Proof. If (2.1) holds, pairwise coprimality gives \operatorname{lcm}(a,b)=\operatorname{lcm}(a,c)=\operatorname{lcm}(b,c)=txyz prime by prime.

Conversely, let L be the common value in (1.1). Fix a prime p and write A=v_p(a),\qquad B=v_p(b),\qquad C=v_p(c),\qquad M=v_p(L). The maximum of each pair among A,B,C is M. Therefore at least two of A,B,C equal M. Define x=\frac{L}{c},\qquad y=\frac{L}{b},\qquad z=\frac{L}{a}. If, for example, p divided both x and y, then C<M and B<M, which would contradict \max(B,C)=M. The same argument for the other two pairs shows that x,y,z are pairwise coprime.

Finally put t=abc/L^2. At the prime p its exponent is A+B+C-2M\ge0, because at least two of A,B,C equal M. Hence t is an integer. A direct valuation calculation now gives v_p(txy)=A,\qquad v_p(txz)=B,\qquad v_p(tyz)=C. Since this holds for every prime, (2.1) follows. ◻

Proof. The polynomial-method bound of Ellenberg and Gijswijt [3] gives |A|\le3\min_{0<t<1} \bigl((1+t+t^2)t^{-2/3}\bigr)^H. Taking t=1/2 gives (1+t+t^2)t^{-2/3}=3\kappa_{\mathrm{cap}}. Also \kappa_{\mathrm{cap}}^3=4\left(\frac7{12}\right)^3 =\frac{343}{432}<1, which proves the displayed strict inequality. The same exact calculation is reproduced by the companion verifier. ◻

Proof. Fix P throughout the proof. Every positive integer has a unique factorization n=mq with q\in\mathcal M_P and (m,\prod_{p\in P}p)=1. If a safe subset of [N] is restricted to one fixed m-fibre, the resulting set of q’s is safe: here m is coprime to every q, so \operatorname{lcm}(mq_i,mq_j)=m\operatorname{lcm}(q_i,q_j). It follows that \begin{equation} \label{eq:fibre-bound} f(N)\le \sum_{\substack{m\le N\\(m,\prod_{p\in P}p)=1}}b_P(N/m). \tag{3.4} \end{equation}

Finite inclusion–exclusion gives A_P(x):=\#\{m\le x:(m,\textstyle\prod_{p\in P}p)=1\} =\delta_Px+E_P(x), \qquad E_P(x)=O_P(1). The function t\mapsto b_P(N/t) is nonincreasing, and its total variation on [1,N] is at most b_P(N). Stieltjes summation therefore gives \begin{align} \sum_{\substack{m\le N\\(m,\prod_{p\in P}p)=1}}b_P(N/m) &=\int_{1^-}^{N}b_P(N/t)\,\,\mathrm dA_P(t)\notag\\ &=\delta_P\int_1^N b_P(N/t)\,\,\mathrm dt+O_P(b_P(N)). \label{eq:stieltjes}\tag{3.5}\end{align} For completeness, the error term follows by integration by parts: the two endpoint terms are O_P(b_P(N)), and the remaining integral is bounded by \|E_P\|_\infty times the total variation of b_P(N/t).

In the main integral set u=N/t. Then \begin{equation} \label{eq:envelope-integral} \delta_P\int_1^N b_P(N/t)\,\,\mathrm dt =\delta_PN\int_1^N b_P(u)\frac{\,\mathrm du}{u^2}. \tag{3.6} \end{equation} By (3.2), b_P(N)=o(N) for fixed P. Divide (3.4)–(3.6) by N and let N\to\infty to obtain (3.3). ◻

Proof. Every q\in\mathcal M_P has a unique decomposition \begin{equation} \label{eq:unit-decomposition} q=m\,d(S), \qquad S=U_E(q), \qquad U_E(m)=\varnothing, \qquad S\subseteq E\setminus\operatorname{supp}(m). \tag{3.13} \end{equation} Indeed, d(S) removes exactly one copy of each prime whose exponent in q is one. Hence every prime of E has exponent either zero or at least two in m.

Fix m in (3.13) and consider a safe family in \mathcal M_P\cap[1,T]. Its m-section is admissible. Otherwise three supports satisfying (3.7) would give \operatorname{lcm}(md(S_i),md(S_j))=m\,d(S_i\cup S_j) for each pair, contradicting safety. Therefore b_P(T)\le \sum_{\substack{m\in\mathcal M_P\\U_E(m)=\varnothing}} A_{E\setminus\operatorname{supp}(m)}^{\mathrm{sf}}(T/m). Insert this inequality into (3.1). Positivity allows termwise integration. For each eligible m, the substitution T=mx and the convention (3.8) give \begin{align*} \int_1^\infty A_{E\setminus\operatorname{supp}(m)}^{\mathrm{sf}}(T/m)\frac{\,\mathrm dT}{T^2} &=\frac1m\int_{1/m}^\infty A_{E\setminus\operatorname{supp}(m)}^{\mathrm{sf}}(x)\frac{\,\mathrm dx}{x^2}\\ &=\frac1m\int_1^\infty A_{E\setminus\operatorname{supp}(m)}^{\mathrm{sf}}(x)\frac{\,\mathrm dx}{x^2} =\frac1m I_{E\setminus\operatorname{supp}(m)}^{\mathrm{sf}}. \end{align*} This proves (3.11).

It remains to normalize the right side. For an eligible m, put R=E\setminus\operatorname{supp}(m) and define \begin{equation} \label{eq:deletion-law} \mathcal W(m):=\delta_P\frac{Z_R}{m}. \tag{3.14} \end{equation} We verify prime by prime that \mathcal W is a probability law. If p\notin E, summing all valuations gives (1-p^{-1})\sum_{k\ge0}p^{-k}=1. If p\in E, valuation zero contributes (1-p^{-1})(1+p^{-1})=1-p^{-2}, while all permitted positive valuations, namely k\ge2, contribute (1-p^{-1})\sum_{k\ge2}p^{-k}=p^{-2}. Thus the local masses sum to one. They also show that the events \{p\notin R\} are independent and satisfy \begin{equation} \label{eq:deletion-error-law} \mathbb P_{\mathcal W}(p\notin R)=p^{-2}\qquad(p\in E). \tag{3.15} \end{equation}

The right side of (3.11) is exactly \mathbb E_{\mathcal W}\left[\frac{I_R^{\mathrm{sf}}}{Z_R}\right]. On \{R=E\} the random ratio equals I_E^{\mathrm{sf}}/Z_E; off that event it lies in [0,1] by (3.10). Hence C(P)\le\frac{I_E^{\mathrm{sf}}}{Z_E} +\mathbb P_{\mathcal W}(R\ne E) \le\frac{I_E^{\mathrm{sf}}}{Z_E}+\sum_{p\in E}p^{-2}, which is (3.12). ◻

Proof. Apply Proposition 3.2 with P=P_y and E=E_y. Since \min E_y\to\infty, \sum_{p\in E_y}p^{-2} \le\sum_{n\ge\min E_y}n^{-2}\longrightarrow0. Thus C(P_y)\to0. Proposition 3.1 gives, for each fixed y, \limsup_{N\to\infty}\frac{f(N)}N\le C(P_y). Letting y\to\infty proves the corollary. ◻

Proof. The word map is injective: if two words first differ in coordinate i, then one of the nonempty petals x_i,y_i,z_i occurs in exactly one of the two supports.

Consider a nonconstant affine line \omega,\qquad \omega+v,\qquad \omega+2v \quad\text{in }\mathbb F_3^H. In a coordinate where v_i=0, all three words make the same choice. In a coordinate where v_i\ne0, the three states are 0,1,2 in some order. The pairwise union of any two corresponding coordinate contributions is then x_i\cup y_i\cup z_i. It follows coordinate by coordinate that the three full supports have identical pairwise unions. Consequently the preimage of an admissible family contains no nonconstant affine line. Proposition 2.3 bounds that preimage by 3\kappa_{\mathrm{cap}}^H3^H, proving the lemma. ◻

Proof. We separate the construction into four steps.

