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\subsection{Constants of Section~\ref{sec:5}}\label{S:const5}

\paragraph{Theorem~\ref{thm:5.1}.} The constants of \eqref{eq:5.2} are
\[
c_T'':=1+\log(64\kpl),\qquad \kpl:=12e,\qquad C_3:=\Cfree+20(2C_\star'+4\Cresp+1)\quad(\text{Proposition~\ref{prop:5.36}}),
\]
with $\Cst$, $\Cfree$ and $\Cresp$ as in item~10 of Section~\ref{ssec:D.1}. The quantity $\eps''$ is defined in \eqref{eq:5.epspp}, with $R'=R_*+2\Rresp=R_*+20\ell$ and $r'=2R_*+2$. The conditions used in the proof are those of item~(a) of Section~\ref{ssec:D.5}, which hold for large $\tau$ because $|\log\eps''|\ge\gamma_T(K)\tau-O(\log\tau+\log r_0)$ with $\gamma_T(K)>0$ for $K>K_*$ (Lemma~\ref{lem:5.16}), the constants $C_3$ and $C_{\Bfr}$ are free of $\beta$ and $K$, the quantities $\epsD$, $M_m$ and $\RNF^{-1}$ are doubly exponentially small in $\tau$, and $\log d_0=O(\log\tau)$. The constants $\Cresp$, $C_{\Bfr}$ and $C_3$ are fixed before $K$, and hence they affect $\beta_0(K)$ but not $K_*$. As for the order of $\etaT$, we have $\Cfree=O(C_\star'(1+1/c_s))$, hence $C_3=O((C_\star'+\Cresp)(1+1/c_s))$, and Lemma~\ref{lem:2.6} shows that $\gamma_T(K)$ is of the order of $\log^2K$ as $K\to\infty$.

\paragraph{Lemma~\ref{lem:5.16}.} Since $R_*\ge w'\ge12\ell$ by Theorem~\ref{thm:4.42}, we have $2R_*+1\le3R_*$ and $2r'+1=4R_*+5\le5R_*$. With the bound $q_m\le24r_0^3\delta^{1/2}$ of Lemma~\ref{lem:5.11} we obtain
\begin{align*}
\log\epsc&\le\log(8e)+3\log3+3\log5+\log24+6\log R_*+3\log r_0-\tfrac12|\log\delta|\\
&\le14.4+6\log R_*+3\log r_0-\tfrac12|\log\delta| .
\end{align*}
By Theorem~\ref{thm:4.42}, $R_*=O((\ell/c_s)\log(\ell/c_s))$ (Section~\ref{S:C1}), and $\ell\le\ell_D=O(\mt^{-1})$ by Definition~\ref{def:3.1}. With $\log(1/\mt)=\pi^2\vCb(0)\,\tau+O(\log\tau)$ and $\pi^2\vCb(0)<2.4943551$, these bounds and the bound on the entropy exponent in Section~\ref{ssec:D.3} give $6\log R_*\le14.9662\,\tau+O(\log\tau)$. Also $\tfrac12|\log\delta|=c_\delta(K)\beta/2=\tau c_\delta(K)/(2K)$. This is the first inequality of \eqref{eq:5.9}. By Lemma~\ref{lem:2.6}, $c_\delta(K)\ge c_\delta^{\mathrm{sp}}(K)-1/(4a_*)-O(\log\beta/\beta)$. The function $\gamma_T(K):=(c_\delta^{\mathrm{sp}}(K)-1/(4a_*))/(2K)-14.9662$ is increasing in $K$, and its zero is given, together with the margins, in Section~\ref{ssec:D.3}. Hence $\gamma_T(K)>0$ for $K>K_*$. The $O(\log\beta/\beta)$ term contributes $O(\log\tau)$ after multiplication by $\tau/(2K)$. For $\eps''$ we use $\Rresp=10\ell$ from Lemma~\ref{lem:5.32}, $L_c\le\ell$ and $R_*=O(\ell\tau)$. Hence $\log(2R'+4L_c+1)\le\log(1/\mt)+O(\log\tau)$, and the same computation applies.

