Proof coverage and consumer interfaces

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This is part of the examination report, dated 2026-10-15. Line locators refer to the retained frozen excerpts, not to later versions of their native pages. The three longest independent rederivations are in weak steps. The source methods are OpenAI's; the reconstruction being examined and its uniformity accounting are Plasma AI's.

E1: statement defect and the entire operative proof

The unrestricted statement in block_sieve.md 19–53 has the following counterexample. Choose the prime set empty, $z=1/2$, $\eta_0=1/2$, $C=1$, and one object with weight 1. There are no bad events. The only squarefree product is $d=1$; set $g(1)=1$, $X=0$, and $r(1)=1$. Every required intersection count is $Xg(1)+r(1)=1$. All prime-density and block-sum conditions are vacuous. Avoidance has weight $\mathcal S=1$, and $G=1$. For every positive even $h$, including every possible $h\ge h_0$,

$$c_1e^{-h}XG+\sum_{d\le z^{4h+2}}|r(d)|=0.$$

The displayed estimate would say $1\le0$. Its fixed-$h=2$ upper bound fails for the same reason. The polynomial clause also fails: a polynomial in no indicators is a constant, but its constant monomial has $d_\emptyset=1>z^{4h+1}$, so the stated support condition forces $U=0$, incompatible with $U\ge1$. This is an exact witness, requiring no machine computation.

Neither the supplied source's conventions nor Lemma 2.9 explicitly imposes a lower bound on $z$. If the usual sieve convention $z\ge2$ is intended, the witness is outside that convention. The distinction is material to the literal statement and immaterial to the five applications, whose cutoffs all tend to infinity. The report does not silently add that convention to the frozen statement.

Here is the deduction-by-deduction examination for $z\ge1$, the entire range needed by every consumer. At $z=1$ the prime set is empty, the constant polynomials $U=V=1$ are supported at $d=1$, and the conclusion follows from the $r(1)$ term. For $z>1$:

1. The alternating-binomial identity follows by applying Pascal's relation to adjacent partial sums. If the truncation index exceeds the available number of conditions, the binomial coefficient is zero. At $t=0$ the partial sum is 1, as required.
2. In block $j$, the even truncation $h_j=(j+1)h$ gives $U_j\ge I_j\ge0$. Its error is $\binom{t-1}{h_j}\le\binom t{h_j+1}=E_j$ for $t\ge1$; both sides of the error comparison vanish for $t=0$.
3. Product telescoping is valid with arbitrary dependent bad events. Nonnegativity permits replacement of each trailing $I_k$ by $U_k$, giving $0\le U-I\le\sum_jE_j\prod_{k\ne j}U_k$. Thus $V=U-\sum_jE_j\prod_{k\ne j}U_k\le I\le U$.
4. There is at most one way to obtain a given monomial in $U$. A correction monomial has degree exactly $h_j+1$ in one block and degree at most $h_k$ in all others. Correction supports are mutually disjoint and disjoint from $U$'s support. Hence $U,V,U-V$ all have coefficients in $\{-1,0,1\}$. Pointwise nonnegativity of $U-V$ follows from its expression as a sum of products of nonnegative functions, not from the signs of its expanded coefficients.
5. Since $\sum_{j\ge0}(j+1)2^{-j}=4$, a monomial in $U$ has product at most $z^{4h}$. A correction adds at most one prime of size $z$, so $z^{4h+1}$ suffices. The stated remainder cutoff $z^{4h+2}$ is therefore valid for $z\ge1$.
6. Independent Bernoulli events with probabilities $g(p)$ are introduced only to evaluate a polynomial's prescribed main term. This does not assume independence of the actual objects. Their block avoidance probability $a_j$ is at least $\exp(-C/\eta_0)$, and $e_j\le C^{h_j+1}/(h_j+1)!$.
7. For $n\ge e^2C$, $C^n/n!\le e^{-n}$. Consequently $R_h=\sum e_j/a_j\le2e^{C/\eta_0-1}e^{-h}$ for all sufficiently large even $h$, irrespective of the number of nonempty blocks. Increase $h_0$ until $R_h\le1$. The upper expectation differs from $G$ by at most $(e^{R_h}-1)G\le eR_hG$. Since $V\le I$, its deficit from $G$ is at most $\mathbb E(U-V)\le R_he^{R_h}G$. These prove the advertised $c_1=2e^{C/\eta_0}$ bound.
8. At $h=2$ the relevant factorial indices are $3,5,\ldots$. Their sum is bounded by $\sinh C$, so $R_2\le e^{C/\eta_0}\sinh C$. The stated upper constant $c_2=\exp(e^{C/\eta_0}\sinh C)$ follows. It is large, but finite and independent of every later extraction parameter.
9. Evaluation in the actual weighted population gives $X\mathbb EP+\sum c_Sr(d_S)$ for either polynomial. Each integer product identifies its squarefree set uniquely, and $|c_S|\le1$ gives the exact unit-coefficient remainder sum. This also covers $X=0$; no division by $X$ occurs.