Step 1: a joint cube–cutoff measure. For a cube c, put D(c):=\max_{\omega\in\mathbb F_3^H}d(S_\omega(c)). Define a finite measure on the cube space times [1,\infty) by \begin{equation} \label{eq:joint-measure} \,\mathrm d\Lambda(c,x):= \mathcal G(c)\,\mathbf 1_{\{x\ge D(c)\}} \frac{\,\mathrm dx}{x^2}\,\,\mathrm d\lambda(c). \tag{4.4} \end{equation} Since D(c)\ge\mathcal G(c), while (4.2) gives D(c)\le\mathrm e^\zeta\mathcal G(c), \begin{equation} \label{eq:joint-mass} \mathrm e^{-\zeta}\le \|\Lambda\| =\mathbb E_\lambda\frac{\mathcal G(c)}{D(c)} \le1. \tag{4.5} \end{equation}

Step 2: the weighted mass of one fibre. Fix a word \omega and a support S\subseteq\mathcal R. Whenever S_\omega(c)=S, the balance assumption implies \begin{equation} \label{eq:fibre-balance} \mathrm e^{-\zeta}d(S)\le\mathcal G(c)\le\mathrm e^\zeta d(S), \qquad D(c)\le\mathrm e^{2\zeta}d(S). \tag{4.6} \end{equation} The second inequality follows because every word product is at most \mathrm e^\zeta\mathcal G(c) and \mathcal G(c)\le\mathrm e^\zeta d(S). By (4.1), \nu_\omega(S)=\frac{G_\omega(S)}{Z_{\mathcal R}d(S)}. Consequently the weighted fibre mass M_\omega(S):= \int_{\{c:S_\omega(c)=S\}}\mathcal G(c)\,\,\mathrm d\lambda(c) satisfies the pointwise bounds \begin{equation} \label{eq:weighted-fibre} \frac{\mathrm e^{-\zeta}G_\omega(S)}{Z_{\mathcal R}} \le M_\omega(S)\le \frac{\mathrm e^\zeta G_\omega(S)}{Z_{\mathcal R}}. \tag{4.7} \end{equation}

Step 3: comparison with the canonical prefix law. We use the following convention. If \sigma is a finite signed measure, then \|\sigma\|_{\mathrm{var}}:=|\sigma|(\Omega) denotes its total variation norm. For probability measures P,Q we write d_{\operatorname{TV}}(P,Q):=\sup_E|P(E)-Q(E)| =\frac12\|P-Q\|_{\mathrm{var}}. Let \Lambda_\omega be the pushforward of \Lambda under (c,x)\mapsto(S_\omega(c),x). With respect to \rho:=\sum_{S\subseteq\mathcal R}\delta_S\otimes\frac{\,\mathrm dx}{x^2} \quad\text{on }2^{\mathcal R}\times[1,\infty), its density is L_\omega(S,x):= \int_{\{c:S_\omega(c)=S\}} \mathcal G(c)\mathbf 1_{\{D(c)\le x\}}\,\,\mathrm d\lambda(c). Let \Pi be the probability measure whose \rho-density is \begin{equation} \label{eq:canonical-prefix-law} \frac{\,\mathrm d\Pi}{\,\mathrm d\rho}(S,x) :=\frac1{Z_{\mathcal R}}\mathbf 1_{\{x\ge d(S)\}}. \tag{4.8} \end{equation} Indeed, \Pi(2^{\mathcal R}\times[1,\infty)) =\frac1{Z_{\mathcal R}}\sum_{S\subseteq\mathcal R}\frac1{d(S)}=1.

Fix S\subseteq\mathcal R and abbreviate G=G_\omega(S) and M=M_\omega(S). If G>0, (4.7) permits us to write M=\frac{\theta_SG}{Z_{\mathcal R}}, \qquad \mathrm e^{-\zeta}\le\theta_S\le\mathrm e^\zeta. If G=0, then (4.7) gives M=0, and the same formula holds after setting \theta_S=1. Since 0\le\zeta\le1, |\theta_S-1|\le\mathrm e^\zeta-1\le2\zeta. On the fibre S_\omega(c)=S, one has D(c)\ge d(S) and, by (4.6), D(c)\le\mathrm e^{2\zeta}d(S). Consequently L_\omega(S,x)=0 for x<d(S) and L_\omega(S,x)=M_\omega(S) for x\ge\mathrm e^{2\zeta}d(S). On the latter region, \begin{align} &\sum_{S\subseteq\mathcal R} \int_{\mathrm e^{2\zeta}d(S)}^\infty \left|L_\omega(S,x)-\frac1{Z_{\mathcal R}}\right| \frac{\,\mathrm dx}{x^2}\notag\\ &=\mathrm e^{-2\zeta}\sum_{S\subseteq\mathcal R} \frac{|M_\omega(S)-Z_{\mathcal R}^{-1}|}{d(S)}\notag\\ &\le \sum_{S\subseteq\mathcal R} \frac{|G_\omega(S)-1|+2\zeta G_\omega(S)} {Z_{\mathcal R}d(S)} \le\varepsilon+2\zeta. \label{eq:high-region}\tag{4.9}\end{align} Here the last line uses \mathbb E_{\mu_{\mathcal R}}|G_\omega-1|\le\varepsilon and \mathbb E_{\mu_{\mathcal R}}G_\omega=1.

It remains to control the transition strip \mathcal T:=\{(S,x):d(S)\le x<\mathrm e^{2\zeta}d(S)\}. Its \Pi-mass is exactly \begin{equation} \label{eq:strip-pi} \Pi(\mathcal T) =(1-\mathrm e^{-2\zeta}) \sum_{S\subseteq\mathcal R}\frac1{Z_{\mathcal R}d(S)} =1-\mathrm e^{-2\zeta}\le2\zeta. \tag{4.10} \end{equation} For the other marginal, Tonelli’s theorem and nonnegativity give \begin{align} \Lambda_\omega(\mathcal T) &=\int \mathcal G(c) \int_{d(S_\omega(c))}^{\mathrm e^{2\zeta}d(S_\omega(c))} \mathbf 1_{\{D(c)\le x\}}\frac{\,\mathrm dx}{x^2}\,\,\mathrm d\lambda(c)\notag\\ &\le(1-\mathrm e^{-2\zeta}) \int\frac{\mathcal G(c)}{d(S_\omega(c))}\,\,\mathrm d\lambda(c)\notag\\ &\le\mathrm e^\zeta(1-\mathrm e^{-2\zeta}) \le6\zeta. \label{eq:strip-lambda}\tag{4.11}\end{align} The penultimate inequality follows from (4.2); the last uses \mathrm e^\zeta\le\mathrm e<3 and 1-\mathrm e^{-2\zeta}\le2\zeta. On \mathcal T the variation norm of the difference is at most the sum of the two masses, while below x=d(S) both measures vanish. Combining (4.9)–(4.11) yields \begin{equation} \label{eq:unnormalized-prefix-TV} \|\Lambda_\omega-\Pi\|_{\mathrm{var}} \le\varepsilon+10\zeta. \tag{4.12} \end{equation}

Put a_\Lambda:=\|\Lambda\|. Every \Lambda_\omega has mass a_\Lambda, and (4.5) gives \mathrm e^{-\zeta}\le a_\Lambda\le1. The word marginal \overline\Lambda_\omega of \overline\Lambda:=a_\Lambda^{-1}\Lambda is a_\Lambda^{-1}\Lambda_\omega. Therefore \begin{align} \|\overline\Lambda_\omega-\Pi\|_{\mathrm{var}} &\le \|a_\Lambda^{-1}\Lambda_\omega-\Lambda_\omega\|_{\mathrm{var}} +\|\Lambda_\omega-\Pi\|_{\mathrm{var}}\notag\\ &=(1-a_\Lambda)+\|\Lambda_\omega-\Pi\|_{\mathrm{var}} \le\varepsilon+11\zeta, \label{eq:normalized-prefix-TV}\tag{4.13}\end{align} because 1-a_\Lambda\le1-\mathrm e^{-\zeta}\le\zeta. Thus, uniformly in \omega, \begin{equation} \label{eq:normalized-prefix-distance} d_{\operatorname{TV}}(\overline\Lambda_\omega,\Pi) \le\frac{\varepsilon+11\zeta}{2}. \tag{4.14} \end{equation}