\paragraph{Lemma~\ref{lem:5.25}.} In the proof of (c), with $\rho=\rho(s)$, we have $|x_c-x_Q|_\infty\le\lceil\rho\rceil/2\le\rho/2+1$. In (ii), the ends of the other cut bond are at distance more than $73\rho-1$ from $x_c$ by (U2), hence more than $73\rho-\rho/2-2>9\rho$ from $x_Q$. In (iii), two bonds of $B_\infty(x_c,10\rho)$ have ends within coarse distance $20\rho$, and $\sqrt3\rbar(20\rho+1)<128\rbar\rho=s$, so (U1) applies. For the last claim of (c), the two chosen bonds have ends within coarse distance $\rho+2$, and $\sqrt3\rbar(\rho+3)\le s$. The range $27L_c\le\rho\le\mathrm{side}_{\min}/24$ holds. The lower bound follows from $\RNF\ge3456\rbar L_c$. For the upper bound, $\varphi_s\ge2$ forces $s$ below the lattice diameter of $\Lc$, which is at most $\sqrt3\rbar\,\mathrm{side}_{\max}\le2\sqrt3\rbar\,\mathrm{side}_{\min}$, the last inequality by Lemma~\ref{lem:3.3}(d), and hence $\rho<2\sqrt3\,\mathrm{side}_{\min}/128$.

\paragraph{Lemmas~\ref{lem:5.31} and~\ref{lem:5.32}.} In the proof of Lemma~\ref{lem:5.31}, summing $121(m_q+m_{q'})$ over the six faces of each block $q$ of $\Delta_0$, we count at most $121(6m(\Delta_0)+6m(U))\le1452\,m$ seal bonds with an end in $\Delta_0$ and, summing over the blocks of $U$, at most $121(6m(U)+6m(U_2(\Delta_0)))\le1452\,m$ seal bonds with both ends in $U$. With $u_3\le c_s^{-1}(i+2904m+2n)$ we obtain
\[
\mathrm{units}(\Delta_0)\le1452m+2u_3+i+2\cdot1452m+2n+6m\le\Bigl(4362+\frac{5808}{c_s}\Bigr)m+\Bigl(2+\frac4{c_s}\Bigr)n+\Bigl(1+\frac2{c_s}\Bigr)i,
\]
since $1452+2904+6=4362$ and $2\cdot2904=5808$. At $c_s=1$, $(C_a,C_b,C_c)=(10170,6,3)$, and the loads of Lemma~\ref{lem:5.32} are $125C_a=1.27\cdot10^6$, $125C_b=750$ and $125C_c=375$. Thus
\[
\Cresp:=375C_a+125C_c\Cst=3.81\cdot10^6+375\cdot2.2141\cdot10^7=8.307\cdot10^9 ;
\]
the term $125C_c\Cst$, the load of the Phase-1 insulation owned by a part, dominates. The range is $\Rresp=10\ell\ge9\ell+1$.

\paragraph{Proposition~\ref{prop:5.34}.} A touched $Y_\Delta$ costs at most $2|\partial\Delta|+2|\mathrm{ins}^1\cap E(\Delta)|+4.5|\partial\Delta|=6.5|\partial\Delta|+2|\mathrm{ins}^1\cap E(\Delta)|$. A touched $Y=Y_{\Delta'}$ costs $2|\partial Y|+|Y|/\ell$, which by the bound on $|\partial Y|$ is at most $4\,\#\{\text{bonds of types (a) to (c) with an end in }\Delta'\}+13|Y|/\ell$, and $13|Y|/\ell\le(26L_c/\ell)m(\Delta')$. Since a bond has two ends, it is charged by at most two cells of the third kind, at most $4$ each. A unit $(\Delta,b)$ is charged at most $6.5$ by $Y_\Delta$ and at most $4$ by the cell of the other end of $b$. A bond of $\mathrm{ins}^1$ inside a removed cell is charged at most $2$ by $Y_\Delta$. All coefficients are at most $20$.

