Proof. Under the law \(u(q)/(qS)\), primes in \(Q\) are independently included with probabilities \(4/(p+4)\). Equation (4) and partial summation give \[\mathbb E_q\log q\le4L,\qquad \mathbb E_q\omega(q)\le4\log L\] for large \(L\). Markov’s inequality therefore excludes \(\log q>98L\) with probability at most \(4/98\), and excludes \(\omega(q)>100\log L\) with probability at most \(4/100\). Every remaining \(q\), together with every \(d\in\mathcal D\), satisfies \(\log(dq)\le100L\) by (5), and so lies in one of the intervals (7). Their probability exceeds \(1/2\). Summing the independent harmonic weights of \(d\), whose total is \(V_*\), proves the lower bound. Dropping both restrictions proves the upper bound. ◻

Proof. Expand by exact active sets of size at most \(m_Q\), with scalar coefficient \(5^{|Q_a|}\). The total coefficient sum is \(\exp(O(L\log L))\), so 7 applies. The product-law expectation is at most \(\mathbb Ew=S\), by independence of the prime coordinates. Finally \(S\ge1\). The bound is uniform in the offset, and summing it proves the last assertion. ◻

Proof. Put \(y=\exp(L^{0.99})\). We need upper bounds for the number of integers in \([D,2D)\) avoiding every prime at most \(y\), and for the number of triples in \([D,2D)^3\) for which all four forms \[x_1,\quad x_2,\quad x_3,\quad x_1+x_2-x_3\] avoid those primes. The last form is positive throughout this integer box. In the first case put \(m=1\), and in the second put \(m=4\). For each prime \(p\), the proportion \(\nu_m(p)\) of residue vectors where at least one form vanishes satisfies \[ \nu_m(p)=\frac mp+O(p^{-2}). \tag{20}\] For the four forms, every pair of defining hyperplanes is independent over every prime field, including the field with two elements. Inclusion and exclusion of pairs proves (20). Moreover \(\nu_m(p)<1\) for every \(p\): the residue \(1\) in the first case and the vector \((1,1,1)\) in the second avoid all the forms.

For completeness, a finite inclusion–exclusion argument suffices here. Let \(r_0=2\lceil500\log L\rceil\) and truncate inclusion–exclusion of the bad-prime events at this even order. This gives an upper bound for the proportion avoiding all the events. For any selected prime set of product \(r\), the Chinese remainder theorem gives residue density \(\prod_{p\mid r}\nu_m(p)\). Its normalized count in the interval or box differs from this density by \(O(r/D)\) when \(r\le D\). Indeed, each residue coordinate occurs \(D/r+O(1)\) times; summing the product of these counts over at most \(r\) or \(r^3\) residue vectors proves the assertion. There are at most \((r_0+1)(1+\pi(y))^{r_0}\) terms, each with \(r\le y^{r_0}\). The total boundary error is therefore at most \[ \frac1D\exp\bigl(O(L^{0.99}\log L)\bigr), \tag{21}\] which is smaller than every fixed negative power of \(L\), by the lower bound on \(D\).

In the independent residue model the difference between the even truncated sum and the full product is at most the next elementary symmetric sum. Partial summation of the prime number theorem and (20) give \[\sum_{p\le y}\nu_m(p)\le(4+o(1))\log L.\] Thus that difference is at most \[\frac{\bigl(\sum_{p\le y}\nu_m(p)\bigr)^{r_0+1}}{(r_0+1)!} \le \left(\frac{e\sum_{p\le y}\nu_m(p)}{r_0+1}\right)^{r_0+1} \ll L^{-100}.\] The full product satisfies \[\prod_{p\le y}(1-\nu_m(p))\ll(\log y)^{-m}=L^{-0.99m};\] the finitely many small primes contribute a positive constant, and the remaining factors follow by taking logarithms in (20). Together with (21), this proves the respective counts \(O(DL^{-0.99})\) and \(O(D^3L^{-3.96})\).

The first count bounds \(\sum_{z\in\mathcal Z}1/z\) and hence \(\|B\|_\infty\). Orthogonality expresses the fourth moment as the sum of \((z_1z_2z_3z_4)^{-1}\) over \(z_1+z_2=z_3+z_4\) with all four variables in \(\mathcal Z\). The factor \(h\) does not change this equality. Each triple determines \(z_4\), and each weight is at most \(D^{-4}\). The second count proves the fourth moment bound. ◻

Proof. For each integer \(1\le v\le Y\) define \[\begin{align*} F_v(\theta)&=\sum_{m=1}^D \lambda(v+m)\mathbf 1_{v+m\equiv b_0\pmod l}\mathrm e(\theta m),\\ G_v(\theta)&=\sum_{a=1}^{(2h+1)D} \lambda(v+a)\mathrm e(-\theta a). \end{align*}\] We first claim that, uniformly in \(\theta\), \[ \sum_{1\le v\le Y}|F_v(\theta)|\ll YD L^{-0.8}. \tag{23}\] For integer \(D\), every \(x\in(v,v+1)\) has precisely \(v+1,\ldots,v+D\) as the integers in \([x,x+D]\). Apply Theorem 11 with \(H=D\) and \(Z=\lfloor Y\rfloor+1\), discarding the unused initial unit interval. The restriction to the residue class follows from \[\mathbf 1_{n\equiv b_0\pmod l} =\frac1l\sum_{a=0}^{l-1}\mathrm e\left(\frac{a(n-b_0)}l\right).\] Its coefficients have total absolute value one. After the unimodular factor arising from \(n=v+m\) is removed, this applies the same theorem to the frequencies \(\theta+a/l\). In particular, no assumption that \(b_0\) is coprime to \(l\) is needed. The scale hypotheses imply \(10\le D\le Z\) and \[\frac{\log\log D}{\log D}+(\log Z)^{-1/700} \ll L^{-0.8},\] which proves (23).

Average the sum on the left of (22), before taking its absolute value, over translations by \(m=1,\ldots,D\). Each translation changes at most \(2m\) prefix terms for each \(z\). By Lemma 10, the normalized endpoint error is at most \[O\left(\frac D Y\sum_{z\in\mathcal Z}\frac1z\right)=O(D/Y).\] Let \(B\) be the polynomial in Lemma 10. Orthogonality now expresses the translated average as \[ \frac1{YD}\sum_{1\le v\le Y} \int_0^1B(\theta)F_v(\theta)G_v(\theta)\,d\theta. \tag{24}\] Indeed, expanding the integral imposes \(a=m+hz\). Since \(1\le m\le D\) and \(D\le z<2D\), this value of \(a\) lies in the full range used to define \(G_v\).

On the set where \(|B|\le L^{-1.05}\), Parseval and Cauchy–Schwarz give \[\int_0^1|F_vG_v|\le\|F_v\|_2\|G_v\|_2 \le\sqrt{D}\sqrt{(2h+1)D}\ll_h D.\] Its contribution to (24) is therefore \(O_h(L^{-1.05})\). The complementary set, denoted by \(\mathcal E\), satisfies \[|\mathcal E|\le L^{4.2}\int_0^1|B|^4 \ll D^{-1}L^{0.24}.\] On this set use \(\|B\|_\infty\ll L^{-0.99}\), \(|G_v|\le(2h+1)D\), and (23). Its contribution is at most \[\frac1{YD}|\mathcal E|\,L^{-0.99}(2h+1)D\,(YDL^{-0.8}) \ll_h L^{0.24-0.99-0.8}=L^{-1.55}.\] Finally \(D/Y\le\exp(2L-L^A/2)\) is negligible. This proves (22). ◻

Proof of 9. The full-divisibility contribution is (9). For every other term in the expansion of (8), let \(I\subset\{1,\ldots,J\}\) be the nonempty set of prime supplies whose factors have been replaced by \(-1/p\). Put \[z=\prod_{i\in I}p_i,\qquad d'=d/z.\] For fixed \(q,d',j\), substitute \(n=qd'v\) and put \(Y=T_j/(qd')\). After extracting the sign \((-1)^{|I|}\) and the factor \(u(q)/(qd')\), the remaining expression is \[ \frac1Y\sum_{1\le v\le Y}\sum_z\frac1z \mathbf 1_{v\equiv bz\pmod l}\lambda(v)\lambda(v+hz), \tag{25}\] where \(z\) ranges over the products from the supplies indexed by \(I\) that also satisfy the bin condition. This substitution is valid in every residue class, since \(qd'\) is invertible modulo \(l\).

The allowed \(z\) lie in an interval of logarithmic width \(\eta<\log2\). Split them by residue modulo \(l\) and by the at most two dyadic intervals \([D,2D)\) that meet this interval. In each resulting piece the residue \(bz\pmod l\) is constant. Unique prime factorization and the disjointness of the supplies show that each integer \(z\) occurs at most once. Since \(I\) is nonempty, all its prime factors are at least \(H_0=\exp(L^{0.995})\), and \[D\ge H_0/2,\qquad D\le\exp(2L).\] Furthermore, if the piece is nonempty, its bin condition implies \(Y>Xe^{-\eta}z\), so \(\log Y\ge L^A/2\) for large \(L\). Lemma 12 therefore bounds (25) by \(O_{h,l}(L^{-21/20})\).

For each fixed \(I\) and bin, the sum of the extraction weights is at most \[\sum_q\frac{u(q)}q\sum_{d'}\frac1{d'} \le S\prod_{i\notin I}V_i\le SV_*,\] where the last inequality uses \(V_i>1\). There are \(O(L/\eta)\) bins and fewer than \(2^J\) nonempty sets \(I\). Hence the total contribution of the partial centering terms is \[O_{h,l}\bigl(SV_*\,2^J\eta^{-1}L^{-1/20}\bigr).\] Together with the full divisibility term this proves (18). ◻

Proof. Conditionally on the primes of \(Q\) dividing \(n\), choose \(q_1,q_2\) independently with probabilities \[\mathbb P(q_a=q\mid n)=\frac{u(q)\mathbf 1_{q\mid n}}{w(n)} \qquad(a=1,2).\] Each available prime is selected with probability \(4/5\). Under the combined tilted law let \(Z=\log(q_1/q_2)\). A prime \(r\) contributes \(+\log r\) or \(-\log r\) with probability \(4/[5(r+4)]\) each, and contributes zero otherwise. The contributions are independent, so its characteristic function, with angular frequency \(t\), is \[ \phi(t)=\mathbb E^*e^{itZ} =\prod_{r\in Q}\left(1-\frac8{5(r+4)} (1-\cos(t\log r))\right). \tag{28}\] Every factor is nonnegative: even at \(r=2\) it is at least \(7/15\). The inequality \(1-x\le e^{-x}\) therefore gives \[0\le\phi(t)\le \exp\left(-\frac85\sum_{r\in Q} \frac{1-\cos(t\log r)}{r+4}\right).\] Replacing \(r+4\) by \(r\) changes the exponent by \(O(1)\), uniformly in \(t\). For \(|t|\le2\), partial summation of the fixed-modulus prime number theorem in (4) gives \[\begin{align*} \sum_{r\in Q}\frac{1-\cos(t\log r)}r &=\frac34\int_{\log2}^L\frac{1-\cos(ty)}y\,dy+O_{h,l}(1) \\ &=\frac34\log(1+L|t|)+O_{h,l}(1). \tag{29}\end{align*}\] For completeness, in the partial summation the derivative of \((1-\cos(t\log x))/(x\log x)\) is \(O(1/(x^2\log x))\) uniformly for \(|t|\le2\). Its product with the prime-number-theorem error is integrable on \([2,\infty)\). Deleting the finitely many prime divisors of \(lh\) also costs \(O_{h,l}(1)\). To verify the second equality, set \(u=|t|y\). The integral over \(u\le1\) is bounded, and integration by parts bounds \(\int_1^U\cos u\,du/u\) uniformly in \(U\ge1\). The lower endpoint \(|t|\log2\) stays bounded. These observations also cover \(t=0\). It follows that \[ 0\le\phi(t)\ll_{h,l}(1+L|t|)^{-6/5}\qquad(|t|\le2). \tag{30}\]

Conditionally on \(n\), the probability that both divisors lie in the \(j\)th bin is \(\rho_j(n,d)^2\). Divisors in the same bin have \(|Z|\le\eta\le1\). Consequently the left side of (27) is at most \(\mathbb P^*(|Z|\le1)\). The function \(\kappa(x)=(\sin x/x)^2\), with \(\kappa(0)=1\), satisfies \[\mathbf 1_{[-1,1]}(x)\le\frac{\kappa(x)}{\sin^2 1},\qquad \kappa(x)=\frac14\int_{-2}^2(2-|t|)e^{itx}\,dt.\] Taking expectations, using (30), and integrating gives \[\mathbb P^*(|Z|\le1) \ll_{h,l}\int_{-2}^2(1+L|t|)^{-6/5}\,dt \ll_{h,l}L^{-1}.\] The common translation \(\log d\) of the two padding logarithms disappeared from \(Z\), which proves the asserted uniformity. ◻

Proof. We first bound the deletion costs in the auxiliary product model, after dropping the Liouville factors and the progression indicator, and replacing the absolute centered product by \[a_d^+(n)=\prod_{p\mid d}\left(\mathbf 1_{p\mid n}+\frac1p\right).\] At the end of the proof we transfer precisely these nonnegative costs to integer averages by Lemma 6. No assertion about untruncated integer moments of \(w\) is needed.

Proof. There are \(O(R\log L)\) slots, since each tuple has \(J=O(\log L)\) factors and every eligible padding has at most \(100\log L\) factors. A partition of \(N\) slots has at most \(N^N\) descriptions. Signs, factor counts and all recorded indices therefore cost \(\exp(O(R\log^2L))\). Each numerical class has reciprocal mass \(O(\log L)\), by the prime harmonic bounds, and \(Lu(q_i)\le L^{1+100\log4}\). This proves the first assertion. Dropping bin or numerical distinctness restrictions enlarges these positive sums. If distinct abstract classes are assigned the same numerical prime in such an enlargement, they retain their separate reciprocal factors.

For the second assertion the coefficient is invertible modulo \(p\), so \(z\) lies in one residue class. Enlarge from primes to integers and compare the reciprocal sum along that progression with its integral: \[\sum_{\substack{H_0\le n\le e^L\\n\equiv c'\!\!\pmod p}}\frac1n \le\frac1{H_0}+\frac Lp\le\frac{1+L}{H_0}.\] Fix the nonselected variables and sum the selected ones in reverse order. Every remaining earlier relation is independent of the variable currently being summed. Its own coefficient test is retained; if this test fails, the admissible inner sum is zero. The displayed bound is uniform in the remaining variables, and backward induction proves multiplication of the savings. After these sums, the other prime variables are bounded by the first assertion. ◻

Proof. We may count all forward prohibited positive words, without requiring minimality. For a fixed numerical word, positivity either gives inconsistent residue conditions or fixes one residue at each distinct prime in its steps. Its probability is accordingly zero or the product of their reciprocals.