Step 4: the moving extremizer. Fix a lexicographic order on the finite set 2^{\mathcal R}, and use the induced lexicographic order on its subfamilies. For each x\ge1, choose the first admissible family \mathcal F_x\subseteq\{S:d(S)\le x\} of maximum size. Only finitely many prefixes occur, so x\mapsto\mathcal F_x is piecewise constant and hence measurable. Set \mathcal E:=\{(S,x):S\in\mathcal F_x\}. Because \mathcal F_x is contained in the x-prefix, Tonelli’s theorem gives \begin{align} \Pi(\mathcal E) &=\frac1{Z_{\mathcal R}} \sum_{S\subseteq\mathcal R}\int_{d(S)}^\infty \mathbf 1_{\{S\in\mathcal F_x\}}\frac{\,\mathrm dx}{x^2}\notag\\ &=\frac1{Z_{\mathcal R}}\int_1^\infty |\mathcal F_x|\frac{\,\mathrm dx}{x^2} =\frac{I_{\mathcal R}^{\mathrm{sf}}}{Z_{\mathcal R}}. \label{eq:prefix-occupancy}\tag{4.15}\end{align}

For \overline\Lambda-almost every (c,x), one has x\ge D(c), so all 3^H word supports lie in the same x-prefix. Since \mathcal F_x is admissible, Lemma 4.1 implies \frac1{3^H}\sum_{\omega\in\mathbb F_3^H} \mathbf 1_{\{S_\omega(c)\in\mathcal F_x\}} \le3\kappa_{\mathrm{cap}}^H. Integrating this pointwise inequality and using that \overline\Lambda_\omega is the (S_\omega,x)-pushforward of \overline\Lambda, we obtain \begin{align} \frac1{3^H}\sum_{\omega\in\mathbb F_3^H} \overline\Lambda_\omega(\mathcal E) &=\int\frac1{3^H}\sum_{\omega\in\mathbb F_3^H} \mathbf 1_{\{S_\omega(c)\in\mathcal F_x\}} \,\,\mathrm d\overline\Lambda(c,x)\notag\\ &\le3\kappa_{\mathrm{cap}}^H. \label{eq:word-average-upper}\tag{4.16}\end{align} On the other hand, (4.14) and (4.15) imply, for every \omega, \overline\Lambda_\omega(\mathcal E) \ge\frac{I_{\mathcal R}^{\mathrm{sf}}}{Z_{\mathcal R}} -\frac{\varepsilon+11\zeta}{2}. Averaging this inequality over all words and comparing with (4.16) gives \frac{I_{\mathcal R}^{\mathrm{sf}}}{Z_{\mathcal R}} \le3\kappa_{\mathrm{cap}}^H+\frac{\varepsilon+11\zeta}{2}, which proves (4.3). ◻

Proof. Put \epsilon_T=T^{\vartheta-1}. Since \frac{w}{\epsilon_T}=\frac{\eta}{T^\vartheta}=o(1), we may enlarge I, only in an available direction at a band endpoint, to an interval [r,s]\subseteq[\epsilon_T,1] such that \begin{equation} \label{eq:interval-enlargement} I\subseteq[r,s],\qquad w\le s-r\le(C_0+2)w,\qquad r\ge\frac12\inf I. \tag{5.8} \end{equation} Set x=\mathrm e^{Tr} and \Delta=\mathrm e^{Ts}-\mathrm e^{Tr}. Because T(s-r)\asymp_{C_0}\eta, \Delta\asymp_{C_0}x\eta,\qquad \Delta\le x,\qquad \log\Delta=Tr+\log\eta+O_{C_0}(1). Now Tr\ge T^\vartheta/2, while |\log\eta|=o(\log T)=o(T^\vartheta). Hence, for large T, \log\Delta\ge\frac14T\inf I \quad\text{and}\quad \Delta\ge2. Brun–Titchmarsh, Proposition 2.2, gives \pi(x+\Delta)-\pi(x) \ll_{C_0}\frac{x\eta}{T\inf I}. Allowing one possible prime at a closed lower endpoint, \begin{align*} \sum_{\substack{p\in\mathcal B(T)\\u_p\in I}}\frac1p &\le\frac{\pi(x+\Delta)-\pi(x)+1}{x}\\ &\ll_{C_0}\frac{\eta}{T\inf I}+\mathrm e^{-T^\vartheta} \ll_{C_0}\frac{w}{\inf I}. \end{align*} The last absorption uses \mathrm e^{-T^\vartheta}=o(w). The enlargement in (5.8) was made for every I, including a degenerate interval. Equivalently, a degenerate I contains at most one normalized prime weight, whose reciprocal contribution is at most \mathrm e^{-T^\vartheta}=o(w); thus single-atom intervals cause no exception.

For (5.7), subtract (2.2) at \mathrm e^{Tt} and \mathrm e^{T(t+c_0w)}. Uniformly for t\in[r_0,r_1] this gives \sum_{\substack{p\in\mathcal B(T)\\t\le u_p\le t+c_0w}}\frac1p = \log\frac{t+c_0w}{t} +O(\mathrm e^{-c\sqrt{Tr_0}})+O(\mathrm e^{-Tr_0}). Both error terms are o(w) because \eta=T^{-o(1)}. Finally, 0\le \sum_{p\in\mathcal B(T)} \left(\frac1p-\frac1{p+1}\right) \le\sum_{n>\mathrm e^{T^\vartheta}}\frac1{n^2} \ll\mathrm e^{-T^\vartheta}=o(w), which proves the versions with 1/(p+1). ◻

Proof. Fix p and write h=h_p. Under the Poisson law, the total number of points at p is Poisson with mean 4h. The empty configuration and each of the four singleton configurations have probabilities \mathrm e^{-4h} and h\mathrm e^{-4h}, respectively. Therefore \begin{align*} 2d_{\operatorname{TV}}(\mathsf Q_p,\widetilde{\mathsf Q}_p) &\le |1-4h-\mathrm e^{-4h}| +4h(1-\mathrm e^{-4h}) +\mathbb P(\operatorname{Poisson}(4h)\ge2). \end{align*} The elementary inequalities 1-x\le\mathrm e^{-x}\le1-x+\frac{x^2}{2}, \qquad 1-\mathrm e^{-x}\le x \quad(x\ge0) bound the first two terms by 8h^2 and 16h^2. If N\sim\operatorname{Poisson}(4h), then \mathbb P(N\ge2)\le\frac{\mathbb E[N(N-1)]}{2}=8h^2. Thus the local total-variation distance is at most 16h^2.

Couple the arrays prime by prime. The probability that any local coupling fails is at most the sum of the local distances, which proves the first inequality in (5.17). Since h_p\le1/(3p), \sum_{p\in\mathcal B(T)}h_p^2 \le\frac19\sum_{n>\mathrm e^{T^\vartheta}}\frac1{n^2} \ll\mathrm e^{-T^\vartheta}. Finally, \log(1/w)=\log T+\log(1/\eta)=(1+o(1))\log T, whereas T^\vartheta/\log T\to\infty. Hence \mathrm e^{-T^\vartheta}=o(w^A) for every fixed A. ◻

Proof. Let H_+(x):=\sum_{p\le x}\frac1{p+1}. For 0\le t\le\lambda_T-R_0, \begin{align} 3\Lambda_T((R_0,R_0+t]) &= H_+(\mathrm e^{T\mathrm e^{-R_0}}) -H_+(\mathrm e^{T\mathrm e^{-(R_0+t)}}) +O(\mathrm e^{-T^\vartheta}). \label{eq:time-change-difference}\tag{5.24}\end{align} The logarithms of both prime cutoffs are at least T^\vartheta. By (2.3), the error in (5.24) is O(\mathrm e^{-cT^{\vartheta/2}}) uniformly in t. Since \log\log(\mathrm e^{T\mathrm e^{-s}})=\log T-s, the main term in (5.24) is t. This proves (5.20). The same subtraction over a fixed depth interval proves (5.23).

For the weighted tail, positivity and (2.4) give \begin{align*} \mathbb EV_{\ell,T} &=\frac1{3T} \sum_{\substack{\mathrm e^{T^\vartheta}<p<\mathrm e^{T\mathrm e^{-R_0}}}} \frac{\log p}{p+1}\\ &\le\frac1{3T} \sum_{p\le\mathrm e^{T\mathrm e^{-R_0}}}\frac{\log p}{p-1} =\left(\frac13+o(1)\right)\mathrm e^{-R_0}. \end{align*} The last expression is at most \mathrm e^{-R_0} for all sufficiently large T, proving (5.22). ◻

Proof. The inequality \mathrm e^{-x}\le1-x+x^2/2 gives q\gamma+\mathrm e^{-\gamma}-1 \le(q-1)\gamma+\frac{\gamma^2}{2} =-\frac3{5000}<0. It follows from independent increments that M(v):=\exp\!\bigl(\gamma(qv-\Pi(v))\bigr) is a nonnegative supermartingale.