The $h$-uniform bound has therefore been checked for arbitrarily many blocks and growing $h$, not inferred from a finite trial.

All five sieve instances

For $g(p)=1/(p-1)$ on odd primes, the block-mass bound follows from that for $1/p$, plus the convergent sum of $1/[p(p-1)]$; its constants are absolute.

E2, E3, and E4: remaining clauses and source scope

The three independent derivations supply the main analytic details. The following records clauses which might otherwise be hidden by a successful final substitution.

E5: every derived Section 2 tool

The harmonic character estimate is proved in the first keeper. It uses the stated Siegel–Walfisz theorem, without a primitive-character restriction. The block sieve is examined above.

For fixed $k\ge0$, the tail $\sum_{a\ge2}(a+1)^kp^{-a}$ is at most $c_kp^{-2}$ uniformly for primes $p\ge2$, by comparison with the convergent series at $p=2$. Positivity and unique factorization therefore give, for $Y\ge1$,

$$\sum_{n\le Y}\frac{\tau(n)^k}{n} \le\prod_{p\le Y}\left(1+\frac{2^k}{p}+\frac{c_k}{p^2}\right) \ll_k(\log(2Y))^{2^k}.$$

Multiplication by $Y$ bounds the ordinary sum. This works for real $k\ge0$, so includes the integer-$k$ source lemma. For the mixed shifted sums, Cauchy–Schwarz uses the second moments of $\tau(u),\tau(2u-1)$, and the second and sixth moments in $\tau(v)\tau(v+1)^3$. Each shifted variable ranges in a set of distinct integers of size $O(x)$; extending to all integers up to that size is a positive upper bound.

A convolution of $b$ fixed factors of logarithmically bounded coefficients is bounded pointwise by $L^C\tau(n)^{b-1}$: choose the first $b-1$ divisors, which determine the last factor. Its square sum on a dyad is $UL^{O(1)}$, and a reciprocal product index changes this to $L^{O(1)}/U$. This checks the source Lemma 2.5 consequence as well as the explicitly consumed bounds. In the Type II endpoint use, $\tau(z_*)^4\le\tau(z)^4$; the varying group-free part cannot defeat the ordinary moment bound.

For Fourier separation, choose the coordinate with largest frequency and integrate by parts $j$ times in it. Along with the zero-frequency bound this gives $|\widehat F(\xi)|\ll_j L^{C(j+1)}(1+|\xi|)^{-j}$. With $j=b+1$, integration gives mass $O(L^{C(b+2)})$. Outside a ball of radius $L^{2C}$, the $j$-derivative estimate gives tail

$$O_j(L^{-Cj+C(2b+1)}).$$

Choose $j\ge2b+1+A/C$. Thus the frequency exponent $2C$ precedes the desired accuracy $A$. A coordinate box truncation has no larger outside region than a comparable ball, so the source's coordinatewise statement follows with fixed constants. Fourier inversion separates the variables. Mellin inversion for fixed $\Psi$ follows by applying this to the smooth compactly supported function $\Psi\circ\exp$; its integrated absolute cost is constant. Finite sums permit all these integral exchanges.

The earlier Type II slot split has $N_s=O(T)$ factors, so it must not be treated as an application of the fixed-dimensional lemma with a growing dimension. Instead separate each fixed one-variable transition function individually. If $C_F$ bounds their Fourier masses, the sum over the $2^{N_s}$ patterns has mass at most $(2C_F)^{N_s}=L^{O(1)}$. The complement of the frequency box $|\xi_i|\le L$ is a union of $N_s$ one-coordinate tails. For every fixed $M$, its mass is at most $2^{N_s}N_sC_MC_F^{N_s-1}L^{-M}$, a fixed logarithmic power times $L^{-M}$. Selecting $M$ after that power gives any required saving. Also $\sum_i1/s_{j(i)}\ll1/T$, by summing the geometric scales up to the final scale $\asymp T$. The retained slot twists are therefore $O(L/T)$. The mixed divisor bound already proved controls the total absolute error before the target accuracy is selected. Thus the growing number of slots introduces no unproved dimension-uniform version of Fourier separation. The later three-variable cutoff does satisfy the fixed-dimensional hypothesis directly.

The squarefree reciprocal-$\varphi$ sum satisfies

$$\sum_{d\le y}\frac{\mu^2(d)}{\varphi(d)} \le\prod_{p\le y}\left(1+\frac1{p-1}\right) =V(y)^{-1}\ll\log(2y) \quad(y\ge1).$$

For the balanced remainders, $3^{\omega(n)}\le d_3(n)$ follows at each prime power from $\binom{a+2}{2}\ge3$ for $a\ge1$. Counting triples and summing two harmonic series yields $\sum_{n\le D}3^{\omega(n)}\le D(1+\log D)^2$. If $\nu(d)\le d3^{\omega(d)}$, then $\sum_{d\le D}\nu(d)/d\le D(1+\log D)^2$ and $\sum_{d\le D}\nu(d)\le D^2(1+\log D)^2$. The reconstruction's weaker common formulation of these two bounds is therefore valid.