At the last transition choose a column whose prime changes. Its last prime \(z\) occurs nowhere earlier, by the interval condition. In (36), the contribution involving \(z\) is exactly the last step. Thus its coefficient is \(\varepsilon_m hq_m d_m/z\). The controlling prime \(p\) is absent from \(d_m\), from \(h\) and from every padding, so this coefficient is a unit modulo \(p\). It is independent of \(z\). Fixing the equality pattern, padding and other tuple values, 16 supplies one saving \((1+L)/H_0\). The remaining count, summed over \(3\le m\le s\), is at most \[\frac{1+L}{H_0}\exp(O(s\log^2L)) \le\exp\bigl(-\tfrac12L^{1-\delta}\bigr).\] This proves the probability bound.

Each eligible numerical pair satisfies \(dq\le\exp(100L+1)\). The disjoint prime supports recover \(d\) and \(q\) uniquely from their product. Thus the number of numerical words of length at most \(s\) is \(\exp(O(sL))\). Preselect those satisfying the numerical prohibition and minimality conditions. Positivity of one such word is a conjunction of \(O(s\log L)\) residue equalities, one per prime occurrence. Taking their disjunction gives the stated depth and size bounds. Translation merely shifts the same residue coordinates. This verifies all the hypotheses on the deleted events in 14. ◻

Proof. For a uniform residue \(Z\) modulo \(p\), a fixed residue \(a\), and any function \(F\) of that coordinate, \[\mathbb E\left[\left(\mathbf 1_{Z=a}-\frac1p\right)F(Z)\right] =\frac1p\mathbb E\bigl[F(a)-F(Z)\bigr].\] Apply this identity successively to the independent singleton coordinates. At a nonsingleton, incompatible lit tests give zero; otherwise they force one residue and supply \(1/p\), while every unlit choice supplies the scalar \(-1/p\). These operations produce exactly the factor \(b_{\mathcal L}\) and the mixed difference, up to the signs of the unlit scalars. The factor \(R\) is unchanged in every operation. Taking the absolute value after these integrations proves the bound. The final assertion follows because \(G\) takes values in \(\{0,1\}\). ◻

Proof. If a nonsingleton has a lit occurrence, each of its unlit occurrences supplies an extra reciprocal beyond its basic \(1/p\). If all its \(m\ge2\) occurrences are unlit, it supplies \(p^{-m}\) and hence at least \(m/2\) extra reciprocal powers beyond the basic one. Thus at least \(U/2\) extra powers remain, all at primes at least \(H_0\). The total is at most \[\exp(O(L\log^2L))H_0^{-U/2} \le\exp\bigl(O(L\log^2L)-\tfrac12L^{1.015}\bigr) \le\exp(-L^{1+\delta}).\] ◻

Proof. Let \(E(c,y)\) denote the divisibility event in the statement, and let \(y'\) be an independent copy of \(y\), independent also of \(c\). Cauchy–Schwarz, applied to the conditional probability with \(c\) fixed, gives \[\begin{align*} \mathbb P(E(c,y))^2 &=\left(\mathbb E_c\mathbb P_y(E(c,y)\mid c)\right)^2\\ &\le \mathbb E_c\mathbb P_y(E(c,y)\mid c)^2\\ &=\mathbb P(E(c,y)\cap E(c,y'))\\ &\le\mathbb P\bigl(c_i\mid (B(y-y'))_i\text{ for every }i\bigr). \end{align*}\] The two copies have the same controlling primes \(c_i\), so subtraction removes the entire vector \(f(c)\).

Write \(v=B(y-y')\). For a subset \(Z\subseteq\{1,\ldots,r\}\) of size \(a\), the rows of \(B\) indexed by \(Z\) have rank \(a\). Choose an invertible \(a\)-column minor in those rows. Conditional on \(y'\) and on the coordinates of \(y\) outside that minor, the equations \(v_i=0\) for \(i\in Z\) determine at most one vector of values for its \(a\) remaining coordinates. Independence and the atom bound imply \[ \mathbb P(v_i=0\text{ for all }i\in Z)\le\alpha^a. \tag{45}\] This includes \(a=0\), with the empty condition having probability one.

Now fix \(y,y'\). A nonzero integer of absolute value at most \(M\) has at most \(d_M\) distinct prime divisors. Each nonzero coordinate \(v_i\) therefore satisfies \(c_i\mid v_i\) with conditional probability at most \(d_M\alpha\). The controls are independent of each other and of \(v\). If the exact zero set of \(v\) is \(Z\), their joint conditional probability is thus at most \((d_M\alpha)^{r-|Z|}\). Using (45) and summing over the possible exact zero sets gives \[\mathbb P(c_i\mid v_i\text{ for every }i) \le\sum_{Z\subseteq\{1,\ldots,r\}} \alpha^{|Z|}(d_M\alpha)^{r-|Z|} =\left[\alpha(1+d_M)\right]^r.\] Taking the square root proves (44). ◻

Proof. Use the positive reciprocal majorant and enumeration of 16. Thus we may discard the denominators and cutoffs, retain one reciprocal for every distinct prime label, and include the factors \(Lu(q_a)\) and \(2^{S_1}\) in the crude count. We retain the lit-consistency conditions. The total unrestricted count, including external dimension factors, is \(\exp(O(L\log^2L))\).

Fix a column, its equality pattern, and the other-column and padding data. Choose \(r\) jointly independent pairs, and denote their controlling labels by \(z_1,\ldots,z_r\) and their difference vectors by \(w_1,\ldots,w_r\). The controlling labels are distinct. Modulo \(C=\operatorname{span}(e_{z_1},\ldots,e_{z_r})\), the images of the \(w_i\) remain independent: a dependence would express a nontrivial linear combination of the \(w_i\) as a combination of the \(e_{z_i}\), contrary to joint independence. Therefore the row matrix of the \(w_i\), after deleting all controlling columns, has rank \(r\). Choose an invertible \(r\)-column minor \(B_0\) of that matrix.

The entries of the entire row matrix, and hence the choice of a minor by any fixed deterministic rule, are independent of all the numerical primes in column \(\nu\). Fix the prime variables in that column outside the controlling coordinates and the minor. Write \(c_i=p_{z_i}\) for the controls and \(y\) for the vector of primes in the minor. The selected lit-consistency conditions become \[ c_i\mid (B_0y+A_0c+d_0)_i\qquad(1\le i\le r), \tag{48}\] with fixed integer matrices \(B_0,A_0\) and a fixed integer vector \(d_0\). No invertibility assertion modulo \(c_i\) is being made.

Enlarge the positive sum by dropping numerical distinctness among abstract labels, as well as bin and closure restrictions. Continue to attach one factor \(1/p\) to each abstract label: if two coordinates now take the same prime, their product weight is \(1/p^2\). The enlarged reciprocal sum is consequently a product sum. Its normalized coordinates have independent law \[\mu_\nu(p)=\frac{1}{V_\nu p}\quad(p\in P_\nu), \qquad \max_p\mu_\nu(p)\le\frac{1}{V_\nu H_0} \le\frac1{H_0}.\] This independence allows numerical coincidences. In particular, after subtraction in 21, the controls remain independent of \(B_0(y-y')\), even on a sample space that permits such coincidences.

We check the size hypothesis after these enlargements. The fixed padding values came from eligible steps, so \(q_a\le e^{100L+1}\). The product of the primes in the other columns is at most \(e^{2L}\) by the geometric spacing of their bands. Hence \[|t_a|\le h e^{102L+1}.\] Each row of the difference matrix has sum of absolute coefficients at most \(2kh e^{102L+1}\). Since every prime in the resampled column is at most \(e^L\), we obtain \[|(B_0(y-y'))_i|\le 2kh e^{103L+1}=e^{O(L)}.\] For fixed \(h\) and sufficiently large \(L\), the constant in this last exponent is absolute. In 21 we may therefore take \(d_M=O(L)\), \(\alpha=(V_\nu H_0)^{-1}\), and \(f(c)=A_0c+d_0\). The probability of (48) is at most \[\left(\frac{O(L)}{V_\nu H_0}\right)^{r/2} \le \exp(O(r\log L))H_0^{-r/2}.\] Multiplying by the unrestricted harmonic masses restores the unnormalized reciprocal sum with this same relative saving. The estimate is uniform in every fixed outside variable, so these variables may now be summed.

The column, selected pairs and selected minor have at most \(J(2k)^{O(r)}\) descriptions. Their cost is absorbed in \(\exp(O(r\log L))\). Together with the crude enumeration, the total is bounded by \[\exp\left(O(L\log^2L)+O(r\log L)-\frac r2\log H_0\right).\] Since \(\log H_0=L^{199/200}\) and \(r=L^{1/50}+O(1)\), its negative term is \((\tfrac12+o(1))L^{203/200}\) and dominates both positive terms. This proves the lemma. ◻

Proof. For the fixed numerical word, let \(\mathscr W\) be the finite set of all possible witness tests at its departure sites. Write \(I_w\) for the indicator that the preselected numerical word \(w\) is positive at its assigned departure. Then \[G=\prod_{w\in\mathscr W}(1-I_w).\] A repeated test in this product is harmless because its factors take values in \(\{0,1\}\). Expand this finite product algebraically and apply the mixed difference in \(\mathcal U\). If the union of the prime supports of a subfamily \(\mathscr A\subseteq\mathscr W\) omits a prime \(p\in\mathcal U\), then \(\prod_{w\in\mathscr A}I_w\) is independent of the \(p\) coordinate. Its full mixed difference is therefore zero. Consequently \(\Delta G\ne0\) implies that some subfamily whose supports cover \(\mathcal U\) has a nonzero mixed difference. Its intersection indicator is then one in at least one hybrid. All witnesses in that subfamily are positive in this single configuration.

Choose an inclusion-minimal subfamily covering \(\mathcal U\) there. Each member contains a prime of \(\mathcal U\) occurring in no other member, since otherwise that member could be removed. A witness has at most \(sJ\) distinct \(P\) labels, so this subfamily has at least \(S_1/(sJ)\) members. For sufficiently large \(L\), \[\frac{S_1}{sJ}>\frac{L^{3/20}}{J}>T.\] Keep any \(T\) members and one private prime in each. Their private primes remain private in the selected subfamily. Finally, \[2k+sT\le 2L+L^{11/60}\le4L.\] The algebraic expansion above has been used only to deduce existence. No bound is obtained by summing the absolute values of its terms, and no enumeration of all its subfamilies is charged. ◻

Proof. Consecutive tuples differ, and each label occupies an interval of step indices. Thus some label \(z\) enters for the first time on the last step. Let \(c\) be the first-step label controlling the prohibited suffix of the word. Since \(c\) is absent from the last tuple and divides neither \(h\) nor a padding factor, \(c\nmid E_m\). The contribution of \(z\) to the prohibited suffix is exactly \(E_m\), so \(z\) is active. The same argument shows that every label used only on the last step is active.

Now suppose that \(y\) occurs before the last step. Restricting attention to steps \(1,\ldots,m-1\), its occurrences form a nonempty interval \([a,b]\), where \(b<m\); \(y\) may also occur on step \(m\). If \(E_a+\cdots+E_b\) is nonzero modulo \(z\), then (50) holds at \(i=m\). Otherwise \[ z\mid E_a+\cdots+E_b. \tag{51}\] Every step before the last is nonzero modulo \(z\), so \(b>a\). Reverse the contiguous subword \(E_a,\ldots,E_m\), negating all its steps. Its first step contains \(z\), its last does not, and its suffix consisting of the reversed steps \(E_b,\ldots,E_a\) has sum divisible by \(z\). This suffix starts strictly after the first step because \(b<m\), and strictly before the last step because \(b>a\). The interval appearances and unequal consecutive tuples persist under reversal. The reversed subword is therefore forward prohibited. Minimality forces \(a=1\).

Choose a label \(z'\) newly entering on step \(2\), which is possible because the first two tuples differ. Let \(b'\) be the smaller of \(b\) and the last index carrying \(z'\). Its contribution to (51) is \(E_2+\cdots+E_{b'}\). If that contribution were zero modulo \(z\), then \(b'>2\), since \(z\nmid E_2\). The reversed subword \(E_m,\ldots,E_2\) would then be forward prohibited, with controlling prime \(z\) and suffix \(E_{b'},\ldots,E_2\). It is shorter than the original word, contradicting minimality. Hence \(z'\) is active through (51). Since \(y\) occurs on step \(1\) and \(z'\) does not, its contribution before step \(2\) is \(E_1\not\equiv0\pmod{z'}\). This proves (50) with \(i=2\) and \(z'\) in place of \(z\). ◻

Proof. Whenever the integrand is nonzero, Lemma 23 supplies \(T\) witnesses positive in one common hybrid, with marked private singletons. Order them as \(w_1,\ldots,w_T\) by their attachment indices \(a_1\le\cdots\le a_T\) on the main word; order ties arbitrarily. Write \(t(y)\) for the unique main step carrying a main singleton \(y\). Thus the main position of \(y\) has been passed at attachment \(a\) exactly when \(t(y)<a\). We will select at least \(\lfloor T/8\rfloor\) prime variables and order their relations so that each relation is independent of all later selected variables. In the first two cases we select active labels, which need not be the marked singletons. In the remaining case we select the private marked singletons themselves.

Internal relations. Suppose at least \(T/8\) witnesses have an active label absent from every earlier witness. Choose one such label from each of these witnesses, and record an internal relation establishing its activity. The chosen labels are distinct. Each earlier relation involves only labels in its own witness, so it is independent of every later chosen label, including as a controlling modulus. Sum the chosen prime values in reverse witness order. When one is summed, all earlier relations are independent of that variable, while its own relation has an invertible coefficient. This gives at least \(\lfloor T/8\rfloor\) successive reciprocal-prime savings. If instead at least \(T/8\) witnesses have an active label absent from every later witness, apply the same argument with the order reversed.

Comparison relations. Assume neither alternative holds. Apart from fewer than \(T/4\) witnesses, every active label occurs in both an earlier and a later witness. In each remaining witness \(w_j\), its private marked singleton \(y_j\) is not active. By Lemma 24, choose an active label \(z_j\) and a departure within \(w_j\) whose prefix has nonzero \(y_j\)-contribution modulo \(z_j\).