We include the short maximal argument, including the harmless strictness at the boundary. Fix L<\infty and \epsilon>0. On a finite time grid in [0,L], stop M at the first grid point at which it reaches \mathrm e^{\gamma(b+\epsilon)}. Conditioning on whether the stopping time has already occurred shows directly that the stopped process is still a nonnegative supermartingale. Its expectation is therefore at most M(0)=1, so the grid crossing probability is at most \mathrm e^{-\gamma(b+\epsilon)}. Apply this to nested dyadic grids. Right continuity detects every strict crossing, and hence \mathbb P\bigl(\exists v\in[0,L]:qv-\Pi(v)>b+\epsilon\bigr) \le\mathrm e^{-\gamma(b+\epsilon)}. Let first \epsilon\downarrow0 and then L\to\infty. We obtain \mathbb P\bigl(\exists v\ge0:qv-\Pi(v)>b\bigr)\le\mathrm e^{-\gamma b}. The complement is exactly the event in (5.25). ◻

Proof. We construct under \widetilde{\mathsf Q}_T an event contained in B_T^{\mathrm P} whose probability is at least a fixed multiple of w^2. The whole-band coupling will then transfer it to the categorical law.

Step 1: fixed numerical choices. Set \begin{equation} \label{eq:first-moment-constants} \delta:=\frac1{100}, \qquad B:=350, \qquad m:=\lceil\alpha R_0\rceil+B=374. \tag{5.27} \end{equation} These constants satisfy the three inequalities needed below: \begin{equation} \label{eq:constant-certificate} 4\mathrm e^{-\gamma B/2}<\frac18, \qquad m\mathrm e^{1-R_0}<\frac{\delta}{2}, \qquad \frac{8\mathrm e^{-R_0}}{\delta}<\frac18. \tag{5.28} \end{equation} We record exact witnesses. First, \mathrm e^{7/2}> \sum_{n=0}^{7}\frac{(7/2)^n}{n!} =\frac{66007}{2048}>32, which proves the first inequality. Next, 374\mathrm e^{-74} <\frac{374\cdot4!}{74^4} =\frac{561}{1874161}<\frac1{200}. Finally, 800\mathrm e^{-75} <\frac{800\cdot3!}{75^3} =\frac{64}{5625}<\frac18. Only positive terms of the exponential series were used. These calculations are also checked exactly in the companion verifier.

Step 2: finite-depth buffers. Let \mathcal K:=\left[\frac9{20},\frac{11}{20}\right), \qquad \mathcal A:=\left[\frac{12}{25},\frac{13}{25}\right). For a half-open weight interval \mathcal I, write D_{\mathcal I}:=\{-\log u:u\in\mathcal I\}; its image is again taken with the corresponding half-open endpoint convention. Choose the four explicit, pairwise disjoint depth intervals \begin{align*} Q_c&=(R_0-1,R_0-7/8],& Q_x&=(R_0-3/4,R_0-5/8],\\ Q_y&=(R_0-1/2,R_0-3/8],& Q_z&=(R_0-1/4,R_0-1/8]. \end{align*} All lie in (R_0-1,R_0).

Let F_{\mathrm{fix}} be the following finite-depth event:

1. label c has exactly m points in Q_c and no other point of depth at most R_0;

2. label z has exactly m points in Q_z, exactly one point in D_{\mathcal A}, and no other point of depth at most R_0;

3. each label \ell\in\{x,y\} has exactly m points in Q_\ell, no point in [0,R_0]\setminus(Q_\ell\cup D_{\mathcal K}), and no condition imposed on its process in D_{\mathcal K}.

Since D_{\mathcal A},D_{\mathcal K}\subset(0,1) whereas every Q_\ell lies in (R_0-1,R_0)=(74,75), the anchor-depth regions are disjoint from all four buffer intervals. We now specify the finite atoms used to evaluate this event. For label c, take Q_c itself and the nonempty interval components of [0,R_0]\setminus Q_c; prescribe counts m and 0, respectively. For label z, take Q_z, D_{\mathcal A}, and the nonempty interval components of their complement in [0,R_0]; prescribe counts m, 1, and 0. For each \ell\in\{x,y\}, take Q_\ell and the nonempty interval components of [0,R_0]\setminus(Q_\ell\cup D_{\mathcal K}); prescribe count m on Q_\ell and count 0 on every listed component. The region D_{\mathcal K} is deliberately absent from the constrained atom family for these two labels. Endpoint assignments are inherited from the half-open conventions defining the intervals, so within each label the listed atoms are disjoint and describe exactly the conditions in (i)–(iii). Let k_{\ell,D} denote the prescribed count. When k_{\ell,D}=0, the corresponding factor below is simply \mathrm e^{-\Lambda_T(D)}. Poisson independent increments give the exact product \begin{equation} \label{eq:fixed-event-product} \widetilde{\mathsf Q}_T(F_{\mathrm{fix}}) = \prod_{(\ell,D)\ \mathrm{constrained}} \mathrm e^{-\Lambda_T(D)} \frac{\Lambda_T(D)^{k_{\ell,D}}}{k_{\ell,D}!}. \tag{5.29} \end{equation} By finite additivity and (5.23), every intensity in this finite product has a finite limit. Each atom on which a positive count is required has positive length and hence a strictly positive limiting intensity. It follows that, for all sufficiently large T, \begin{equation} \label{eq:fixed-event-lower} \widetilde{\mathsf Q}_T(F_{\mathrm{fix}}) \ge c_{\mathrm{fix}}>0, \tag{5.30} \end{equation} where c_{\mathrm{fix}} is independent of T.

Step 3: survival through the entire remaining depth. For a label \ell, define M_{\ell,T}(t):= \sum_{\substack{p\in\mathcal B(T)\\R_0<s_p\le R_0+t}}K_{\ell,p}, \qquad 0\le t\le\lambda_T-R_0. This independent-increment process can be realized as \begin{equation} \label{eq:poisson-time-change} M_{\ell,T}(t) =\Pi_\ell(\Lambda_T((R_0,R_0+t])), \tag{5.31} \end{equation} where the \Pi_\ell are independent rate-one Poisson processes. Put \epsilon_T:=\sup_t|\varepsilon_T(t)|. For all sufficiently large T, q\epsilon_T\le B/2.

Suppose that \begin{equation} \label{eq:survival-event} \Pi_\ell(v)+B/2\ge qv \quad\text{for every }v\ge0. \tag{5.32} \end{equation} With v=\Lambda_T((R_0,R_0+t]), the time-change estimate gives t/3\le v+\epsilon_T. Therefore \begin{align} M_{\ell,T}(t)+B &\ge qv+B/2 \ge \alpha t, \label{eq:profile-growth}\tag{5.33}\\ m+M_{\ell,T}(t) &\ge \alpha(R_0+t). \label{eq:full-profile-growth}\tag{5.34}\end{align} The second line uses m\ge\alpha R_0+B. By Lemma 5.4 and the first inequality in (5.28), the probability that (5.34) fails for at least one of the four labels is less than 1/8.

The weighted tails are also small at fixed positive cost. By (5.22), Markov’s inequality, and the last inequality in (5.28), \begin{equation} \label{eq:tail-union} \mathbb P\left( \max_{\ell\in\{c,x,y,z\}}V_{\ell,T}>\delta/2 \right) \le\frac{8\mathrm e^{-R_0}}{\delta}<\frac18. \tag{5.35} \end{equation} All depth->R_0 processes are independent of F_{\mathrm{fix}}. Let \mathcal E_T be the event that (5.34) holds for all four labels and all 0\le t\le\lambda_T-R_0, and that \begin{equation} \label{eq:small-tails} V_{\ell,T}\le\delta/2 \qquad(\ell=c,x,y,z) \tag{5.36} \end{equation} holds. The preceding two union bounds show that \begin{equation} \label{eq:profile-tail-success} \widetilde{\mathsf Q}_T(\mathcal E_T\mid F_{\mathrm{fix}}) \ge\frac34. \tag{5.37} \end{equation} No error depending on w has been subtracted.