\paragraph{Lemma~\ref{lem:5.35} and Proposition~\ref{prop:5.36}.} The constants are
\begin{gather*}
C_{\Bfr}:=30C_\star'+170\Cresp+30,\qquad M_m:=16\epsD(1+8L_c)(C_\star'+3\Cresp+6),\\
C_3:=\Cfree+20(2C_\star'+4\Cresp+1).
\end{gather*}

To obtain $C_{\Bfr}$, we consider the proof of Lemma~\ref{lem:5.35}(c). There the bonds of $\mathrm{ins}^1$ number at most $C_\star'N_{\mathcal Q}$, as in (b), the cut bonds number at most $A_{\mathcal F}$, and the units number at most $2\Cresp(N_{\mathcal Q}+A_{\mathcal F})$ by (a), each of these being charged by at most two blocks. The marks involved, at most $N_{\mathcal Q}$, belong to parts of $\Qatt$, each charged by at most seven blocks. Summing, we obtain
\[
|\Bfr_{\mathcal F}|\le2\bigl(C_\star'N_{\mathcal Q}+A_{\mathcal F}+2\Cresp(N_{\mathcal Q}+A_{\mathcal F})\bigr)+7N_{\mathcal Q}\le(2C_\star'+4\Cresp+10)N_{\mathcal Q}
\]
by Lemma~\ref{lem:5.35}(b), and $2C_\star'+4\Cresp+10\le C_{\Bfr}$.

For $M_m$, we have $m\le2A_{\mathcal F}\le16\epsD(1+8L_c)(C_\star'+3\Cresp+6)N_{\mathcal Q}=M_mN_{\mathcal Q}$ by Lemma~\ref{lem:5.35}(d) and~(b).

\emph{The size of $\Cresp$.} We have $\Cresp\ge125C_c\Cst\ge125\Cst$, in particular $\Cresp\ge C_\star'+20$.

For $C_3$, we note that $20(u_{\mathrm{touch}}+c_{\mathrm{touch}}+i_{\mathrm{touch}})\le20(\Cresp+C_\star')N_{\mathcal Q}+20(\Cresp+1)A_{\mathcal F}$ in the proof of Proposition~\ref{prop:5.36}, and that $26L_c/\ell\le26/308<1/11$. With $A_{\mathcal F}\le N_{\mathcal Q}/(48\Cresp+2)$, all terms with $A_{\mathcal F}$ together are at most $N_{\mathcal Q}$, and
\[
\#(\Xi'\cap\Ffr)/\rbar^{2}\le\bigl(\Cfree+20C_\star'+20\Cresp+2\bigr)N_{\mathcal Q}\le C_3N_{\mathcal Q}.
\]

\emph{The binding condition.} The first condition of \eqref{eq:5.17} reads $|\log\eps''|\ge3C_{\Bfr}\ln(2^{30}\cdot26e)=75.16\,C_{\Bfr}$. At the illustrative values $(c_g,c_s,r_0)=(1/4,1,1)$, the constants of item~10 are $\Cst=2.2141\cdot10^7$, $\Cfree=5.536\cdot10^7$ and $\Cresp=8.307\cdot10^9$, hence $C_3=6.655\cdot10^{11}$ and $C_{\Bfr}=1.413\cdot10^{12}$, and this condition becomes $|\log\eps''|\ge1.06\cdot10^{14}$. The condition $\etaT\le1$ needs $|\log\eps''|\ge3c_T''\rbar^{2}C_3=1.73\cdot10^{13}\rbar^{2}$, with $c_T''=8.644$, and $\etaT^{(1)}\le1$ needs $|\log\epsc|\ge c_T\rbar^{2}\Cfree=3.63\cdot10^8\rbar^{2}$. Each of these is a lower bound on $\tau$, by Lemma~\ref{lem:5.16}.