First consider those \(j\) for which \(t(y_j)\ge a_j\). Choose an earlier witness \(w_h\), \(h<j\), containing \(z_j\), and a departure there using \(z_j\). Let \(\beta_j\) and \(\beta_h\) be the offsets of these selected departures from the respective starts of \(w_j\) and \(w_h\). Positivity in the common hybrid gives \[ z_j\mid(\xi_{a_j}+\beta_j)-(\xi_{a_h}+\beta_h) =\sum_{a_h\le v<a_j}D_v+\beta_j-\beta_h. \tag{52}\] The main segment omits \(y_j\) because \(t(y_j)\ge a_j\), and \(w_h\) omits it by privacy. Its only contribution is therefore through \(\beta_j\), where it has the nonzero prefix contribution given by Lemma 24.

For two selected witnesses \(w_j,w_i\) with \(j<i\), the private label \(y_i\) occurs in neither witness prefix in (52). It is also absent from its main segment, since \[t(y_i)\ge a_i\ge a_j.\] Thus the comparison belonging to \(w_j\) is independent of every later selected variable \(y_i\). This includes its controlling prime: \(z_j\) occurs in both \(w_j\) and \(w_h\), whereas each selected \(y_i\) is private. Eliminating the selected labels in decreasing attachment order is therefore valid. Equal attachment indices cause no difficulty, as the main segment still ends strictly before that index.

For the witnesses with \(t(y_j)<a_j\), choose instead a later witness containing \(z_j\). The comparison uses the main segment starting at \(a_j\) and ending just before that later attachment, so it omits \(y_j\). Order this group by decreasing attachment index. A selected label from a smaller attachment has its unique main position still smaller than that attachment, and hence lies before the start of every earlier comparison segment in this order. Privacy removes it from the two witness prefixes as well. The same reverse-elimination argument applies. One of these two groups has at least \(T/8\) witnesses for sufficiently large \(T\): there are more than \(3T/4\) candidates before dividing them according to whether their main position has passed.

Summing the relations. In all cases, record the chosen relations, their order, their controlling labels, and the nonvanishing coefficient tests. Fix the equality pattern, the numerical padding factors, and all nonselected numerical labels. Every selected congruence is linear in its selected prime \(x\); its coefficient is a unit modulo its controlling prime \(p\). The latter is unselected or belongs to an earlier relation in the chosen order. The reciprocal sum is bounded by \[\sum_{\substack{H_0\le x\le e^L\\x\equiv c\pmod p}}\frac1x \le\frac1{H_0}+\frac Lp \le\frac{1+L}{H_0},\] where enlarging from primes to integers only increases the sum. Retain the nonvanishing coefficient condition during each elimination; if it fails for the remaining fixed data, that inner sum is zero. After summing a selected variable, drop its already used relation. The explicit dependencies above ensure that all still retained relations are independent of the variable just eliminated.

It remains to justify the residue cost and the number of records. Bound \(|\Delta G|\) by \(2^{S_1}\), drop denominators and cutoffs from \(R\), and retain its padding divisibilities. The factor \(b_{\mathcal L}\) already supplies one reciprocal for every main \(P\) label. Each new witness \(P\) label and each distinct main or witness \(Q\) label has a positivity test in its original residue coordinate, which no hybrid changes. Their joint average supplies one reciprocal per distinct such label. Constraints on overwritten main coordinates may be dropped. In particular no choice of hybrid needs to be counted: these retained residue tests are unchanged throughout the hybrids, and the recorded numerical relations are necessary consequences of positivity in the common one.

The total recorded length is at most \(4L\). Lemma 16 therefore bounds the equality patterns, numerical reciprocal sums before the selected savings, signs, and weights by \(\exp(O(L\log^2L))\). Attachments, marked labels, internal interval endpoints, partner-witness indices, and elimination orders have only polynomially many options per recorded slot, so their inclusion preserves this bound. The factors \(2^{S_1}\) and \(|\mathcal D|^2M=\exp(O(L))\) do so as well. We have proved the bound \[\exp(O(L\log^2L)) \left(\frac{1+L}{H_0}\right)^{\lfloor T/8\rfloor}.\] Finally, \(T\log H_0\asymp L^{1/12+199/200}=L^{647/600}\). This dominates both \(L\log^2L\) and \(L^{21/20}\), proving the assertion. ◻

Proof. If the interval assertion fails, choose across all columns two consecutive occurrence groups having the shortest gap. Let \(i\) be the last position in the first group and \(i'\) the first in the next, and let \(p\) be their common prime. The prime \(p\) is absent from \(i+1,\ldots,i'-1\). Every label in the substring \(i,\ldots,i'-1\) has interval-shaped uses, since otherwise it would give a shorter gap. Positivity at the departures of steps \(i\) and \(i'\) gives \[p\mid \sum_{a=i}^{i'-1}D_a, \qquad\text{hence}\qquad p\mid \sum_{a=i+1}^{i'-1}D_a.\] The length \(i'-i\) cannot be two: the latter sum would be one displacement whose tuple and padding both omit \(p\), while \(p\nmid h\). The substring therefore has length at least three. It is a forward prohibited word, with its suffix starting at its second step. Pass to a minimal prohibited contiguous subword, allowing reversal. Positivity survives these operations, and its starting vertex is one of the main vertices. This contradicts \(\sigma_j=1\) there.

For the second assertion a single displacement is nonzero. In any subinterval of length at least two, the first and last tuples differ. Indeed, if they agreed in every column, the interval assertion would make every tuple in that subinterval equal, contrary to the nonbacktracking condition. Choose a first-step prime \(p\) absent from the last tuple. If the total displacement were zero, deleting the first step would leave a suffix sum divisible by \(p\). In a two-step subinterval this is impossible, because \(p\) does not divide the last displacement. In length at least three it makes the subinterval forward prohibited. Minimal descent gives the same contradiction. A reversed subword may begin at the end of the block; this endpoint is also a main vertex. At the end of the closed word it is its initial vertex, so the survival condition still applies. ◻

Proof. Fix one realization and one column. Let \(\mathcal Z\) be its set of distinct labels and let \(\mathcal V\) be the real vector space with basis \((e_z)_{z\in\mathcal Z}\). If the label at position \(a\) is \(z_a\), define its formal departure offsets by \[v(i)=\sum_{a<i}t_a e_{z_a},\qquad t_a=D_a/p_{z_a}.\] The coefficients \(t_a\) do not use the prime values in this column. Choose an inclusion-maximal collection of pairs \((e_z,v(i)-v(i'))\), where \(i,i'\) are lit occurrences of \(z\), whose combined vectors are independent. If there are \(m\) pairs, their span \(D_0\) has dimension \(2m<2r\).

Choose a basis of \(\mathcal V/D_0\) from the images of the coordinate vectors. Call the corresponding labels regular and the other labels omitted. There are exactly \(2m<2r\) omitted labels. Write \(\bar v\) and \(\bar e_z\) for images in the quotient. For a regular label \(z\), all lit starts lie on one affine line parallel to \(\bar e_z\). Otherwise some difference \(v(i)-v(i')\) would be outside \(D_0+\mathbb R e_z\); since \(e_z\notin D_0\), adjoining that pair would contradict maximality. The regular directions are nonzero and jointly independent. For each regular label \(z\) with a lit occurrence, let \[\ell_z=\bar v(i)+\mathbb R\bar e_z, \qquad i\text{ any lit occurrence of }z.\] The preceding argument makes this line independent of the choice of \(i\). The lines so defined are distinct, since their directions are independent.

Compress each constant run in each perfect block to one run-entry. By 26, a given label occurs in at most one run per block. A regular run of label \(z\) has quotient increment \[ \left(\sum_{a\text{ in the run}}D_a/p_z\right)\bar e_z\ne0. \tag{55}\] The inequality uses the nonzero-subinterval assertion of 26. Cut the runs at omitted entries and at block boundaries. This leaves at most \((2r+1)B\) nonempty regular segments and at most \(2rB\) omitted run-entries. The parameter bounds and (54) give \[ (2r+1)B\ll L^{0.92}. \tag{56}\]

Each regular run of label \(z\) starts and ends on \(\ell_z\), and (55) says that its endpoints are distinct. A transition from a \(z\)-run to a \(z'\)-run is therefore a common point of \(\ell_z\) and \(\ell_{z'}\). We encode these incidences by a simple bipartite graph: one vertex represents each line used by the regular segments, and one vertex represents each distinct projected transition point. Join a transition point to the two lines of its transition, identifying vertices whenever the same line or point recurs. Include an isolated line vertex if it occurs only in one-run segments. There are at most \(2k\) line vertices and at most \(2k\) point vertices.

This graph is a forest. A simple cycle would pass through distinct line vertices, and thus through distinct independent directions. On each of its lines the two adjacent point vertices are distinct. The displacements around the cycle would consequently give a linear relation among the regular directions with every coefficient nonzero, a contradiction.

A regular segment gives a walk in this forest. It has no immediate reversal: a walk line–point–same line would repeat an adjacent run label, and a walk point–line–same point would contradict (55). A walk without immediate reversal in a forest is the unique simple path between its endpoints. Therefore the ordered pair of line endpoints recovers all run labels of the segment. Equal endpoints encode a one-run segment.

Figure 1 illustrates this decoding. The graph records only incidences; the numerical positions of the points will not enter the code.

We describe explicitly a code for the whole column pattern.

1. Record imperfect positions, block boundaries, run boundaries, and the omitted or regular status of each run. These are binary data on \(O(k)\) positions and have \(\exp(O(k))\) possibilities.

2. Record the equality partition among the omitted run-entries. Its cost is at most \((2k)^{O(rB)}\).

3. Give the abstract forest, rooted and ordered in any manner, with its line or point vertex types. The parenthesis traversal of a rooted ordered forest with at most \(4k\) vertices, together with the type bits, has \(\exp(O(k))\) possibilities. The traversal numbers its vertices.

4. Give the ordered pair of line-vertex numbers for every regular segment. By the unique-path property this recovers its run sequence. The cost is at most \((4k)^{O((2r+1)B)}\).

5. At every imperfect position, give a representative occurrence of its label. A label already represented in a perfect position points to such a position; otherwise use its first imperfect occurrence. This costs at most \((2k)^I\).

These data determine a unique equality partition. Omitted labels cannot equal regular labels. Their recorded partition determines all omitted equalities, while the forest vertices determine all regular equalities. The last step supplies every remaining equality. Neither coordinates of transition points nor numerical coefficients are needed in this decoding.

All constants in these code counts are absolute. Their logarithms sum to at most \[C_0k+C_0L^{0.92}\log(4k) +C_0(S_1+U)\log(2k)\le Ck\] for an absolute \(C\) and sufficiently large absolute \(L\). Here \(k=\lfloor L\rfloor\) and both exponents \(0.92\) and \(1/4\) are strictly less than one.

The code universe just described depends only on the length and the displayed numerical bounds. Every realizable pattern, whatever the coefficients, has a code in that same universe. For example, choose the first valid code in a fixed ordering; the decoder shows that two different patterns cannot receive the same code. Thus the bound counts the union over all coefficients, not just the patterns at one fixed coefficient choice. Applying this universal bound in each of the \(J\) columns proves the last assertion. ◻

Proof. Let \(c_i(n,q)\) denote the \(i\)th factor in the product. The definition of \(\rho_j\) allows all squarefree padding products in the bin; the extra bound on \(\omega(q)\) in \(\mathcal S_j\) only reduces their sum. Consequently, at every site \(n\), \[\sum_q c_i(n,q) \le L\rho_j(n,d_i)\mathbf 1_{\rho_j(n,d_i)\le K/L}\le K.\] Set \(F_{m+1}(n)=1\) and recursively define \[F_i(n)=\sum_q c_i(n,q) F_{i+1}(n+\varepsilon_i hqd_i).\] Backward induction, using a bound uniform in the starting site at each stage, gives \(F_i(n)\le K^{m-i+1}\). In particular \(F_1(n_1)\le K^m\). The shifts caused by earlier padding choices therefore introduce no additional factor. ◻

Proof of 18. First work in the product residue law. The contributions with \(U>L^{1/50}\), with a high-rank column, or with \(S_1>L^{1/4}\) are \(o(1)\) in total by [q:many-unlit,q:rank,q:singletons], including the external dimension factors of the trace expansion. We sum the remaining contributions using 19.

By 27, the combined column patterns lie in a single family of at most \(\exp(CJk)\) possibilities, independently of all padding choices. Fix one such pattern, its numerical column labels, the signs, and the lit or unlit designations. Replace \(|\Delta G|\) by \(2^{S_1}\) and retain just one reciprocal per distinct column prime in \(b_{\mathcal L}\). Every nonsingleton provides at least one such reciprocal: either it has a lit occurrence, or all of its at least two occurrences are unlit. Dropping the other reciprocals and lit consistency enlarges the nonnegative majorant.

In \(R\), drop closure, progression and block restrictions, arrival cuts, and the low-\(Q\) cuts. Retain the padding divisibilities, bin conditions, departure weights, and departure \(\rho_j\) cuts. The closed-word identity for its denominators was established before this enlargement. Conditional on the non-\(P\) environment, 28 with \(m=2k\) bounds the full padding sum by \(K^{2k}\). All its factors are unchanged by the \(P\)-coordinate overwrites, and the bound is uniform in the environment. Taking its expectation preserves that bound.

Let \(m_i\) be the number of distinct labels in column \(i\), and put \(m_P=\sum_i m_i\). Among the \(2kJ\) column occurrences, exactly \(S_1\) belong to singleton labels and every other label occurs at least twice. Hence \[2kJ\ge S_1+2(m_P-S_1),\qquad m_P\le Jk+S_1/2.\] After the uniform padding bound, the numerical label sum is at most \[ \prod_{i=1}^J V_i^{m_i} \le (2W)^{Jk+S_1/2} \le (2W)^{Jk+L^{1/4}/2}. \tag{58}\] Here dropping distinctness among prime classes is simply an enlargement of a positive reciprocal sum. The classes are already identified by their first occurrences, so no further permutation factor is introduced.

There are at most \(2^{2k}\) sign choices and \(2^{2kJ}\) lit or unlit designations. The mixed difference contributes \(2^{S_1}\). The external indices cost at most \(|\mathcal D|^2M\le\exp(108L)\) for large \(L\), since \(|\mathcal D|\le e^{2L}\) and \(M=\lceil e^{103L}\rceil\). Combining these bounds with the forest count and (58) gives \[ \mathbb E\|H^k\|_{\mathrm{HS}}^2 \le K^{2k}\exp(C_4Jk)(2W)^{Jk+L^{1/4}/2}+o(1), \tag{59}\] where \(C_4\) is absolute. The factor \(\exp(108L)\) is absorbed here using \(J\ge1\) and \(k\ge L/2\).