Step 4: two adaptive correcting anchors. Let \mathcal H_T be the \sigma-field generated by the restrictions of all four labelled Poisson arrays except for the restrictions of the x- and y-arrays to D_{\mathcal K}. The event F_{\mathrm{fix}} is \mathcal H_T-measurable because it imposes no condition on those two omitted restrictions. The event \mathcal E_T is also \mathcal H_T-measurable: it is determined entirely by the four depth->R_0 processes. By independence across labels and independent increments on disjoint depth sets, conditionally on \mathcal H_T the two omitted restrictions remain independent Poisson processes, each with its original intensity \Lambda_T restricted to D_{\mathcal K}.

Let U_z\in\mathcal A be the normalized weight of the unique z-point in D_{\mathcal A}. For \ell\in\{x,y,z\}, let E_\ell be the total weight of the m points in Q_\ell plus V_{\ell,T}. Every point in Q_\ell has weight at most \mathrm e^{1-R_0}, so (5.28) and (5.36) imply \begin{equation} \label{eq:error-weights} 0\le E_\ell\le\delta. \tag{5.38} \end{equation}

Work conditionally on \mathcal H_T, on F_{\mathrm{fix}}\cap\mathcal E_T. The variables U_z,E_x,E_y,E_z are then fixed. Define the \mathcal H_T-measurable centers \begin{equation} \label{eq:anchor-centers} t_x:=U_z+E_z-E_x, \qquad t_y:=U_z+E_z-E_y \tag{5.39} \end{equation} and target intervals \begin{equation} \label{eq:anchor-intervals} I_x:=\left[t_x-\frac w8,t_x+\frac w8\right], \qquad I_y:=\left[t_y-\frac w8,t_y+\frac w8\right]. \tag{5.40} \end{equation} From U_z\in[12/25,13/25) and (5.38), t_x,t_y\in\left[\frac{47}{100},\frac{53}{100}\right]. Since w\to0, both intervals in (5.40) lie in \mathcal K for all sufficiently large T.

Require the x-process to have exactly one point in I_x and no other point in D_{\mathcal K}, and impose the analogous requirement on the y-process. Let \mathcal A_T denote this two-anchor event on F_{\mathrm{fix}}\cap\mathcal E_T, and declare it empty off that event. Conditionally on \mathcal H_T, the probability of the x-requirement is exactly \begin{equation} \label{eq:anchor-probability} \left( \sum_{\substack{p\in\mathcal B(T)\\u_p\in I_x}}h_p \right) \exp\!\left( -\sum_{\substack{p\in\mathcal B(T)\\u_p\in\mathcal K}}h_p \right). \tag{5.41} \end{equation} The interval I_x has the form [t,t+w/4] with t=t_x-w/8; for large T, this left endpoint lies in the fixed compact interval [23/50,27/50]\subset(0,1). Thus Lemma 5.1, with c_0=1/4, gives, almost surely on F_{\mathrm{fix}}\cap\mathcal E_T and uniformly over every possible conditional realization, \sum_{\substack{p\in\mathcal B(T)\\u_p\in I_x}}h_p =\frac13\sum_{\substack{p\in\mathcal B(T)\\u_p\in I_x}}\frac1{p+1} \ge c_1w. Although I_x is random before conditioning, it is \mathcal H_T-measurable and hence is a fixed deterministic interval under each conditional realization. The lower bound in Lemma 5.1 is uniform over all left endpoints in [23/50,27/50], so the same constant c_1 applies simultaneously to every such realization. The identical observation applies to I_y. Moreover, (2.3) gives \sum_{\substack{p\in\mathcal B(T)\\u_p\in\mathcal K}}h_p \longrightarrow\frac13\log\frac{11}{9}. Consequently the exponential factor in (5.41) is at least a fixed c_2>0. The same bounds hold for I_y. Independence of the two omitted processes therefore gives both correcting anchors with conditional probability at least \begin{equation} \label{eq:two-anchor-lower} c_3w^2. \tag{5.42} \end{equation}

Step 5: verification of the target event and transfer. Let U_x,U_y be the weights of the two correcting anchors. On the event just constructed, X=U_x+E_x,\qquad Y=U_y+E_y,\qquad Z=U_z+E_z. By (5.39)–(5.40), \begin{equation} \label{eq:balanced-totals} |X-Z|\le w/8,\qquad |Y-Z|\le w/8. \tag{5.43} \end{equation} Thus the range of X,Y,Z is at most w/4. Also Z\in[48/100,53/100] by (5.38); since w\to0, (5.43) places all three totals in the fixed closed interval J for large T.

At depth R_0, every label has at least m points. For s=R_0+t, (5.34) gives N_\ell^{\mathrm P}(s) \ge\alpha s\ge\lfloor\alpha s\rfloor \qquad(\ell=c,x,y,z). The constructed event is therefore contained in B_T^{\mathrm P}. The conditional estimate (5.42) is uniform on F_{\mathrm{fix}}\cap\mathcal E_T. The tower property, (5.30), and (5.37) therefore give \begin{align*} \widetilde{\mathsf Q}_T( F_{\mathrm{fix}}\cap\mathcal E_T\cap\mathcal A_T) &=\mathbb E_{\widetilde{\mathsf Q}_T}\!\left[ \mathbf 1_{F_{\mathrm{fix}}\cap\mathcal E_T} \widetilde{\mathsf Q}_T(\mathcal A_T\mid\mathcal H_T) \right]\\ &\ge c_3w^2\, \widetilde{\mathsf Q}_T(F_{\mathrm{fix}}\cap\mathcal E_T) \ge\frac34c_{\mathrm{fix}}c_3w^2. \end{align*} The event on the left is contained in B_T^{\mathrm P}, so \begin{equation} \label{eq:poisson-first-moment} \widetilde{\mathsf Q}_T(B_T^{\mathrm P}) \ge\frac34c_{\mathrm{fix}}c_3w^2=:c_4w^2. \tag{5.44} \end{equation} Finally, the defining property of total-variation distance gives \left| \mathsf Q_T(B_T^{\mathrm P}) -\widetilde{\mathsf Q}_T(B_T^{\mathrm P}) \right| \le d_{\operatorname{TV}}(\mathsf Q_T,\widetilde{\mathsf Q}_T). Lemma 5.2 with A=2 makes the right side o(w^2). Hence (5.19) and (5.44) give \beta_T =\mathsf Q_T(B_T^{\mathrm P}) \ge\widetilde{\mathsf Q}_T(B_T^{\mathrm P})-o(w^2) \ge c_4w^2-o(w^2) \ge\frac{c_4}{2}w^2 for all sufficiently large T. This proves (5.26). ◻

Proof. Expand the expectation on the left. For every ordered tuple appearing in the expansion, make the bijective substitution A=S\setminus\{p_1,\ldots,p_d\}. The deleted points are pairwise distinct and absent from A, so \mu(A\cup\{p_1,\ldots,p_d\}) =\mu(A)\prod_{j=1}^d\frac{r_{p_j}}{1-r_{p_j}}. Substitution gives (6.6). Every sum is finite, so no limiting interchange is involved. ◻

Proof. The inequality K\ge a\lambda_T is (6.7) at s=\lambda_T. Put s_{(k)}=-\log u_{(k)}. If s_{(k)}\le R_0, then u_{(k)}\ge\mathrm e^{-R_0}\ge\ell_k. Otherwise take R_0\le s<s_{(k)} and let s\uparrow s_{(k)}. At depth s there are at most k-1 points, so as\le k-1. Hence s_{(k)}\le(k-1)/a, which is stronger than (6.8). ◻

Proof. Write \mathcal P_T(S):= \{N_S(s)\ge as\text{ for every }R_0\le s\le\lambda_T\}. For brevity, write \mu:=\mu_{\mathcal B(T)}. For each realization put K:=|S|. All ranks below refer to decreasing order of the weights u_p. For a marked realization \xi=(\xi_p)_{p\in S}, define r(\xi):=\dim_{\mathbb R}\operatorname{span}\{\xi_p:p\in S\}. The span of the empty set is understood to be \{0\}. Thus the cases r(\xi)=0,1,2 are disjoint and exhaustive.

