm NF}$.
\end{lemma}

\begin{proof}
(a) A $\Gp$-bond with both ends in a block $Q$ lies in $\Gp\cap\hat Q(Q)$, and hence its ends lie in one cell. Therefore a bond of $\partial_{\Gp}U$ joins cells of different blocks, and all bonds across $F'$ between a cell of $U$ and a cell outside $U$ lie in $\partial_{\Gp}U$. (b) The cut consists of all $\Gp$-bonds between $U_j$ and $\Kc_j\setminus U_j$, where $\Kc_j$ is a component of $\Gp$ at time $t_j$, and $\Gp$ only shrinks by (H4) of Proposition~\ref{prop:4.4}. Consequently, each later component lies in $U_j$, in $\Kc_j\setminus U_j$, or outside $\Kc_j$. If $j_a$ fires before $j_b$, the sides of $j_b$ lie in $\Kc_{j_b}$, a component at time $t_{j_b}$, so each of them lies in a side of $j_a$ or misses both. As the two sides of one cut are disjoint, any two sides are disjoint or nested. (c) By (a) the cut contains a whole interface $I$, and (S-face) holds at $t_j$ by Theorem~\ref{thm:4.rest}(b). Thus $|\partial_{G'}U_j|\ge\good(I)\ge c_gL_c^2$, where the second inequality comes from Lemma~\ref{lem:4.17}.
(d) Cut bonds are grid bonds with the hull weight, and they therefore enter Corollary~\ref{cor:4.10} and Lemma~\ref{lem:4.11} as grid-type terms, as seal bonds do. A tube crosses no other block plane, by (G5), and this proves the last assertion of (d).
(e) If $J(U)>0$, then $\varphi_s(\partial_{G'}U)\ge2$ for some $s=2^jL\ge R_{\mathrm{NF}}$, and interfaces in different $s$-clusters are more than $s$ apart.
\qed
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:3.19}]
Termination is Proposition~\ref{prop:4.4}(b). At the terminal state a full pass changes nothing. Thus the state is hull-closed, which by Lemma~\ref{lem:4.2} implies (S-hull), and no removal cell is over threshold, which is (S-rem). For the same reason, no surviving block has auxiliary material, by Lemma~\ref{lem:4.aux}(a), so (S-face) and (S-plug) hold by Theorem~\ref{thm:4.rest}. Finally, the cut rule has no candidate at the terminal state, which is (S-NF) and contains the form used in Section~\ref{ssec:7.8}.

We now assume that the Euclidean diameter of $\Lamo$ is at most $R_{\mathrm{NF}}$. Then no pair is fragmented, since the right side of \eqref{eq:3.NFstar} is positive and, by Lemma~\ref{lem:3.18}(e), $J(U)>0$ would give two interfaces more than $R_{\mathrm{NF}}$ apart. Since, moreover, by Theorem~\ref{thm:4.30} the Phase-1 output is hull-closed, satisfies (S-rem) and has no auxiliary material, Phase~2 changes nothing in this case.
\qed
\end{proof}