For every fixed \(W\), once \(L\) is sufficiently large, \((2W)^{L^{1/4}/2}\le e^k\). This changes the required lower threshold on \(L\), but not the absolute constant in the exponential. An absolute choice of \(C_2\) therefore bounds the right-hand side of (59) by \[\bigl[K(C_2\sqrt W)^J\bigr]^{2k}.\] Finally, the finite-law comparison for the trace expansion changes the expectation by at most \(\exp(-L^9)\). Increasing the same absolute constant \(C_2\) absorbs this error as well and proves the integer-interval assertion of 18. In particular, \(C_2\) is fixed before \(W\) is selected; dependence on the fixed parameters is confined to how large \(X\) must be. ◻

Proof. Put \(m=\max\{a,b,\rho(\mathscr B)\}\). If \(m=0\), all \(B_i\) vanish. Otherwise set \(t=(2m)^{-1}\). For real \(|u|\le t\), the operators \(I-u\mathscr B\) and \(I+uB_i\) are invertible. Define \[F(u)=I-\sum_{i=1}^N uB_i(I+uB_i)^{-1}.\] Each summand is self-adjoint, since \(B_i\) commutes with its own resolvent. If \(F(u)v=0\), set \(z_i=(I+uB_i)^{-1}v\). Then \[(u\mathscr Bz)_i =\sum_{j\ne i}uB_jz_j =v-uB_i z_i=z_i.\] A nonzero \(v\) would give a nonzero \(z\), contradicting the invertibility of \(I-u\mathscr B\). Hence \(F(u)\) is invertible throughout \([-t,t]\). It is positive definite there by continuity, because \(F(0)=I\).

The resolvent identity gives \[F(u)=I-u\sum_i B_i+Q(u),\qquad Q(u)=u^2\sum_i B_i^2(I+uB_i)^{-1}.\] Since \(|u|\|B_i\|\le1/2\), the inverses in the last expression are positive and bounded above by \(2I\). For \(v=Pv\), \[0\le\langle v,Q(u)v\rangle \le2u^2\sum_i\|B_i v\|^2 \le2u^2b^2\|v\|^2.\] Positivity of \(F(t)\) and \(F(-t)\) therefore yields \[\left|\left\langle v,\sum_i B_i v\right\rangle\right| \le \frac{1+2t^2b^2}{t}\|v\|^2 \le3m\|v\|^2.\] Taking the supremum over unit vectors in the range of \(P\) proves (60). No step commutes two distinct \(B_i\). ◻

Proof. For each of the two edge orientations, the weighted absolute row sum at \(n\) is at most \[a_d(n)\sum_q\frac{Lu(q)}{w(n)}\mathbf 1_{q\mid n} \mathbf 1_{(d,q)\in\mathcal S_j}\mathbf 1_{\rho_j(n,d)\le K/L} \le K a_d(n).\] Indeed the sum without the last indicator is at most \(L\rho_j(n,d)\). For an incoming edge, \(q\) divides one endpoint if and only if it divides the other. All the other edge restrictions can be discarded in this nonnegative bound. Consequently \[ \sum_{n'}|A_d(n,n')| \frac{\sqrt{w(n')}}{\sqrt{w(n)}}\le2K a_d(n). \tag{63}\]

The weighted Schur inequality for a real symmetric matrix \(B\) follows by applying \[2|v_n v_{n'}|\le |v_n|^2\frac{\sqrt{w(n')}}{\sqrt{w(n)}}+ |v_{n'}|^2\frac{\sqrt{w(n)}}{\sqrt{w(n')}}\] to its quadratic form. Apply it to \(A_d\) on each level set of \(a_d\); these level sets are invariant under \(A_d\). The resulting bound is \[ \|A_d v\|^2\le4K^2\sum_n a_d(n)^2|v_n|^2. \tag{64}\] In particular, \(a_d(n)\le1\) proves (61).

At a site retained by \(E\), the total P-degree is at most \(6WJ\). Since \(V_i\le2W\), the arithmetic-geometric mean inequality gives \[\begin{align*} \sum_d a_d(n)^2 &\le\sum_d a_d(n) \le\prod_{i=1}^J\bigl(\omega_{P_i}(n)+V_i\bigr)\\ &\le\left(\frac{\omega_P(n)+\sum_iV_i}{J}\right)^J \le(8W)^J. \end{align*}\] Sum (64) over \(d\). This proves (62) without requiring \(E\) to commute with \(A_d\). ◻

Proof. For every finite matrix, \(\rho(H)^{2k}\le\|H^k\|_{\mathrm{HS}}^2\). Thus 18 and Markov’s inequality show that \[\rho(H)\le eK(C_2\sqrt W)^J\] except on a fraction at most \(e^{-2k}\) of the origins. On each remaining block apply 29 with \[a=2K,\qquad b=2K(8W)^{J/2},\qquad P=E.\] For \(J\ge1\) and \(W\ge10\), the choice \[ C_3=3\max\{2,2\sqrt8,eC_2\} \tag{66}\] gives (65). In particular, \(C_3\) is independent of \(W,h,l\). ◻

Proof. Write \(T=T_j\) and \(N=\lfloor T\rfloor\), and average over blocks \(I_t=\{t+1,\ldots,t+M\}\), \(0\le t<N\). For large \(X\), \(N\ge\exp(L^A/2)\), so the interval of origins is admissible in [q:finite-law,q:trace]. On each block put \[f_t(n)=\sqrt{w(n)}\lambda(n) \mathbf 1_{\omega_Q(n)\le400\log L},\qquad g_t=Ef_t.\] We first check the averaged norm needed for testing: \[ \frac1N\sum_{t=0}^{N-1}\|g_t\|^2 \le\frac1N\sum_{t=0}^{N-1}\|f_t\|^2\le2MS. \tag{68}\] For each of the \(M\) block positions, 8 applied to the corresponding translated interval of \(N\) origins bounds the average truncated weight by \(2S\). Summing proves (68). This uses the truncated comparison already established, without an untruncated integer moment of \(w\).

On blocks satisfying (65), the averaged absolute quadratic form is at most \(2MSK(C_3\sqrt W)^J\). On every block, including exceptional ones, (63) gives the deterministic bound \[ \left|\left\langle g_t,\sum_d A_d g_t\right\rangle\right| \le 2KM\,5^{400\log L}(8W)^J. \tag{69}\] Here we bound the form by its entrywise absolute value, use \(|g_t(n)|\le\sqrt{w(n)}\), and then sum the weighted row bounds at sites satisfying both degree cutoffs. The right side divided by \(M\) is a fixed power of \(L\) for fixed \(W\). Multiplication by the exceptional fraction \(e^{-2k}\) makes its contribution, after division by \(2LM\), at most \(e^{-L}\) for large \(L\). Therefore \[ \left|\frac1{2LMN}\sum_{t=0}^{N-1} \left\langle g_t,\sum_d A_d g_t\right\rangle\right| \ll \frac{S K(C_3\sqrt W)^J}{L}+e^{-L}. \tag{70}\]

It remains to compare this form with the original prefix sum. Denote by \(c_j(n;d,q)\) the signed summand belonging to \((d,q)\) in \(T C_j^\circ\), including all its endpoint cutoffs. Use the same formula to define it for every positive \(n\), and put \(r=hqd\). These cutoffs are predicates of the ambient sites, independent of the choice of block; restricting to a block only removes edges leaving it. Testing an increasing matrix entry cancels its two square-root weights and gives \(L c_j(n;d,q)\); symmetry gives the same term in the opposite orientation. Thus the form on the left of (70), before taking absolute values, is exactly \[ \frac1{MN}\sum_{n,d,q}m_N(n,r)c_j(n;d,q),\qquad m_N(n,r)= \bigl[\min(N-1,n-1)-\max(0,n+r-M)+1\bigr]_+, \tag{71}\] where \([x]_+=\max(x,0)\). This is the number of blocks containing both endpoints. All contributing \(n\) are positive.

Let \[R_L=h e^{100L+1},\qquad D_L=\frac KL\,5^{400\log L}(8W)^J.\] Every displacement is at most \(R_L\), and the one-orientation row bound proves \(\sum_{d,q}|c_j(n;d,q)|\le D_L\) for every \(n\). For large \(L\) we have \(N\gg M>R_L\). If \(M\le n\le N\), then \(m_N(n,r)=M-r\). The lower boundary and the extra sites \(N<n<N+M\) contain at most \(2M\) sites, and always \(0\le m_N(n,r)\le M\). Consequently the difference between (71) and \(C_j^\circ=T^{-1}\sum_{n\le N,d,q}c_j(n;d,q)\) is at most \[ O\left(D_L\left(\frac{R_L}{M}+\frac MN+\frac1T\right)\right). \tag{72}\] The last term accounts for replacing \(N^{-1}\) by \(T^{-1}\). Since \(M=\lceil e^{103L}\rceil\), \(T\ge X=e^{L^A}\) and \(D_L\) is polynomial in \(L\), (72) is \(O(e^{-L})\) for sufficiently large \(L\). This also bounds edges whose terminal endpoint exceeds \(T\) and the floor errors. Combining with (70) proves (67). ◻

Proof of 4. There are \(O(L/\eta)\) bins. Sum (67), use \(S_0\ge SV_*/2\) and \(V_*\ge W^J\), and then apply the centering and deletion estimates of [q:center,q:cuts]. We obtain \[\begin{align*} |F(X)|\ll_{h,l,W}{}& \eta+\eta^{-1}2^J L^{-1/20} +2^J\bigl(K^{-1}+L^{-100}+e^{-2WJ}\bigr)\\ &+\frac K\eta\left(\frac{C_3}{\sqrt W}\right)^J +e^{-L^{0.9}}+X^{-1}. \tag{73}\end{align*}\] The deletion error in 14 is already summed over the bins. The new error \(O((L/\eta)e^{-L})\) from 32 is absorbed by \(e^{-L^{0.9}}\), because \(L/\eta\) is polynomial in \(L\) and \(SV_*\ge1\).

Choose the absolute constants in the following order. First fix \(A\) as required by 6; in particular \(A\ge1000\) and \(A>C_1+20\), where \(C_1\) is the absolute exponent in that comparison. The trace proof gives an absolute \(C_2\), and (66) then gives an absolute \(C_3\). Now choose \(W\ge10\) so large that \[ e^5 C_3/\sqrt W\le e^{-1}. \tag{74}\] Only after fixing these constants do we increase the threshold for \(X\), allowing it to depend on \(h,l,W\).

Recall \(\eta=e^{-J}\), \(K=e^{4J}\) and \(J\le\delta\log L/(6W)\) with \(\delta=1/200\). The spectral term in (73) is at most \(e^{-J}\) by (74). The other terms satisfy \[\begin{align*} 2^JK^{-1}&=e^{-(4-\log2)J}\le e^{-J},\\ 2^Je^{-2WJ}&=e^{-(2W-\log2)J}\le e^{-J},\\ e^J2^J L^{-1/20}&\le e^{-J},\qquad 2^J L^{-100}\le e^{-J}. \end{align*}\] For the last line, it is enough to note that \[\frac{\delta(2+\log2)}{6W}<\frac1{20},\qquad \frac{\delta(1+\log2)}{6W}<100.\] Also \(e^{-L^{0.9}}+X^{-1}\ll e^{-J}\) for large \(X\). Hence \(|F(X)|\ll_{h,l}e^{-J}\). Finally, \[e^{-J}\le e L^{-\delta/(6W)} =e(\log X)^{-c},\qquad c=\frac{\delta}{6WA}>0.\] All three constants \(A,W,c\) are absolute. The argument applies to every sufficiently large real \(X\), with no excluded scales. On the remaining bounded range \(3\le X\le X_0(h,l)\), the estimate follows from \(|F(X)|\le1\) after enlarging the implied constant. This proves 4. ◻

Proof. Write \(v_p(n)\) for the exponent of \(p\) in \(n\). For a finite set \(\mathcal P\) of primes, multiplicativity gives \[|f(n)|\le \prod_{\substack{p\in\mathcal P\\v_p(n)=1}}|f(p)|.\] The majorant is periodic modulo \(\prod_{p\in\mathcal P}p^2\). Its ordinary mean is \[\prod_{p\in\mathcal P} \left(1-(1-|f(p)|)\left(\frac1p-\frac1{p^2}\right)\right).\] These products tend to zero as \(\mathcal P\) increases through the primes: the sum of the subtracted quantities diverges, whereas \(\sum_p p^{-2}<\infty\). First take the long average with \(\mathcal P\) fixed, and then increase \(\mathcal P\). ◻

Proof. Normalize the summands of (88) to a probability law on divisors, and denote its expectation and probability by \(\mathbb E_{\mathrm{div}}\) and \(\mathbb P_{\mathrm{div}}\). Each prime \(p\) in band \(a\) is included independently with probability \(A_a/(p+A_a)\). Prime harmonic estimates give \[\mathbb E_{\mathrm{div}}\omega(d)\ll_A\log B, \qquad \mathbb E_{\mathrm{div}}\log d\le A\sum_{p\le e^B}\frac{\log p}{p}\ll_A B.\] Choose \(C_0\) large enough that the probabilities of \(\omega(d)>\lceil C_0\log B\rceil\) and \(\log d>C_0B\) are each at most \(1/8\), by Markov’s inequality.

For \(r_p=|f_1(p)f_2(p)|\), the inequality \(1-\prod_j r_j\le\sum_j(1-r_j)\), valid when \(0\le r_j\le1\), gives \[\begin{align*} \mathbb E_{\mathrm{div}}\bigl(1-|f_1(d)f_2(d)|\bigr) &\le A\sum_{p>P_0}\frac{1-|f_1(p)f_2(p)|}{p}\\ &\le A\sum_{p>P_0}\frac{(1-|f_1(p)|)+(1-|f_2(p)|)}p=o(1). \end{align*}\] Here squarefreeness permits the use of multiplicativity. Another application of Markov’s inequality shows that the probability of \(|f_1(d)f_2(d)|<1/2\) tends to zero. Also \(\mathbb P_{\mathrm{div}}(d=1)=L_0^{-1}\to0\). Thus the remaining divisors carry at least \(L_0/2\) of the unnormalized mass for large \(B\).