Mark-counting sublemma. For v\in\mathcal M\setminus\{0\}, put L(v):=\mathcal M\cap\mathbb Rv=\{-v,0,v\}. Thus |L(v)|=3 and |\mathcal M\setminus L(v)|=6. On a support of size K, the total probability of all mark sequences whose first nonzero mark has rank i and whose first subsequent mark outside its real span has rank j>i is \begin{equation} \label{eq:rank-two-mark-mass} 9^{-K}\cdot8\cdot3^{j-i-1}\cdot6\cdot9^{K-j} =16\,3^{-(i+j)}. \tag{6.12} \end{equation} The total probability of all rank-one mark sequences whose first nonzero mark has rank i is \begin{equation} \label{eq:rank-one-mark-mass} 9^{-K}\cdot8\cdot3^{K-i} =8\,3^{-(K+i)}. \tag{6.13} \end{equation} Indeed, there are eight choices for the mark at rank i. For (6.12), the ranks i+1,\ldots,j-1 have the three choices in L(v), rank j has the six choices outside L(v), and every later rank has nine choices. For (6.13), every later mark must belong to L(v). Multiplication and division by the 9^K equally likely mark sequences prove both formulas.

Rank two. For a rank-two marked support, let i be the rank of its first nonzero mark and let j>i be the rank of its first mark outside the span of the mark at rank i. Let p and q be the primes at these ranks. This choice is canonical.

We spell out the data remaining after deletion of p and q. Put A=S\setminus\{p,q\} and K=|A|+2. Order A as r_1,\ldots,r_{K-2} with b_1:=u_{r_1}>\cdots>b_{K-2}:=u_{r_{K-2}}, \qquad b_0:=+\infty,\quad b_{K-1}:=0. When A=\varnothing, the displayed ordered list is empty and only the two endpoint conventions apply. For 1\le i<j\le K, let \mathcal G_{ij}(A) be the set of ordered pairs of distinct primes p,q\in\mathcal B(T)\setminus A such that \begin{equation} \label{eq:rank-two-gaps} \begin{cases} b_{i-1}>u_p>u_q>b_i, & j=i+1,\\ b_{i-1}>u_p>b_i\ \text{ and }\ b_{j-2}>u_q>b_{j-1}, & j\ge i+2. \end{cases} \tag{6.14} \end{equation} These inequalities say exactly that p and q occupy ranks i and j after insertion. For v\ne0, let \mathcal C_{ij}(A;v) be the set of mark vectors \zeta=(\zeta_1,\ldots,\zeta_{K-2}) on this ordered copy of A satisfying \begin{equation} \label{eq:rank-two-residual-marks} \zeta_k=0\ (k<i),\qquad \zeta_k\in L(v)\ (i\le k\le j-2),\qquad \zeta_k\in\mathcal M\ (k\ge j-1), \tag{6.15} \end{equation} where an empty index range imposes no condition. Put D_A(\zeta):=\sum_{k=1}^{K-2}b_k\zeta_k.

We now verify explicitly that deletion and insertion preserve every canonical index. Let t_1,\ldots,t_K denote the decreasing-weight ordering after p,q are inserted. For (p,q)\in\mathcal G_{ij}(A) this ordering is, and can only be, \begin{equation} \label{eq:rank-two-reconstruction} (t_1,\ldots,t_K)= (r_1,\ldots,r_{i-1},p,r_i,\ldots,r_{j-2},q, r_{j-1},\ldots,r_{K-2}), \tag{6.16} \end{equation} where every sublist with lower index larger than its upper index is empty. If j\ge i+2, the two insertion conditions in (6.16) are precisely b_{i-1}>u_p>b_i,\qquad b_{j-2}>u_q>b_{j-1}. If j=i+1, both points lie in the same residual gap and their required order is precisely b_{i-1}>u_p>u_q>b_i. Thus (6.16) is equivalent to (p,q)\in\mathcal G_{ij}(A), including the endpoint ranks under the sentinel conventions.

For an ordered marked support (S,\xi), define \mathcal E_{ij}^{v,v'}(S,\xi;p,q) to be the event \begin{gathered} p=t_i,\qquad q=t_j,\qquad \xi_{t_k}=0\quad(k<i),\\ \xi_p=v\ne0,\qquad \xi_{t_k}\in L(v)\quad(i<k<j),\qquad \xi_q=v'\notin L(v). \end{gathered} There is no restriction on marks after rank j. Pointwise in (S,\xi), \begin{equation} \label{eq:rank-two-canonical-identity} \mathbf 1_{\{r(\xi)=2\}} = \sum_{\substack{p,q\in S\\p\ne q}} \sum_{1\le i<j\le K} \sum_{v\in\mathcal M\setminus\{0\}} \sum_{v'\in\mathcal M\setminus L(v)} \mathbf 1_{\mathcal E_{ij}^{v,v'}(S,\xi;p,q)}. \tag{6.17} \end{equation} Indeed, if r(\xi)=2, the sole nonzero summand uses the prime carrying the first nonzero mark and the prime carrying the first later mark outside its span. If r(\xi)<2, no summand can occur. In particular, exchanging p and q never creates a second summand: their canonical ranks are different, and (6.16) assigns each rank to one specific insertion gap.

Under (6.16), deletion of the two pivot marks turns the full-rank conditions in \mathcal E_{ij}^{v,v'} into exactly (6.15). Namely, residual indices k<i retain full rank k; indices i\le k\le j-2 retain full rank k+1 and lie strictly between the pivots; and indices k\ge j-1 retain full rank k+2 and occur after the second pivot.

To apply Lemma 6.1 without suppressing the mark average, for distinct p,q\in S define \begin{align*} F_2(S;p,q) &:= 9^{-|S|}\sum_{\xi\in\mathcal M^S} \mathbf 1_{\mathcal P_T(S)} \mathbf 1_{\{\sum_{s\in S}u_s\xi_s\in Q\}}\\ &\qquad\times \sum_{1\le i<j\le |S|} \sum_{v\in\mathcal M\setminus\{0\}} \sum_{v'\in\mathcal M\setminus L(v)} \mathbf 1_{\mathcal E_{ij}^{v,v'}(S,\xi;p,q)}. \end{align*} Equation (6.17) gives the exact identity \mathbb P(r(\xi)=2,\mathcal P_T(S),D\in Q) = \mathbb E_{S\sim\mu}\sum_{\substack{p,q\in S\\p\ne q}}F_2(S;p,q). For p,q\notin A, the support and mark coefficient after deletion is \begin{align*} \mu(A\cup\{p,q\})9^{-(|A|+2)} &=\mu(A) \frac{1/(p+1)}{1-1/(p+1)} \frac{1/(q+1)}{1-1/(q+1)} 9^{-(|A|+2)}\\ &=\frac{\mu(A)}{pq}\,9^{-(|A|+2)}. \end{align*} Thus the two-point case of Lemma 6.1, followed by the unique decomposition of the marks into (v,v',\zeta) described above, gives the exact equality \begin{align} &\mathbb P(r(\xi)=2,\ \mathcal P_T(S),\ D\in Q)\notag\\ &=\sum_{A\subseteq\mathcal B(T)}\mu(A) \sum_{1\le i<j\le |A|+2}9^{-(|A|+2)} \sum_{v\in\mathcal M\setminus\{0\}} \sum_{v'\in\mathcal M\setminus L(v)} \sum_{\zeta\in\mathcal C_{ij}(A;v)}\notag\\ &\quad\times \sum_{(p,q)\in\mathcal G_{ij}(A)}\frac1{pq} \mathbf 1_{\mathcal P_T(A\cup\{p,q\})} \mathbf 1_{\{D_A(\zeta)+u_pv+u_qv'\in Q\}}. \label{eq:rank-two-exact}\tag{6.18}\end{align} Here factorial insertion changes the support mass from \mu(A\cup\{p,q\}) to \mu(A)/(pq); (6.14) restores the two deleted ranks; and (6.15) restores exactly the residual mark restrictions. Conversely, every tuple counted on the right reconstructs a rank-two marked support whose canonical data are (i,j,p,q,v,v'). Thus (6.18) has neither omission nor multiplicity.

Fix a term through the choice of A,i,j,v,v',\zeta. On the profile event, Lemma 6.2 gives \begin{equation} \label{eq:pivot-lower-weights} u_p\ge\ell_i,\qquad u_q\ge\ell_j. \tag{6.19} \end{equation} Let M_{v,v'} be the matrix with columns v,v'. It has nonzero integer determinant. Every cofactor has absolute value at most one, while the determinant has absolute value at least one; hence every entry of M_{v,v'}^{-1} has absolute value at most one. Define \mathfrak R:=M_{v,v'}^{-1}(Q-D_A(\zeta)), \qquad J_h:=\pi_h(\mathfrak R)\quad(h=1,2). Each J_h is an interval of length at most 2C_Qw. Indeed, if z,z'\in Q and \rho_h is row h of M_{v,v'}^{-1}, then |\rho_h(z-z')| \le\|\rho_h\|_1\max_{k=1,2}|z_k-z'_k| \le2C_Qw.