There are at most \(1+C_0B/\log\tau\) intervals \((\tau^r,\tau^{r+1}]\), \(r\ge0\), meeting \((1,e^{C_0B}]\). One of them carries at least \(cL_0/B\) of that mass, for a fixed \(c>0\). Take its lower endpoint as \(H\) and retain the divisors already satisfying the preceding conditions. ◻

Proof. Put \[V_{\mathcal D}=\sum_{d\in\mathcal D}A^{\omega_C(d)},\qquad K_{\mathcal D}=\sum_{d\in\mathcal D} |f_1(d)f_2(d)|^2A^{\omega_C(d)}\ge\tfrac14 V_{\mathcal D}.\] For a fixed \(d\), write \(x=dm\). Unless \(\gcd(d,m)>1\) or \(\gcd(d,m+h)>1\), ordinary multiplicativity gives \[a_df_1(dm)f_2(d(m+h)) =|f_1(d)f_2(d)|^2f_1(m)f_2(m+h).\] The exceptional set has ordinary density at most \(2\sum_{p\mid d}p^{-1}\le2J/P_0\). At fixed \(B\) there are only finitely many \(d\), so all associated residue-counting errors vanish as \(j\to\infty\).

For \(p\in\mathcal C\) not dividing \(d\), divisibility of \(dm\) by \(p\) is equivalent to divisibility of \(m\). Hence the mean and variance of \(k_d(dm)\) under uniform residues are \[\mu_d=v_C-\sum_{\substack{p\mid d\\p\in\mathcal C}}\frac1p, \qquad \operatorname{Var}(k_d(dm)) =\sum_{\substack{p\in\mathcal C\\p\nmid d}} \frac1p\left(1-\frac1p\right)\le v_C.\] The same formulas hold with \(m+h\) in place of \(m\). For large \(B\), \(|\mu_d-v_C|\le J/P_0\le(T/4)\sqrt{v_C}\). Chebyshev’s inequality and a union bound show that the proportion on which either cutoff is not \(1\) is at most \(32/T^2\). Independence of the two endpoints is not needed.

The length \(X_j/d\) lies between \(N_j/\tau\) and \(N_j\). Comparing each shorter sum with the sum up to \(N_j\) therefore gives \[ \begin{split} |R_B(X_j)-K_{\mathcal D}S(N_j)| \le N_jV_{\mathcal D}\left( \tau-1+\frac{32}{T^2}+\frac{4J}{P_0}+o_{j\to\infty}(1)\right). \end{split} \tag{94}\] The factor \(4J/P_0\) allows a difference of size at most \(2\) on each coprimality failure. The endpoint rounding errors are included in \(o(1)\).

Choose \(\tau-1\le\gamma/64\) and \(32/T^2\le\gamma/64\), and then take \(B\) and \(j\) large enough for the other two errors to be at most \(\gamma/64\) each. The reverse triangle inequality, valid for the complex number \(S(N_j)\), yields \[|R_B(X_j)|\ge\frac\gamma8N_jV_{\mathcal D}.\] Finally, \(V_{\mathcal D}\ge H\sum_{d\in\mathcal D}A^{\omega_C(d)}/d \ge cHL_0/B\). This proves the result with \(c_\gamma=\gamma c/8\). ◻

Reduction of 2. We deduce 2 from [prop:graph-estimate,prop:analytic-centering]. The zero-function and divergent-defect cases were settled above. Otherwise suppose (86) holds. Choose \(\tau,T,\phi,C_0\) and, for each sufficiently large fixed \(B\), the divisor family of 35. Combining 36 with the two propositions gives \[c_\gamma\frac{L_0}{B} \le C L_0\left(B^{-1-c_*/2}+B^{-1-\varepsilon}+\frac{J}{P_0}\right).\] After dividing by \(L_0/B\), this reads \[ c_\gamma\le C\left(B^{-c_*/2}+B^{-\varepsilon} +BJ\exp(-B^{1-\varepsilon})\right), \tag{96}\] which is impossible for sufficiently large \(B\).

In this comparison the functions, shift, and bias \(\gamma\) are fixed first; then \(\tau,T,\phi,C_0\) are fixed. For each fixed \(B\) the limit is taken along the original sequence \(N_j\), with \(X_j=HN_j\). Only after these inequalities hold does \(B\) increase. Thus the contradiction excludes every alleged biased sequence. ◻

Proof. Outside \(\mathcal P\) the summands in the squared distances coincide. At a prime in \(\mathcal P\) their difference has absolute value at most \(|b(p)-f(p)|/p\le2/p\). Sum this bound, and then take the infimum over \(|t|\le X\) and the minimum over the finite character family. ◻

Proof. We give the sieve bounds including their counting errors. Put \[s=2\lceil100\log B\rceil,\qquad z_0=P_0^{1/(100s)}.\] For every prime \(p<z_0\), forbid \(\nu_p\) residue classes, where \(0\le\nu_p\le2\). If \(r(n)\) is the number of these prime conditions satisfied by \(n\), even inclusion-exclusion gives \[\mathbf 1_{r(n)=0}\le\sum_{j=0}^s(-1)^j\binom{r(n)}j.\] Indeed, for \(r>0\) the sum on the right is \(\binom{r-1}s\ge0\), and for \(r=0\) it is \(1\). Write \(E_j((a_p))=\sum_{|\mathcal I|=j}\prod_{p\in\mathcal I}a_p\) for the elementary symmetric sum over subsets of the primes \(p<z_0\). The Chinese remainder theorem, applied on any interval \(I\) of real length \(V\), now yields \[ \begin{split} \#\{n\in I:n\text{ avoids the forbidden classes}\} \le V\sum_{j=0}^s(-1)^jE_j((\nu_p/p)) +O\left(\sum_{j=0}^sE_j((\nu_p))\right). \end{split} \tag{107}\] The residue-counting error is independent of the position of \(I\). It is at most \[(s+1)(2z_0)^s=(s+1)2^sP_0^{1/100}\le P_0^{1/5}\] for large \(B\).

Let \(\Lambda_0=\sum_{p<z_0}\nu_p/p\). Mertens’ estimate gives \(\Lambda_0\le2\log\log z_0+O(1)\le3\log B\) for large \(B\). Since \(E_j((\nu_p/p))\le\Lambda_0^j/j!\), the difference between the truncated density in (107) and its full Euler product has absolute value at most \[\sum_{j>s}\frac{\Lambda_0^j}{j!} \le\frac{(e\Lambda_0/(s+1))^{s+1}}{1-\Lambda_0/(s+2)} \le B^{-10}.\] Consequently the count in (107) is at most \[ V\left(\prod_{p<z_0}(1-\nu_p/p)+B^{-10}\right)+O(P_0^{1/5}). \tag{108}\]

For one rough integer, take \(\nu_p=1\). The product is \(O(1/\log z_0)=O(B^{-1+\varepsilon}\log B)\). Thus the number \(R\) of \(P_0\)-rough integers in \((M,\tau M]\) satisfies \[ R\ll M B^{-1+\varepsilon}\log B. \tag{109}\] Both error terms in (108) are absorbed, since \(M\ge P_0/\tau\).

The one-point bound proves (105), because \(w>M\). To estimate the fourth moment, extend \(a_w=c_w/w\) by zero outside its support. Fourier orthogonality gives \[\int_{\mathbb R/\mathbb Z}|Q_M(\alpha)|^4\,d\alpha =\sum_k\left|\sum_w a_{w+k}\overline{a_w}\right|^2.\] There is no factor depending on \(h\): multiplication by the nonzero integer \(h\) preserves Haar measure on \(\mathbb R/\mathbb Z\). Let \(R_k\) count pairs \(w,w+k\in(M,\tau M]\) that are both \(P_0\)-rough. Each inner absolute value is at most \(R_k/M^2\). We next bound \(R_k\) for \(k\ne0\); only \(|k|\le2M\) can contribute. The permitted interval for \(w\) is an intersection of two intervals of length less than \(M\). Sieve out the residues \(0,-k\) modulo \(p<z_0\). If \(k\) is odd there are no pairs for large \(B\), because both rough integers must be odd. If \(k\) is even, the density product is \[\begin{align*} \frac12\prod_{3\le p<z_0}(1-2/p) \prod_{\substack{3\le p<z_0\\p\mid k}} \frac{1-1/p}{1-2/p} &\ll\frac1{(\log z_0)^2} \exp\left(\sum_{p\mid k}\frac1p+O(1)\right). \end{align*}\] Here \((1-2/p)/(1-1/p)^2=1-1/(p-1)^2\le1\) and \(\log((1-1/p)/(1-2/p))=1/p+O(p^{-2})\) for \(p\ge3\). The singular factor is uniformly \(O_{C_0}(\log B)\): indeed \[\sum_{p\mid k}\frac1p \le\sum_{p\le B}\frac1p+\frac{\log|k|}{B\log B} \le\log\log B+O_{C_0}(1),\] since \(\log|k|\le C_0B+\log2\). Applying (108) proves \[ R_k\ll M B^{-2+2\varepsilon}(\log B)^3\qquad(k\ne0). \tag{110}\]

There are \(O(M)\) nonzero differences, so their total contribution is \(O(M^{-1}B^{-4+4\varepsilon}(\log B)^6)\) by (110). The term \(k=0\) is at most \((R/M^2)^2\ll M^{-2}\), which is smaller than the same bound because \(M\ge P_0/\tau\). This proves (106). ◻

Completion of 37. Apply (100) with \(c_w=c_w^{(u)}\) to each fixed \(u\) in the expansion at the start of this section. Dividing its contribution by \(X\) contributes the factor \(A^{\omega_C(u)}/u\); the extracted factors \(f_1(u)f_2(u)\) have modulus at most \(1\). The two Fourier inversions cost at most \(\|\widehat\phi\|_1^2\). At fixed \(B\) all sums over \(u,w\) are finite, and the uniform estimate therefore gives, by (98), a total limsup \(O(L_0B^{-1-\varepsilon})\). Adding the earlier coprimality error \(O(L_0J/P_0)\) proves (95). ◻

Proof. Full edge divisibility survives reversal: \(e\mid x\) and \(x'=x\pm he\) imply \(e\mid x'\). For a candidate subpath all residue tests are constant on the original cylinder, since they involve only its fixed prime coordinates. If the current specification is not primitive, take a shorter qualifying contiguous subpath in an allowed orientation. Its support is contained in the preceding support. The positive length strictly decreases, so the procedure terminates. ◻

Proof. It suffices to count all qualifying specifications. For fixed edge labels, signs, and suffix, its displacement \(D=z_m-z_{i_0-1}\) is nonzero because the path vertices are distinct, and \(|D|\le Lh\tau H\). Sum the extra prime first, keeping these numerical edge labels fixed. Since \(p>P_0\), \[\sum_{\substack{p\in\mathcal S\,,\ p\mid D}} \frac1p \le \frac{\log(|D|+2)}{P_0\log 2} \ll \frac{\log(Lh\tau H+2)}{P_0}.\] The support weight for the remaining distinct edge primes is the product of their reciprocals. There are at most \(LJ\) edge-prime slots. Choosing the length, signs, suffix, factor counts, and a partition of these slots into equality classes costs \(\exp(O(LJ\log B))\). Indeed a partition of at most \(LJ\) slots has at most \((LJ)^{LJ}\) descriptions. Summing each remaining class over its prime values costs at most \(v_C+v_Z\), or \(1+v_C+v_Z\) as a uniform upper bound. At this point we may drop all bin and distinctness restrictions in those positive sums. The resulting bound is \[\mathbb P(n\notin Y) \le P_0^{-1} \exp\bigl(O(LJ\log B)+O(LJ\log\log B)+O(\log B)\bigr).\] Since \(LJ\log B=O(B^{1-\rho}\log^2B)=o(\log P_0)\), this proves the assertion. ◻

Proof. Suppose successive occurrence blocks of \(q\) are separated by edges not containing \(q\). The intervening path begins and ends at active vertices for \(q\), since adjacent qualifying \(q\)-edges have divisible endpoints. It is a shorter simple \(q\)-free path with nonzero total displacement divisible by \(q\). With extra prime \(q\) and its whole path as suffix, it contradicts primitivity. This proves the first part.

Within a constant-label block, an adjacent change of sign would repeat a vertex, so all signs agree. If \(b=1\), the prescribed suffix displacement is \(\pm khe_m\) for some \(1\le k\le L\). The prime \(p\) divides neither \(e_m\) nor \(h\), and \(p>L\), so this is impossible. Thus \(b>1\). The unequal squarefree labels \(e_b,e_{b-1}\) lie in one bin of ratio less than \(2\). If every prime of \(e_b\) divided \(e_{b-1}\), their quotient would be an integer at least \(2\), a contradiction. Choose \(r\mid e_b\) absent from \(e_{b-1}\). By the first part it is absent from the entire prefix. Its coefficient in the suffix is \[\pm kh\frac{e_b}{r},\qquad 1\le k\le L.\] This is nonzero modulo \(p\), because \(p>h,L\) and \(p\) divides none of the edge labels. The same reasoning applies to every prime confined to the tail.

Finally, intersect the occurrence interval of \(q\) with the prefix \(1,\ldots,b-1\). The resulting nonempty interval is \(i,\ldots,j\); in particular, \(j=b-1\) if \(q\) also occurs in the tail. The sum of these prefix terms is \(S=z_j-z_{i-1}\ne0\). If \(r\mid S\), reverse the path from \(z_{b-1}\) to \(z_{i-1}\). Its edges avoid \(r\), it starts at a vertex active for \(r\), and its suffix from \(z_j\) to \(z_{i-1}\) has displacement divisible by \(r\). Full divisibility survives reversal. This shorter qualifying specification contradicts primitivity. Since \(q\ne r\) and \(S\) is \(q\) times the asserted coefficient, that coefficient is invertible modulo \(r\). ◻

Proof. The number of slots is at most \[\ell J+2t_0(LJ+1)=O(B\log B),\] since \(4\varepsilon<\rho\). A partition of \(R\) slots has at most \(R^R\) descriptions. All further indices have polynomially many choices in \(B\) per slot or per recorded vertex, giving the stated entropy. Recording the destination of each step among at most \(\ell+1\) vertices includes possible vertex identifications and any chosen tree of recorded steps within the same bound.

Drop the cutoffs and \(W^{-1}\), which lie in \([0,1]\). The powers of \(A\) from core occurrences are at most \(A^{\ell J}\). A lit occurrence of \(p\), or qualification of any attached specification using \(p\), fixes its residue and supplies at most \(1/p\). In the absence of either, \(p\) appears only on unlit line occurrences. An unlit core occurrence vanishes; each unlit center occurrence supplies \(\theta/p\), so at least one factor \(1/p\) is still available. The remaining conditions may be discarded after taking these positive primewise bounds. Independence of the prime residue coordinates proves (118).

If a repeated center prime has a lit occurrence, every unlit occurrence supplies an additional reciprocal beyond the residue probability. If all its \(m\ge2\) line occurrences are unlit, they supply \(p^{-m}\), saving at least \(p^{-(m-1)}\) beyond the displayed majorant; here \(m-1\ge m/2\). Additional qualification conditions can only improve the estimate. At least \(t_0\) repeated-unlit occurrences therefore save \(P_0^{-t_0/2}\).

For each unrestricted prime variable its reciprocal sum is at most \(1+v_C+v_Z=O(1+\log B)\). Thus the total before the extra saving is \(\exp(O(B\log^2B))\), including \(A^{\ell J}\). The saving has logarithm at most \(-\frac12 B^{1+3\varepsilon}\), which proves the last assertion for sufficiently large \(B\). ◻

Proof. Fix the nonselected variables. For the first type of test, \[\sum_{\substack{q\in\mathcal S\,,\ q\mid D_j}}\frac1q \le \frac{\log(|D_j|+2)}{P_0\log 2} \ll \frac{B\log B}{P_0}.\] For the second type, invertibility leaves one residue class modulo \(r_j\). Comparison with the integral of \(1/x\) along that progression gives, even when the sum is enlarged from primes to integers, \[\sum_{\substack{P_0<n\le e^B\\ n\equiv a_0\pmod{r_j}}}\frac1n \le \frac1{P_0}+\frac{B-\log P_0}{r_j} \le \frac{1+B}{P_0}.\] The bound includes the zero residue class.

Sum the selected variables in the order \(q_k,q_{k-1},\ldots,q_1\). When summing \(q_j\), every remaining earlier test is independent of it, and its own test has one of the preceding uniform bounds. The conditions \(D_j\ne0\) and \(a_j\not\equiv0\pmod{r_j}\) are retained through this elimination. If either fails for fixed earlier data, the admissible inner sum is zero. Numerical distinctness may also be retained; dropping it is unnecessary for the progression bound. Backward induction therefore gives a factor \[\left(\frac{O(B\log B)}{P_0}\right)^k.\] The other prime sums and all descriptions cost \(\exp(O(B\log^2B))\). Since \(k\log P_0\ge B^{1+\varepsilon}\), their product satisfies (119).

We verify the final description convention. There are \(O(B\log B)\) slots and polynomially many recorded positions or vertices. A bounded number of paths per test is specified by its endpoints and type, together with a choice of prime variables and order; choosing at most the number of slots many tests costs \(\exp(O(B\log^2B))\). The common integer offset has \(O(\ell H+1)=\exp(O(B))\) possible values. After fixing it we may drop the equation that originally identifies it with a difference of tree offsets, since we are taking an upper bound. Each edge product under arbitrary slot assignments is at most \(e^{BJ}\). Hence every indicated bounded sum of path displacements still has logarithmic size \(O(B\log B)\), even after the original divisor-bin restrictions have been relaxed. Nonzero and invertibility tests are never relaxed. ◻

Proof. Root each component. The top of a path is its vertex nearest that root. Choose a path whose top has maximum depth, and mark its one or two edges incident with the top. Every path sharing an edge with the chosen path must contain a marked edge. Indeed, a path sharing an edge below a mark and avoiding that mark remains wholly in the component below it; its top would be strictly deeper. Remove the paths hit by the marks and repeat. The paths chosen at successive stages are pairwise edge-disjoint. There are fewer than \(t\) stages, so fewer than \(2t\) marked edges suffice. ◻

Proof. An unlit core occurrence makes \(K_{\boldsymbol d}=0\). A primitive specification supported on \(\mathcal F\) and qualifying at a trace vertex makes the corresponding \(Y\) factor zero for every remaining coordinate assignment. Discard these configurations. By 46, those with at least \(t_0\) unlit repeated center occurrences have the asserted negligible total. Each of these decisions uses only the line and \(n_{\mathcal F}\).

Cut every tree edge having an unlit fixed occurrence. There are fewer than \(t_0\) such edges. In the remaining forest a gap for \(p\) is a path of length at most \(L\) with active endpoints and at least one interior vertex, all interior vertices being inactive, for a fixed tree prime \(p\). Such a gap has no \(p\)-edge: an active endpoint of a \(p\)-edge forces its other endpoint to be active, while a \(p\)-edge with inactive endpoints was cut.

Every gap contains a nonfixed prime label. Otherwise all its edge labels are fixed and lit, so the path is fully divisible. Its nonzero total displacement is divisible by \(p\), and its edges avoid \(p\). Taking extra prime \(p\) and the whole path as suffix gives a qualifying specification. By 42, a primitive one with fixed support qualifies at a vertex of that tree path. Every tree vertex is a trace vertex, contrary to the preceding exclusion.

If there are \(t_0\) edge-disjoint gaps, choose a nonfixed prime \(q\) on each. Each chosen prime occurs once on the full line, hence nowhere on the other selected gaps. The displacement of its gap is linear in \(q\) modulo the corresponding fixed prime \(p\), with coefficient \(\pm h\) times the other prime factors of that edge. This coefficient is invertible modulo \(p\), since the gap is \(p\)-free and \(p\nmid h\). The selected constraints do not involve one another’s variables. Their paths and variables have the description cost of 47, which discards this class. The inequality \(t_0\ge B^{2\varepsilon}\) holds for large \(B\).

Otherwise 49 supplies fewer than \(2t_0\) further cuts meeting every gap. For each remaining piece traverse every edge twice and split that traversal into segments of at most \(L\) steps. The edges and vertices in a segment form a connected subtree of diameter at most \(L\); include a singleton block for a piece with no edges. There are \[O(\ell/L+t_0+1)=O(B^\rho)\] blocks. They may overlap. Any two active vertices in a block have their joining path inside that block. An inactive portion on it would give a surviving short gap, impossible after the cuts. Thus activity is connected or empty on each block. Each block meets at most one whole-tree active component, and every such component has a vertex in a block. This proves the component bound. ◻

Proof. Start with arbitrary roots. A connected vertex set in a tree contains the entire path between any two of its vertices. It therefore has a unique nearest vertex to a chosen root. The root-to-\(x_p^j\) path avoids \(p\)-edges. If a retained label \(q\) occurs on this path, both ends of its edge lie in \(A_q^j\); the nearest vertex \(x_q^j\) is then a strict ancestor of \(x_p^j\). Thus dependencies are contained in the two strict ancestor orders. Indices with equal nearest points are incomparable.

Every finite partially ordered set with \(u\) elements has a chain or an antichain of size at least \(\sqrt u\): partition its elements by the length of a longest chain ending there. Apply this to the first ancestor order. If an antichain is obtained, there are no dependencies from block 1 among its indices, so depth order in block 2 suffices. Otherwise apply the same argument in block 2 on the selected chain. An antichain there similarly suffices. In the remaining case we have chains in both blocks with at least \(m^{1/4}\) indices. The Erdős–Szekeres monotone-subsequence theorem (Erdős and Szekeres 1935, 467) shows that two total orders have a common agreeing or reversed subsequence of size at least the square root of their length. To recall the elementary argument, assign to each position the lengths of the longest increasing and decreasing subsequences ending there. Two positions have different pairs, so if both lengths are less than \(\sqrt u\) there cannot be \(u\) positions. We thus retain at least a constant times \(m^{1/8}\) indices. Agreeing orders already give the conclusion.

Suppose the orders are reversed. In block 2 take the segment from its root to the farthest selected \(x_p^2\). The intersection of each \(A_p^2\) with that segment is a closed vertex interval beginning at \(x_p^2\); these left endpoints are distinct. At a vertex of the segment, each positive-length interval containing it contains an incident segment edge. Each of the at most two such edges has at most \(J\) prime factors. At most one interval can consist of that vertex alone. Hence at most \(2J+1\) intervals overlap at a vertex. Greedily coloring closed intervals in order of their left endpoints uses at most \(2J+1\) colors: an existing color is available only if its preceding right endpoint is strictly earlier than the new left endpoint. One color retains at least a \(1/(2J+1)\) fraction of pairwise vertex-disjoint intervals.

Reroot block 2 at the far end of the segment. The nearest point of \(A_p^2\) is now the right endpoint of its interval. Indeed, for any vertex of \(A_p^2\) off the segment, its projection onto the segment lies in that interval, because the path from \(x_p^2\) to the vertex lies in \(A_p^2\). The path from the new root first reaches \(A_p^2\) at the right endpoint. This also proves that off-segment branches cannot change the nearest point.

The right endpoints of vertex-disjoint intervals have the same order as their left endpoints along the original segment. Seen from its other end, that order reverses. The two block orders therefore agree after rerooting. The first paragraph applies to the new root as well and proves the dependency assertion. Each representative stayed in the same whole-tree component for its prime, so the two representatives for that prime are still distinct. ◻

Proof. For every such prime choose two blocks meeting distinct whole-tree active components. There are \(O(B^{2\rho})\) ordered block pairs, so one pair serves \(m\gg B^{1-4\rho}\) primes. Inside each block, the active set is nonempty and connected. Every edge carrying a fixed prime has active endpoints after the earlier cuts, and the converse follows from \(p\nmid h\). Apply 51 and then specify the common integer \[K=b(c_2)-b(c_1),\qquad |K|=O(\ell H).\] Each retained prime gives the test (122). The expression is independent of \(p\), since both local paths avoid it. It involves only earlier retained variables by the lemma. Its nonvanishing follows from distinctness of the two representatives and the distinct integer offsets of tree vertices. The number of tests is \[\gg \frac{B^{(1-4\rho)/8}}{J} =\frac{B^{1/10}}{O(\log B)}>B^{2\varepsilon}\] for large \(B\).

Roots, endpoints, and prime choices can be recorded with the entropy allowed in 47. Choose \(K\) only after the rerooting; its \(O(\ell H+1)\) possibilities cost \(\exp(O(B))\). Although its original value depends on the labels, fixing it and dropping that defining equation enlarges a positive sum. The local path expressions then have the required triangular dependence. Retain their nonzero tests under this enlargement and apply 47. ◻

Proof. For every corrupted prime choose a witnessing vertex outside its edge endpoints, and put \(D=\lfloor B^{1-3\rho}\rfloor\). Distance from a vertex to an edge means its minimum tree distance to the two endpoints. Suppose first that at least half the selected witnesses have distance at most \(D\) from their prime edge. Use the path from the nearer edge endpoint to its witness. This path avoids the prime’s sole edge, so its nonzero displacement is independent of that prime and is divisible by it. The expression contains at most \(JD\) other prime labels.

On the selected primes draw a directed dependency edge \(p\to q\) if the test for \(p\) contains \(q\). Its outdegree is at most \(JD\). Every induced subgraph of the underlying undirected graph has average degree at most \(2JD\). Successively choose a vertex of degree at most \(2JD\) and delete its neighbors. The resulting independent set has size at least a constant times \[\frac{B^{1-2\rho}}{JD+1}\gg\frac{B^\rho}{J}>B^{2\varepsilon}.\] Its tests have no selected-variable dependencies. They are covered by 47.

Otherwise at least half the witnesses have distance greater than \(D\). Cover the full tree by \(O(\ell/D+1)=O(B^{3\rho})\) connected traversal blocks of diameter at most \(D\), as in the earlier construction. Assign each prime edge to a block containing that edge and its witness to a block containing that vertex. One ordered pair serves \(\gg B^{1-8\rho}\) primes. No selected prime edge has an endpoint in the second block: its corresponding witness lies there, so that would put it at distance at most \(D\). Consequently paths inside the second block contain no selected prime label.

Root the two blocks at \(c_1,c_2\). In block 1 let \(x_p^1\) be the nearer endpoint of the prime edge; in block 2 let \(x_p^2\) be its witness. The first local path avoids \(p\). If it contains a selected prime \(q\), the nearer endpoint of the \(q\)-edge is a strict ancestor of \(x_p^1\). Thus the tests (122), after specifying the common root offset, are triangular in the depths of the first-block endpoints. The second local paths introduce no selected variables. Each tested displacement is nonzero because its witness differs from the chosen edge endpoint. Since \(1-8\rho=3/5>2\varepsilon\), there are more than enough tests for 47. All choices of witnesses, blocks, roots, and paths fit its description bound. ◻

Proof of 48. Define \(R_{\boldsymbol d}\) by retaining the conditions obtained in [lem:activity-blocks,lem:few-disconnected,lem:few-corrupted]. Every activity and occurrence status of a fixed label is determined by \(n_{\mathcal F}\); fixed forbidden qualifications use only those coordinates; corruption uses only numerical offsets. Covers and witnesses can be chosen by fixed finite ordering whenever they exist. Thus \(R_{\boldsymbol d}\) depends on no free center residue.

The zero exclusions vanish with \(K_{\boldsymbol d}\prod_j\mathbf 1_Y\). The other excluded classes have the stated absolute contribution by the three lemmas, yielding (121). Their proofs used precisely the majorants of [lem:crude-count,lem:triangular-saving], which allow up to \(2t_0\) attached specifications. Their retained tests depend only on the numerical line and fixed residues, and are unchanged when qualification indicators are added. This proves the additional assertion as well. ◻

Proof. If \(I\le J\), every event containing \(I\) also contains \(J\). Thus every family admitted for the maximum defining \(\operatorname{rk}(I)\) is admitted for \(\operatorname{rk}(J)\), proving monotonicity. Choose an inclusion-minimal family generating \(I\). Each member must have a private coordinate: otherwise all its prescribed values already follow from the other members, since the prescriptions are compatible, and it can be removed. Its size is therefore at most \(\operatorname{rk}(I)\), proving (124) as well.

Fix \(J\in\mathscr P\) and let \(m=|\operatorname{supp}(J)|\). An element of the interval \([\widehat0,J]\) is determined by its support: its values must be the corresponding restrictions of those prescribed by \(J\). Consequently this interval has at most \(2^m\) elements. Along a strict chain from \(\widehat0\) to \(J\), supports grow strictly, so there are at most \(m\) steps. A chain of \(k\) steps is specified by recording, for each coordinate of \(\operatorname{supp}(J)\), the step at which it enters. There are at most \(k^m\) such chains. Expanding the inverse of the poset’s upper triangular incidence matrix, or applying the Möbius recursion repeatedly, gives the alternating sum over these chains. Thus, for \(m\ge1\), \[ |\mu(\widehat0,J)|\le\sum_{k=1}^m k^m\le m^{m+1}; \tag{126}\] the coefficient at \(\widehat0\) is \(1\). This proves the asserted coefficient bound for intersections of rank below \(t\).

For \(x\in\Omega\), let \(J_x\) be the intersection of all events satisfied at \(x\), with \(J_x=\widehat0\) if none are satisfied. An intersection \(I\in\mathscr P\) contains \(x\) if and only if \(I\le J_x\). Indeed any family generating \(I\) then consists of events satisfied at \(x\). It follows that \[ \sum_{I\in\mathscr P}\mu(\widehat0,I)\mathbf 1_I(x) =\sum_{I\le J_x}\mu(\widehat0,I) =\mathbf 1_{J_x=\widehat0}. \tag{127}\] Since no forbidden event is the whole space, the last expression is precisely the avoidance indicator in (125).