Let H_i,H_j denote the two individual rank gaps in (6.14); when j=i+1, both equal (b_i,b_{i-1}), while for j\ge i+2 they are H_i=(b_i,b_{i-1}),\qquad H_j=(b_{j-1},b_{j-2}). Taking closures only enlarges the sets, so define I_i:=\overline{ J_1\cap H_i\cap[\ell_i,1]\cap[T^{\vartheta-1},1]}, \qquad I_j:=\overline{ J_2\cap H_j\cap[\ell_j,1]\cap[T^{\vartheta-1},1]}. If either intersection is empty, the corresponding term in (6.18) is zero. Otherwise I_i,I_j\subseteq[T^{\vartheta-1},1], both have length at most 2C_Qw, and \inf I_i\ge\ell_i,\qquad \inf I_j\ge\ell_j. Every pair retained by (6.18) lies in the Cartesian product determined by I_i,I_j. Hence \begin{align} &\sum_{(p,q)\in\mathcal G_{ij}(A)}\frac1{pq} \mathbf 1_{\mathcal P_T(A\cup\{p,q\})} \mathbf 1_{\{D_A(\zeta)+u_pv+u_qv'\in Q\}}\notag\\ &\quad\le \left(\sum_{\substack{p\in\mathcal B(T)\\u_p\in I_i}}\frac1p\right) \left(\sum_{\substack{q\in\mathcal B(T)\\u_q\in I_j}}\frac1q\right) \ll_{C_Q}\frac{w^2}{\ell_i\ell_j}. \label{eq:two-local-sums}\tag{6.20}\end{align} The product on the right deliberately adds pairs with p=q, primes belonging to A, the wrong order u_p\le u_q in the adjacent-rank case, and points of J_1\times J_2 outside the parallelogram \mathfrak R. It also discards the profile restriction. All summands are nonnegative, so each enlargement has the required upper-bound direction.

The final bound in (6.20) is uniform in A,i,j,v,v',\zeta. The mark-counting sublemma therefore allows the mark sums in (6.18) to be performed before the remaining restrictions are discarded: \begin{align} &\mathbb P(r(\xi)=2,\ \mathcal P_T(S),\ D\in Q)\notag\\ &\quad\ll_{C_Q}w^2 \sum_{A\subseteq\mathcal B(T)}\mu(A) \sum_{1\le i<j\le |A|+2} \frac{16\,3^{-(i+j)}}{\ell_i\ell_j}\notag\\ &\quad\le 16Cw^2 \left(\sum_{k\ge1}\frac{3^{-k}}{\ell_k}\right)^2 \sum_{A\subseteq\mathcal B(T)}\mu(A) \ll_{C_Q}w^2, \label{eq:rank-two-bound}\tag{6.21}\end{align} where (6.10) and \sum_A\mu(A)=1 were used.

Rank one. Let p be the canonical first nonzero pivot and let i be its rank. After deleting p, put A=S\setminus\{p\} and K=|A|+1. Order A with weights b_1>\cdots>b_{K-1} and put b_0=+\infty, b_K=0. The pivot has rank i exactly when u_p\in G_i(A):=(b_i,b_{i-1}). For v\ne0, let \mathcal C_i^{(1)}(A;v) consist of the residual marks \zeta=(\zeta_1,\ldots,\zeta_{K-1}) satisfying \zeta_k=0\quad(k<i),\qquad \zeta_k\in L(v)\quad(k\ge i). When A=\varnothing, this residual list is empty and the endpoint conventions still apply. Put D_A(\zeta):=\sum_{k=1}^{K-1}b_k\zeta_k. Let t_1,\ldots,t_K denote the reconstructed decreasing-weight order. It is uniquely \begin{equation} \label{eq:rank-one-reconstruction} (t_1,\ldots,t_K) =(r_1,\ldots,r_{i-1},p,r_i,\ldots,r_{K-1}), \tag{6.22} \end{equation} which is equivalent to b_{i-1}>u_p>b_i. For an ordered marked support (S,\xi), let \mathcal E_i^v(S,\xi;p) denote the event p=t_i,\qquad \xi_{t_k}=0\ (k<i),\qquad \xi_p=v\ne0,\qquad \xi_{t_k}\in L(v)\ (k>i). Then, pointwise, \begin{equation} \label{eq:rank-one-canonical-identity} \mathbf 1_{\{r(\xi)=1\}} = \sum_{p\in S}\sum_{i=1}^K \sum_{v\in\mathcal M\setminus\{0\}} \mathbf 1_{\mathcal E_i^v(S,\xi;p)}. \tag{6.23} \end{equation} The unique nonzero summand uses the first nonzero marked prime. Deleting that prime from (6.22) makes residual indices k<i the marks before the pivot and indices k\ge i the marks after it. Thus the residual conditions are exactly those defining \mathcal C_i^{(1)}(A;v), and in particular |\mathcal C_i^{(1)}(A;v)|=3^{K-i}.

For p\in S, define \begin{align*} F_1(S;p) &:= 9^{-|S|}\sum_{\xi\in\mathcal M^S} \mathbf 1_{\mathcal P_T(S)} \mathbf 1_{\{\sum_{s\in S}u_s\xi_s\in Q\}} \sum_{i=1}^{|S|} \sum_{v\in\mathcal M\setminus\{0\}} \mathbf 1_{\mathcal E_i^v(S,\xi;p)}. \end{align*} Equation (6.23) gives \mathbb P(r(\xi)=1,\mathcal P_T(S),D\in Q) =\mathbb E_{S\sim\mu}\sum_{p\in S}F_1(S;p). For p\notin A, the coefficient after deletion is exactly \mu(A\cup\{p\})9^{-(|A|+1)} =\mu(A)\frac{1/(p+1)}{1-1/(p+1)}9^{-(|A|+1)} =\frac{\mu(A)}p\,9^{-(|A|+1)}. The one-point case of Lemma 6.1, followed by the unique decomposition of the remaining marks as \zeta, therefore gives the exact equality \begin{align} &\mathbb P(r(\xi)=1,\ \mathcal P_T(S),\ D\in Q)\notag\\ &=\sum_{A\subseteq\mathcal B(T)}\mu(A) \sum_{i=1}^{|A|+1}9^{-(|A|+1)} \sum_{v\in\mathcal M\setminus\{0\}} \sum_{\zeta\in\mathcal C_i^{(1)}(A;v)} \sum_{\substack{p\in\mathcal B(T)\setminus A\\u_p\in G_i(A)}}\frac1p \mathbf 1_{\mathcal P_T(A\cup\{p\})} \mathbf 1_{\{D_A(\zeta)+u_pv\in Q\}}. \label{eq:rank-one-exact}\tag{6.24}\end{align}

Fix A,i,v,\zeta and choose a coordinate h with |v_h|=1. Write Q=Q_1\times Q_2. The last indicator in (6.24) forces u_p\in J_i^0:=v_h(Q_h-D_{A,h}(\zeta)), an interval of length at most C_Qw. On the profile event, Lemma 6.2 also gives u_p\ge\ell_i. Intersect J_i^0 with the rank gap, the profile lower bound, and the prime band, and set I_i^{(1)}:= \overline{ J_i^0\cap G_i(A)\cap[\ell_i,1]\cap[T^{\vartheta-1},1]}. If the intersection is empty, the term is zero. Otherwise I_i^{(1)}\subseteq[T^{\vartheta-1},1] is an interval of length at most C_Qw and \inf I_i^{(1)}\ge\ell_i. Therefore \begin{align} &\sum_{\substack{p\in\mathcal B(T)\setminus A\\u_p\in G_i(A)}}\frac1p \mathbf 1_{\mathcal P_T(A\cup\{p\})} \mathbf 1_{\{D_A(\zeta)+u_pv\in Q\}}\notag\\ &\qquad\le \sum_{\substack{p\in\mathcal B(T)\\u_p\in I_i^{(1)}}}\frac1p \ll_{C_Q}\frac{w}{\ell_i}. \label{eq:one-local-sum}\tag{6.25}\end{align} The right side adds back primes in A and discards the profile and the unused coordinate condition from D_A(\zeta)+u_pv\in Q; nonnegativity gives the displayed upper-bound direction.

The profile event further implies K\ge a\lambda_T. We may now drop all of its other restrictions and sum the residual marks explicitly. The mark-counting sublemma gives \begin{align} &\mathbb P(r(\xi)=1,\ \mathcal P_T(S),\ D\in Q)\notag\\ &\quad\le Cw\sum_{A\subseteq\mathcal B(T)}\mu(A) \mathbf 1_{\{|A|+1\ge a\lambda_T\}} \sum_{i=1}^{|A|+1} \frac{8\,3^{-(|A|+1+i)}}{\ell_i}\notag\\ &\quad\le 8Cw\,3^{-a\lambda_T} \left(\sum_{i\ge1}\frac{3^{-i}}{\ell_i}\right) \sum_{A\subseteq\mathcal B(T)}\mu(A) =o(w^2), \label{eq:rank-one-bound}\tag{6.26}\end{align} by (6.10), (6.9), and \sum_A\mu(A)=1.

Rank zero. All K marks vanish, which has conditional probability 9^{-K}. The profile gives K\ge a\lambda_T, so \begin{equation} \label{eq:rank-zero-bound} \mathbb P(r(\xi)=0,\ \mathcal P_T(S),\ D\in Q) \le9^{-a\lambda_T} =o(w^2) \tag{6.27} \end{equation} by (6.9). The three rank cases prove (6.11). ◻

Proof. On the event in (6.28), choose canonically the selected prime p_* of largest weight among those of depth at most R_0, and delete it. Let \nu be the product law with parameters q_p on \mathcal B(T)\setminus S. The one-point case of Lemma 6.1 gives the exact identity \begin{align} &\mathbb P\bigl(X\in I,\ N_x(R_0)\ge1\mid S\bigr)\notag\\ &\quad= \sum_{A\subseteq\mathcal B(T)\setminus S}\nu(A) \sum_{p\in(\mathcal B(T)\setminus S)\setminus A} \frac{q_p}{1-q_p}\, \mathbf 1_{\{p\ \mathrm{is\ canonical}\}} \mathbf 1_{\{u_p+\sum_{r\in A}u_r\in I\}}. \label{eq:missing-insertion}\tag{6.29}\end{align} There is exactly one canonical point on the event, so no multiplicity is present in this identity.

The canonical restriction gives u_p\ge\mathrm e^{-R_0}. For fixed A, the last indicator in (6.29) confines u_p to a translate of I. Moreover, \begin{equation} \label{eq:missing-odds} \frac{q_p}{1-q_p}=\frac1{3p-1}\le\frac1p. \tag{6.30} \end{equation} Discard the remaining canonical order restriction and apply Lemma 5.1 on [\mathrm e^{-R_0},1]. The inner sum in (6.29) is O_{C_X}(w) uniformly in A,S. Summing \nu(A) proves (6.28).

Given the root data, the two missing-x processes are independent, so the two one-copy estimates multiply. The product already includes the possibility that both copies select the same prime; no diagonal correction is needed. ◻

Proof. Let B_T^{(1)} and B_T^{(2)} be formed from two independent completions conditional on the same root S. Then, conditionally on S, \mathbb P(B_T^{(1)}\cap B_T^{(2)}\mid S) =\mathbb P(B_T\mid S)^2=g_T(S)^2. Taking expectations proves the equality in (6.31).

Let \mathcal F_{\mathrm{root}} :=\sigma\bigl(S,(\xi_p)_{p\in S}\bigr), where the two coordinates of \xi_p record the two complete colourings of the root support as in (6.3). Thus \mathcal F_{\mathrm{root}} reveals the common root and both root colourings, but neither missing-x process. Define the root-good event \mathcal R_T:= \left\{ N_S(s)\ge as\text{ for every }R_0\le s\le\lambda_T \right\} \cap\{D\in[-w,w]^2\}. It is \mathcal F_{\mathrm{root}}-measurable. If both completions satisfy B_T, then, in completion h, the root labels c,y,z partition S. Consequently, for R_0\le s\le\lambda_T, N_S(s) =N_{c,h}(s)+N_{y,h}(s)+N_{z,h}(s) \ge3\lfloor\alpha s\rfloor \ge as by (5.13). Moreover D_h=Y_h-Z_h\in[-w,w]. Hence B_T^{(1)}\cap B_T^{(2)}\subseteq\mathcal R_T, and Lemma 6.3, with Q=[-w,w]^2, gives \begin{equation} \label{eq:root-good-probability} \mathbb P(\mathcal R_T)\le C_0w^2 \tag{6.32} \end{equation} for an absolute constant C_0.

For h=1,2, let \mathcal X_h\subseteq\mathcal B(T)\setminus S be the primes assigned to the missing x-petal in completion h. By (6.2), for every fixed realization of \mathcal F_{\mathrm{root}} the indicators \left( \mathbf 1_{\{p\in\mathcal X_h\}}: p\in\mathcal B(T)\setminus S,\ h\in\{1,2\} \right) are mutually independent and satisfy \mathbb P(p\in\mathcal X_h\mid\mathcal F_{\mathrm{root}})=\frac1{3p}. Indeed, conditional on S, the two completions are independent, and within each completion the colouring variables on S and the missing-x variables on \mathcal B(T)\setminus S are disjoint independent factors. Thus further conditioning on the two root colourings leaves the two missing processes independent and leaves each with exactly the product law used in Lemma 6.4.

The variables Y_h,Z_h are \mathcal F_{\mathrm{root}}-measurable. On \mathcal R_T, define the corresponding \mathcal F_{\mathrm{root}}-measurable interval \begin{equation} \label{eq:missing-target} I_h:=[\max(Y_h,Z_h)-w,\ \min(Y_h,Z_h)+w], \tag{6.33} \end{equation} and define I_h:=\{0\} off \mathcal R_T. On \mathcal R_T one has |Y_h-Z_h|\le w, so I_h is nonempty and |I_h|=2w-|Y_h-Z_h|\le2w. The range condition in B_T^{(h)} forces X_h\in I_h, whereas its x-profile at R_0 forces N_{x,h}(R_0)\ge\lfloor\alpha R_0\rfloor=24\ge1. Put E_h:=\{X_h\in I_h,\ N_{x,h}(R_0)\ge1\}. We have the explicit event inclusion \begin{equation} \label{eq:second-moment-event-inclusion} B_T^{(1)}\cap B_T^{(2)} \subseteq\mathcal R_T\cap E_1\cap E_2. \tag{6.34} \end{equation} All the other window and profile requirements in B_T^{(1)}\cap B_T^{(2)} have been discarded on the right; because an upper bound is sought, this only enlarges the event.

After conditioning on \mathcal F_{\mathrm{root}}, the root S and the two intervals I_h are fixed. On \mathcal R_T their lengths are at most 2w. The uniform one-copy estimate in Lemma 6.4 and the conditional independence of \mathcal X_1,\mathcal X_2 therefore give \mathbb P(E_1\cap E_2\mid\mathcal F_{\mathrm{root}}) =\prod_{h=1}^2 \mathbb P(E_h\mid\mathcal F_{\mathrm{root}}) \le C_1^2w^2 on \mathcal R_T. Finally, (6.34), the tower property, and (6.32) yield \begin{align*} \mathbb P(B_T^{(1)}\cap B_T^{(2)}) &\le \mathbb E\!\left[ \mathbf 1_{\mathcal R_T} \mathbb P(E_1\cap E_2\mid\mathcal F_{\mathrm{root}}) \right]\\ &\le C_1^2w^2\mathbb P(\mathcal R_T) \le C_0C_1^2w^4. \end{align*} This proves the upper bound in (6.31). ◻

Proof. Corollary 6.5 gives a constant C_2 such that the numerator in \mathbb E\left(\frac{g_T}{\beta_T}\right)^2 =\frac{\mathbb Eg_T^2}{\beta_T^2} is at most C_2w^4. Proposition 5.5 gives a constant c_1>0, independent of T and \eta, such that \beta_T\ge c_1w^2. Therefore \mathbb E\left(\frac{g_T}{\beta_T}\right)^2 \le\frac{C_2}{c_1^2}. Take K=C_2/c_1^2. All structural and buffer constants were fixed before T and \eta varied, so this K has the asserted uniformity. ◻