If \(\operatorname{rk}(J_x)<t\), monotonicity shows that the truncated sum contains the entire interval \([\widehat0,J_x]\), so it is exact. Suppose instead that \(\operatorname{rk}(J_x)\ge t\). Given any \(I\le J_x\) of rank below \(t\), successively intersect it with events satisfied at \(x\) until reaching \(J_x\). Let \(K\) be the first intersection whose rank is at least \(t\), and let \(K^-\) be its immediate predecessor. Replace the generating family for \(K^-\) by an inclusion-minimal one. It has at most \(t-1\) members. Adjoining the event that produces \(K\) shows that \(K\) is generated by at most \(t\) events. Thus \[I\le K\le J_x,\qquad K\in\mathscr H_t, \qquad |\operatorname{supp}(K)|\le M.\] This argument allows the rank to jump by more than one. The re-minimization of the predecessor, rather than the length of the successive list, bounds the number of generators.

Every satisfied low-rank \(I\) is therefore below a satisfied member of \(\mathscr H_t\). For any one such member there are at most \(2^M\) possible predecessors \(I\), by the interval bound already proved. The exact avoidance indicator is zero in the present case. Bounding each retained coefficient by (126) now gives (125). ◻

Proof. Order the witnesses by nondecreasing attachment indices \(k_1,\ldots,k_{t_0}\); break ties arbitrarily. For witness \(j\), choose a private prime \(q_j\notin\mathcal F\), write \(p_j\) for its extra prime, and choose a tail prime \(r_j\) as in 45. Thus \(r_j\) occurs only in its final constant block. That lemma supplies two coefficient facts:

1. the intrinsic suffix displacement divisible by \(p_j\) is linear in a tail-only prime, with coefficient nonzero modulo \(p_j\);

2. if a prime occurs before the constant tail, its coefficient in the prefix displacement immediately preceding that tail is nonzero modulo \(r_j\).

All private primes \(q_j\) are distinct. Each occurs at most once as a label on the original line, because it is outside \(\mathcal F\). We divide into three cases to control every dependence on another selected variable.

Proof. At fixed \(\boldsymbol d,n_{\mathcal F}\) with \(R_{\boldsymbol d}=1\), apply 54 to the residual cylinders with \(t=t_0\) and \(w=LJ+1\). Multiply the identity with its pointwise error bound by \(K_{\boldsymbol d}\) and integrate the residual coordinates. For the low-rank sum, take the absolute value after each such integral and use the generating lists already constructed. The coefficient bound gives the main term on the right of (130).

For the error, take the absolute value of \(K_{\boldsymbol d}\) inside the integral and use the generating and witnessing lists described above. After integrating the fixed coordinates, their indicators require joint qualification of at most \(2t_0\) specifications, including the \(t_0\) witnesses with private primes outside \(\mathcal F\). We may drop the retained-data restriction in this positive bound. 55 supplies at least \(B^{2\varepsilon}\) triangular constraints for each nonzero contribution.

Refine the error count by the lit or unlit status of every line occurrence. Joint qualification fixes one residue for each distinct prime used by the specifications, or is inconsistent and contributes zero. For each refinement, 46 gives \(A^{\ell J}\prod_p p^{-1}\), with the product over distinct prime labels, rather than charging a shared residue repeatedly. The line and lists have at most \[\ell J+2t_0(LJ+1)=O(B\log B)\] prime slots. Their equality patterns, occurrence statuses and all choices of witness constraints have cost \(\exp(O(B\log^2 B))\). These are exactly the positive majorant and description budget required by 47. Its reverse elimination applies to the system just supplied by 55 and gives \(O(e^{-B^{1+\varepsilon/2}})\). The additional factor \(\mathfrak C_B\) is absorbed by the same estimate, since (128) gives \(\log \mathfrak C_B=o(B)\) whereas the saving before this final simplification is \(\exp(-\Omega(B^{1+\varepsilon}))\). This proves the proposition. ◻

Proof. At each step record whether its destination is new and record its sign. At each return record one of at most \(\ell+1\) previously visited vertices. These data give the stated bound. A tree-edge label and its sign specify the difference of its endpoint offsets; all offsets are consequently determined, and a return label must be the absolute endpoint difference divided by \(h\). We retain only assignments for which the abstract vertices have distinct offsets and all these return labels belong to \(\mathcal D\). There is no independent choice of a return label.

Let \(q\) be the number of leaves and \(b\) the number of vertices with at least two children. The adding steps occur in at most \(r_*+1\) runs, each a downward path, so \(q\le r_*+1\). Also \(b\le q-1\) and \[\sum_{s(v)\ge2}s(v)=q+b-1.\] The number of edges incident with a leaf or such a branch vertex is therefore at most \(4q\). There are at most \(2r_*\) endpoints of return steps. An endpoint that has not already been counted is unary and is incident with at most two tree edges, giving at most \(4r_*\) further edges. The root adds at most one edge unless it was already counted as a branch vertex. These bounds imply (132). ◻

Proof. Start at a prescribed vertex \(v\). At every reached vertex include each child edge independently with probability \(A_a\beta_a\) and exclude it with probability \(\beta_a\). The resulting connected subtree has top \(v\). A particular outcome \(Q\) has probability \[(A_a\beta_a)^{|E(Q)|} \beta_a^{\sum_{w\in V(Q)}s(w)-|E(Q)|}=u_a(Q).\] The outcome with no edges has probability \(\beta_a^{s(v)}\). Sum the other probabilities, then sum over the possible tops \(v\). ◻

Proof. If adjacent vertices are active for \(p\), their difference \(\pm hd\) is divisible by \(p\). Since \(p\nmid h\), the intervening label contains \(p\). Conversely, a lit occurrence has both endpoints active. Thus the nontrivial active components record exactly the lit tree occurrences. Ghosts record the other occurrences; isolated active vertices create no label. Multiplying the distinct prime values of the tokens covering an edge recovers its label, proving (i).

By 48, the fixed primes with several active components contribute \(O(B^{1-2\rho}B^\rho)=O(B^{1-\rho})\) subtree tokens. Unlit fixed occurrences contribute \(O(t_0)\) ghosts and at most that many newly tagged connected labels. Corrupted nonfixed primes contribute \(O(B^{1-2\rho})\) ghosts. Lists contribute \(O(M_0)\) slots and at most one new token per newly tagged connected tree prime: disconnected primes and primes with unlit occurrences have already been counted. Since \[M_0=O(B^{1-\rho+4\varepsilon}\log B),\qquad 1-\rho+4\varepsilon<1-\rho/2,\] the sum of these bounds is at most \(N_{\mathrm{exc}}\) eventually.

For (iii), such a prime is repeated on the line and all its occurrences are lit. If its tree token consists of a single edge, its connected active set contains only the two endpoints of that edge. A further occurrence cannot be another tree edge, and hence must be a return touching these endpoints. This contradicts goodness. An active return endpoint elsewhere would give another active component and would have tagged the prime. ◻

Proof. Order of integration. The set \(\mathcal F\) does not contain every core prime. Put \(\mathcal C'=\mathcal C\setminus\mathcal F\) and \(\mathcal Z'=\mathcal Z\setminus\mathcal F\). At fixed \(n_{\mathcal F}\) use exactly the inequality \[ \left|\mathbb E_{\mathcal C'}\mathbb E_{\mathcal Z'} K_{\boldsymbol d}I_{\mathfrak L}\right| \le \mathbb E_{\mathcal C'} \left|\mathbb E_{\mathcal Z'} K_{\boldsymbol d}I_{\mathfrak L}\right|. \tag{137}\] In the inner average hold all core coordinates as parameters. The cutoffs can depend on the background core coordinates in \(\mathcal C'\), but never on \(\mathcal Z'\). Apart from these cutoffs, the kernel and each compatible list cylinder separate by prime. The inner center-prime averages therefore factor, with their signs intact. We will subsequently use positive majorants in the other coordinates. No enlargement of the fixed set in the cylinder construction is involved.

Charging the cutoffs. Use the origin cutoff at each return step. For each \(u\in\mathcal U\), use the endpoint cutoff at \(u\) of its incoming tree edge, which exists because a good origin is not the root. These are distinct cutoff occurrences: vertices of \(\mathcal U\) touch no return, and an incoming tree edge has a unique child. A charged cutoff with label \(d\) is at most \[\exp\{-\kappa(v_C-T\sqrt{v_C})\} \exp\left\{\kappa \sum_{\substack{p\in\mathcal C\\p\nmid d}} \mathbf 1_{p\mid n+b(u)}\right\}.\] This inequality holds also off its support. The other cutoff factors are at most one. The constant parts give the first negative term in the exponent of (136).

For a return, allocate the normalization at its origin to that return. For a core prime its multiplier is \(A\beta_C\) when the prime is in the label and lit, zero when it is in the label and unlit, and at most \(\max(1,\beta_Ce^\kappa)=1\) otherwise. Both \(A\beta_C<1\) and \(\beta_Ce^\kappa<1\) follow from \(A=e^{2\kappa}\). Center-prime return multipliers also have absolute value at most one. These bounds will be used except when integrating a good singleton, where we retain the return factors until its signed average has been taken.

Fixed tree primes. After removing the bounded return factors, the core tree factors for an active component \(Q\) containing an edge are \[A^{|E(Q)|}\prod_{v\in V(Q)}\beta_C^{s(v)}=u_C(Q).\] At a charged vertex \(u\in\mathcal U\), activity in such a component either continues through the incoming label, in which case that prime is excluded from this charge, or begins a subtree token at \(u\), which is excluded by the definition of \(\mathcal U\). Thus an additional \(e^\kappa\) can occur only at an isolated active vertex there. Its normalization is \(\beta_C^{s(u)}e^\kappa=\beta_Ce^\kappa\le1\). Other isolated activities also contribute at most one. The remaining bound is the product of the core subtree weights.

A core tree prime has a lit occurrence. Choose one deterministically from its tokens and retain its activity indicator. This is one specified residue modulo \(p\) and supplies the factor \(1/p\) upon integration. For a fixed center tree prime, the same argument gives the product of its active-subtree weights; lit factors \(1-\theta/p\) are at most one, and ghost occurrences are bounded absolutely. If there is a lit tree occurrence, retain one such residue indicator. If there is none, at least one unlit tree occurrence supplies \(\theta/p\le1/p\) directly, without a residue restriction. Ghost weight \(2\) bounds all the remaining ghost factors. These constructions are positive primewise majorants on the assigned group.

Nonfixed tree primes. A tagged nonfixed tree prime has one occurrence. Its absolute integral is at most \(2/p\): the lit residue contributes at most \(1/p\) and the other residues at most \(\theta/p\). If the list fixes its residue, the bound only improves. An untagged nonfixed prime on a nongood edge \(Q\) contributes at most \[\frac{u_Z(Q)}p+\frac\theta p =\frac{\widetilde u_Z(Q)}p.\] Indeed its lit residue includes the normalizations of both endpoints, and any additional activities can only reduce their product.

Now let \(p\) be an untagged nonfixed prime on a good edge \(Q=(u,v)\). It occurs nowhere else on the line and in no list. On its lit residue, noncorruption says that no other tree vertex is active. Each endpoint has exactly one tree departure and no return incidence. Its entire prime multiplier, including return factors, is therefore exactly \[(1-\theta/p)\beta_Z^2=\theta(1-\theta/p).\] Let \(a_0\) be its lit residue, and for \(a\in\mathbb Z/p\mathbb Z\) define \[\nu_p(a)=\#\{0\le i<\ell:a+b_i\equiv0\pmod p\}.\] This counts departures with multiplicity, including return departures. We have \(\nu_p(a_0)=2\). On every other residue the entire prime multiplier is exactly \(-\theta\beta_Z^{\nu_p(a)}/p\). Consequently its signed average is exactly \[\begin{align*} &\frac\theta p(1-\theta/p) -\frac\theta{p^2}\sum_{a\ne a_0}\beta_Z^{\nu_p(a)}\\ &\qquad=\frac\theta{p^2}\left( 1-\theta+ \sum_{\substack{a\ne a_0\\\nu_p(a)>0}} (1-\beta_Z^{\nu_p(a)})\right). \end{align*}\] There are at most \(r_p-1\) terms in the last sum, where \(r_p\le\ell+1\) is the number of residue classes represented by tree vertices. The displayed quantity is nonnegative and at most \[\frac{\theta r_p}{p^2} \le\frac{\theta(\ell+1)}{p^2} \le\frac{\theta}{2p} =\frac{\widetilde u_Z(Q)}p\] for sufficiently large \(B\). Thus \(\theta=\beta_Z^2\) cancels the terms of order \(1/p\) before any absolute value is taken. Removing return factors by an absolute estimate before taking this average would not justify the calculation on the no-activity residues; here they have been retained throughout.

List primes and background primes. For a list prime absent from the tree, qualification supplies one specified residue. All other factors, including charged factors, are bounded by one: any activity in the tree is isolated, and at a vertex of \(\mathcal U\) its factor is again \(\beta_Ce^\kappa\le1\). This yields \(1/p\) for that prime.

For any remaining prime, discard return factors by the absolute bounds already proved. The positive tree multiplier is \[\prod_{v\text{ active}}\beta_a^{s(v)} \exp\{\kappa\mathbf 1_{a=C}\#(\mathcal U\cap\{v\text{ active}\})\} \le1.\] If the tree offsets are pairwise distinct modulo \(p\), its integral is exactly \[ 1-\frac{S_a(\mathscr T) -\mathbf 1_{a=C}\beta_C(e^\kappa-1)|\mathcal U|}{p}. \tag{138}\] For offset collisions we use the bound one. A prime occurring only on a return divides the nonzero difference between that return’s two distinct vertices, so is automatically one of these collision primes. It entails no independently chosen prime slot.

The omitted background primes are tree or list primes, or divide one of the \(O(\ell^2)\) nonzero differences of size \(O(\ell H)\). There are \(O(B\log B)\) tree and list slots. Since every prime is at least \(P_0\) and \(\log(\ell H+2)=O(B+\log B)\), their reciprocal mass, multiplied by \(\ell+1\), is \(o(1)\), uniformly in the valid assignment. For example each nonzero difference has at most \(\log(O(\ell H)+2)/\log P_0\) prime divisors from \(\mathcal S\). Using \(1-x\le e^{-x}\) in (138) and restoring the omitted harmonic masses gives the remaining exponent in (136), with an \(o(1)\) error. The numerator in (138) is nonnegative, since at every charged vertex \(\beta_C e^\kappa\le1\).

Finally, the grouping does not assert independence of conditioned coordinates. After (137) and the signed free-center integrations, use the positive majorants above on the group. Whenever a used-prime probability \(1/p\) is claimed, its chosen activity or list residue indicator is retained. Drop all other group restrictions and integrate the resulting product over the remaining coordinates. This is an enlargement of a nonnegative integral and hence is valid despite any dependencies within the original group. It yields one bound per assigned record, not one bound for each individual fixed-coordinate configuration. Combining the primewise bounds proves the lemma. ◻

Proof. Choose one eligible type at each origin using any fixed order, and then choose its first named slot. Distinct origins choose different slots. Because \(s(u)=1\), a nontrivial subtree with top \(u\) contains the edge leaving \(u\).

Fix all other prime values and order the selected slots ancestor-first by their tops. If the token of a selected slot occurs on the edge leaving \(u\), its top is \(u\) or an ancestor of \(u\). Thus that edge’s product contains its selected prime and only earlier selected primes. The bin requirement has the form \[H<p_i D_i(p_1,\ldots,p_{i-1})\le\tau H,\] where \(D_i\) is positive and all its other factors have been fixed. The harmonic sum of a core prime in such an interval is \(O(B^{-1+\eta})\), uniformly in its position. To see this, write the interval as \((x,\tau x]\). If it meets the core band, then \(x\ge \exp(B^{1-\eta})/\tau\), and the prime-counting upper bound gives \[\sum_{\substack{x<p\le\tau x\\p\in\mathcal C}}\frac1p \le\frac{\pi(\tau x)}x \ll_\tau\frac1{\log x}\ll B^{-1+\eta}.\] Eliminate the selected variables in reverse ancestor order. Each bound is uniform in the remaining earlier values, so induction gives one such factor per selected slot. Since \(v_C\ge1\) for large \(B\), each bound is also \(O(B^{-1+\eta})v_C\). All unselected variables keep their unrestricted harmonic sums. This choice of named pivots requires no enumeration of possible pivots, and the bin condition holds for every slot permutation counted in (140).

The charge contribution is \[-\kappa(v_C-T\sqrt{v_C})r_* -\bigl[\kappa(v_C-T\sqrt{v_C}) -v_C\beta_C(e^\kappa-1)\bigr]|\mathcal U|.\] Both bracketed coefficients exceed \(100\log B\) for large \(B\): \(v_C=(\eta+o(1))\log B\), \(\kappa\eta=400\), and \(\beta_C(e^\kappa-1)<1\). Every non-skipped good edge therefore pays either its pivot saving or the stronger factor \(B^{-100}\) from its origin in \(\mathcal U\). At most \(N_{\mathrm{exc}}\) good edges were skipped. Absorbing the constants for at most \(\ell\) pivots gives (141). ◻

Proof. First apply [lem:trace-integration,lem:trace-pivots] to valid assigned records. The pivot bound is uniform in the detailed record and may be factored out for each topology.

For the tagged/list groups, unrestricted prime sums cost at most \((1+v_C+v_Z)^{O(N_{\mathrm{exc}})}\). A tagged subtree shape can be summed with its weight using (134), at a cost at most \(\ell+1\); a ghost has at most \(\ell\) possible edges and weight \(2\). Ignoring compatibility between token shapes increases these positive sums. Thus all these shape costs are \((O(\ell+1))^{O(N_{\mathrm{exc}})}\). Together with (139), the lost skipped-edge savings and the cylinder prefactor, their logarithm is \[ O(N_{\mathrm{exc}}\log B+M_0\log B)=o(\ell). \tag{142}\]

For the untagged tokens, retain (140) and sum over unrestricted multiplicities. Their total is at most \[\begin{align*} \prod_{a,Q}\sum_{j\ge0} \frac{(v_a\widetilde u_a(Q))^j}{j!} &=\exp\left\{\sum_a v_a\sum_Q\widetilde u_a(Q)\right\}. \end{align*}\] These are finite products of positive convergent series. They cancel the unmodified background terms \(-v_CS_C(\mathscr T)-v_ZS_Z(\mathscr T)\) in (136). By (135), the remaining exponential is \[\exp\left\{-\frac\theta2v_Z|\mathcal G| +\theta v_Z|E(\mathscr T)\setminus\mathcal G|\right\} \le B^{-(\varepsilon/10)|\mathcal G| +\varepsilon|E(\mathscr T)\setminus\mathcal G|}\] for large \(B\), since \((\theta/2)(\varepsilon-\eta)>\varepsilon/10\) and \(\theta(\varepsilon-\eta)<\varepsilon\).

It remains to check the numerical margin including the topology count. Write \(g=|\mathcal G|\), \(b=|E(\mathscr T)\setminus\mathcal G|\) and \(r=r_*\). Then \(\ell=g+b+r\) and \(b\le10(r+1)\). Set \(\delta=\varepsilon/10-\eta=9\eta\). Combining the preceding bounds, (141) and the count in 57, the negative base-\(B\) exponent is at least \[(1+\delta)g+(99-o(1))r-\varepsilon b-o(\ell).\] Here \(4^\ell\) and the fixed constants per pivot contribute \(O(\ell/\log B)=o(\ell)\), and \((\ell+1)^r=B^{(1+o(1))r}\). Substitution of \(g=\ell-b-r\) and \(b\le10(r+1)\) gives the lower bound \[(1+\delta)\ell +(88-11\delta-10\varepsilon-o(1))r -10(1+\delta+\varepsilon)-o(\ell).\] The coefficient of \(r\) is positive, and \(\delta-2c_*=7\eta>0\). The slack absorbs the fixed last term, the displayed \(o(\ell)\) terms, and the sum over at most \(\ell+1\) possible values of \(r\). This proves the claimed bound. ◻

Completion of the proof of 43. The density assertion is 44. Apply 48 to split the trace sum into retained and discarded configurations. For the retained part, the triangle inequality bounds the absolute full expectation by \(\mathbb E_{\mathcal F}R_{\boldsymbol d} |\mathbb E_{\mathcal S\setminus\mathcal F} K_{\boldsymbol d}\prod_j\mathbf 1_Y(n+b_j)|\). Now apply 56. Their discarded contributions are \(O(\exp(-B^{1+\varepsilon/2}))\). The retained conditional main term is bounded by 62. Since \(\ell=2\lfloor B\rfloor\), \[B^{-(1+2c_*)\ell}+O(\exp(-B^{1+\varepsilon/2})) \le B^{-(1+c_*)\ell}\] for sufficiently large \(B\). This proves the trace assertion and completes the theorem. ◻

Proof. The product \(\chi\overline\psi\) is a Dirichlet character modulo the least common multiple of the two moduli. At every prime outside \(S\), \[G(p)\overline{\chi(p)p^{it}} =f(p)\overline{(\chi\overline\psi)(p)p^{it}}.\] This identity holds also at primes where a character vanishes. Each prime in \(S\) contributes at most \(2/p\) to the difference of squared distances, proving (147). Take the infimum over exactly \(|t|\le X\) on both sides of the resulting lower bound. For each fixed \(\chi\), the character \(\chi\overline\psi\) is fixed, and the bounded error cannot prevent divergence to infinity. For conjugation the exact identity is \[D(\overline f,\chi n^{it};X)^2 =D(f,\overline\chi n^{-it};X)^2.\] The interval \(|t|\le X\) is unchanged by \(t\mapsto-t\). ◻

Proof. Let \(P(a)\) be the set of prime divisors of \(a\). For \(p\in P(a)\) set \(\alpha_p=v_p(a)\) and define \[ A_p(k)= \begin{cases} 0,&0\le k<\alpha_p,\\ f(p^{k-\alpha_p}),&k\ge\alpha_p. \end{cases} \tag{149}\] Factorization into pairwise coprime prime powers gives the exact identity \[ F_{a,f}(m)= \prod_{p\in P(a)}A_p(v_p(m)) \prod_{\substack{p\mid m\\p\notin P(a)}}f(p^{v_p(m)}). \tag{150}\] No identity between \(f(p^k)\) and \(f(p)^k\) is used.

The obstruction to multiplicativity in this product is that \(A_p(0)=0\). Resolve it by setting \[ B_p(0)=C_p(0)=1,\qquad B_p(k)=A_p(k),\quad C_p(k)=0\quad(k\ge1). \tag{151}\] Then \(A_p=B_p-C_p\) at every nonnegative exponent. For \(E\subseteq P(a)\) put \[f_{a,E}(m)= \prod_{p\in E}C_p(v_p(m)) \prod_{p\in P(a)\setminus E}B_p(v_p(m)) \prod_{\substack{p\mid m\\p\notin P(a)}}f(p^{v_p(m)}).\] All local factors have modulus at most one and value one at exponent zero. If \(u\) and \(v\) are coprime, at most one of \(v_p(u),v_p(v)\) is positive at each prime; hence \(f_{a,E}(uv)=f_{a,E}(u)f_{a,E}(v)\). Expanding the finite product in (150) yields \[ F_{a,f}(m)= \sum_{E\subseteq P(a)}(-1)^{|E|}f_{a,E}(m). \tag{152}\] Each summand agrees with \(f\) at prime powers outside \(P(a)\), so 38 proves the last assertion. ◻

Proof. If \(d\nmid m\), both sides vanish. If \(d\mid m\), the congruence on the left is equivalent to \(m/d\equiv c\pmod q\). When \(q>1\), the residue \(c\) is a unit modulo \(q\), so character orthogonality on \((\mathbb Z/q\mathbb Z)^\times\) gives \[\mathbf 1_{m/d\equiv c\pmod q} =\frac1{\varphi(q)}\sum_{\chi\bmod q} \overline{\chi(c)}\chi(m/d).\] For a nonunit \(m/d\) every character on the right vanishes. The case \(q=1\) is the trivial identity with the stated convention. The finite expansion now follows from 64. ◻

Proof. We may assume both functions are nonzero. Choose \(j\in\{1,2\}\) for which \(f_j\) is uniformly nonpretentious. Since \[\mathbf 1_{n\equiv b\pmod l} =\mathbf 1_{n+h_j\equiv b+h_j\pmod l},\] 65, using the least nonnegative representative of \(b+h_j\) modulo \(l\), expands the indicator as a function of \(m=n+h_j\) into finitely many bounded multiplicative functions, each agreeing with a fixed Dirichlet character outside a fixed finite prime set. Their products with \(f_j\) are uniformly nonpretentious by 63. Apply 2 to each product at shift \(h_j\) and the unchanged other factor at its distinct shift, and sum the finitely many conclusions. This proves (154) at integer \(X\). For real \(X\), replacing the normalization \(X\) by \(\lfloor X\rfloor\) changes the average by at most \(1/X\), proving the assertion. ◻

Proof of 3. Zero factors again give an immediate conclusion. Otherwise set \[ \Delta=a_1b_2-a_2b_1,\qquad l=a_1a_2,\qquad c_0=\min(a_2b_1,a_1b_2),\qquad h=|\Delta|\ge1. \tag{155}\] For \(\Delta>0\) put \(U=F_{a_2,f_1}\) and \(V=F_{a_1,f_2}\); for \(\Delta<0\) put \(U=F_{a_1,f_2}\) and \(V=F_{a_2,f_1}\). In either case, \[ U(ln+c_0)V(ln+c_0+h) =f_1(a_1n+b_1)f_2(a_2n+b_2). \tag{156}\] Indeed, when \(\Delta>0\) the two arguments on the left are \(a_2(a_1n+b_1)\) and \(a_1(a_2n+b_2)\); when \(\Delta<0\) their order is reversed.

Expand \(U\) and \(V\) by 64. Each pair of multiplicative components has a uniformly nonpretentious factor, inherited from the corresponding original function. Applying 66 termwise with shifts \(0,h\) gives \[ \frac1X\sum_{1\le m\le X} U(m)V(m+h)\mathbf 1_{m\equiv c_0\pmod l}\longrightarrow0. \tag{157}\] Put \(T(m)=U(m)V(m+h)\mathbf 1_{m\equiv c_0\pmod l}\) and \(X=lN+c_0\). The integers in this class satisfying \(c_0<m\le X\) are precisely \(ln+c_0\), \(1\le n\le N\). Therefore \[\begin{align*} \frac1N\sum_{n=1}^N f_1(a_1n+b_1)f_2(a_2n+b_2) &=\frac XN\left(\frac1X\sum_{1\le m\le X}T(m)\right) -\frac1N\sum_{1\le m\le c_0}T(m). \tag{158}\end{align*}\] The first term tends to zero because \(X/N\to l\) and (157) holds. The second has modulus at most \(c_0/N\) and is empty when \(c_0=0\). This proves the corollary, including zero intercepts and coefficients with common factors. ◻

Proof. For a real \(1\)-bounded multiplicative function \(f\), (Matomäki et al. 2015, Lemma C.1) gives \[ D(f,\chi n^{it};X) \ge\frac14\sqrt{\log\log X}-O_\chi(1) \quad(1\le|t|\le X). \tag{160}\] When \(\chi^2\) is nonprincipal the same bound holds throughout \(|t|\le X\). Thus taking \(f=\lambda\) proves the assertion for nonreal \(\chi\) and handles \(1\le|t|\le X\) for all \(\chi\). For real \(\chi\), the other part of the cited lemma gives \[ D(\lambda,\chi n^{it};X) \ge\frac13D(\lambda,\chi;X)-O(1) \quad(|t|\le1). \tag{161}\] It remains to estimate the distance at \(t=0\) for these real characters.

If \(\chi\) is nonprincipal, the prime number theorem in arithmetic progressions for its fixed modulus yields, for some \(c_\chi>0\), \[A_\chi(u):=\sum_{p\le u}\chi(p)\log p =O_\chi\bigl(u\exp(-c_\chi\sqrt{\log u})\bigr).\] Partial summation gives \[\sum_{p\le X}\frac{\chi(p)}p =\frac{A_\chi(X)}{X\log X} +\int_2^X A_\chi(u) \frac{\log u+1}{u^2(\log u)^2}\,du =O_\chi(1),\] since the integral converges absolutely as \(X\to\infty\). As \(\lambda(p)=-1\), Mertens’ estimate now gives \[D(\lambda,\chi;X)^2 =\sum_{p\le X}\frac{1+\chi(p)}p =\log\log X+O_\chi(1).\] If instead \(\chi\) is principal modulo \(q\), then \[D(\lambda,\chi;X)^2 =2\sum_{p\le X}\frac1p -\sum_{\substack{p\mid q\\p\le X}}\frac1p =2\log\log X+O_q(1).\] In both cases (161) is at least \(\frac13\sqrt{\log\log X}-O_\chi(1)\). Combining this with (160) proves (159) on the entire interval \(|t|\le X\). ◻