Source Lemmas 2.6 and 2.7 and Corollary 2.10 are present in the delivered source excerpt but not consumed by E or by the supplied conditional Type II derivation. Their validity is not used here as a way of certifying P. Basic Fourier decay, character orthogonality, CRT, and elementary integral manipulations are the explicit boundaries of the derived tools.

E5: the Type I application, including all conductor losses

Fix $\vartheta<1/2$ and $D>0$. The interesting range is $\vartheta>0$; otherwise the positive-integer modulus sum is empty or contains only $d=1$, whose remainder is zero. Extract the first pure slot of band 0. Its leftover coefficient $b(r)$ is nonnegative and at most $L^{c_b}\tau(r)$: the same exceptional-divisor and band-permutation argument used for $a_Q$ applies after removing one labelled slot. The normalization in band 0 remains $1/r_0!$, so removing that slot cannot increase the permutation bound. Its prime is group-free and $W_1(pr)=W_1(r)$.

Contributing dyads have $PU\asymp x$ and $x^{a_0}\ll P,U\ll x^{1-a_0}$, $a_0=c_1s'>0$. Indeed $p\in[x^{c_1s'},x^{c_2s'}]$, $c_2s'<1/4$, and $r\asymp x/p$. Fix $c'$ bounding both the first and second moments of $b$ before selecting any accuracy.

For odd squarefree $d$, the progression $2u-1\equiv0\bmod d$ is a unit class. A nonprincipal character has primitive conductor $f>1$, with $d=hf$, $(h,f)=1$, and $\varphi(d)=\varphi(h)\varphi(f)$. Its values are those of the primitive character together with $(p,h)=(r,h)=1$. This decomposition is unique. It requires retaining squarefreeness of $h$ when estimating its harmonic mass.

For $R/2<f\le R$, the inequality $1/\varphi(f)\le(2/R)f/\varphi(f)$, Cauchy–Schwarz, and the multiplicative large sieve give, for every Mellin twist,

$$\sum_{R/2<f\le R}\frac1{\varphi(f)} \sum_{\chi\bmod f}^{*}|A_\chi B_\chi| \ll\frac{L^{c'/2}}R \sqrt{(P+1+R^2)(U+1+R^2)PU}.$$

The prime square sum is $O(P)$, without needing a prime-density gain, and the other is $UL^{c'}$. The coprimality masks only reduce these square sums. Expanding the square-root bound and using $PU\asymp x$ yields

$$xL^{c'/2} \left(R^{-1}+P^{-1/2}+U^{-1/2}+R/\sqrt x\right).$$

One may drop squarefreeness of $f$ and $(f,h)=1$, since this is a positive upper bound. Summing at most $O(L)$ conductor dyads, $O(L^2)$ factor dyads, and squarefree $h$ with total weight $O(L)$, then integrating the fixed Mellin transform, gives

$$O\left(xL^{c'/2+4} (L^{-C'}+x^{-a_0/2}+x^{\vartheta-1/2})\right).$$

Choose $C'\ge D+c'/2+4$. The latter two terms save every fixed logarithmic power for the fixed positive gaps involved.

For $f\le L^{C'}$, apply Siegel–Walfisz to the prime sum before Mellin separation. The logarithms of its endpoints are comparable to $L$. Modulus exponent $2C'+1$ and accuracy $K+C'+2$ give a nonprincipal prime sum $O_K(yL^{-K})$ after the loss from summing the $O(L^{C'})$ residue classes. The cutoff $\Psi(yr/x)$, including its band and dyad endpoints, has bounded total variation, uniformly on a contributing dyad. Partial summation therefore gives $O_K(PL^{-K})$ per $r$.

The primes in this interval dividing $h\le x^\vartheta$ are at most $\vartheta/a_0+1$. Removing them costs $O(x^{1-a_0}L^{c'})$ per dyad before summing moduli. The number of characters for each $f$ is at most $\varphi(f)$, cancelled by its reciprocal weight. The sums over $f,h$, and dyads give the complete bound

$$O\left(xL^{c'+C'+3-K} +x^{1-a_0}L^{c'+C'+3}\right).$$

Choosing $K\ge D+c'+C'+3$ finishes the nonprincipal part. A factor 2 in the conductor cutoff, as in the consumer's dyadic notation, changes only a constant and not the choice order.

The principal term has a different, positive error. Every prime dividing a supported $u\le2x$ is at least $L^{20}$, and their number is $O(L/\log L)$. Breaking a squarefree modulus sharing such a prime with $u$ into $ld'$, and overcounting if it shares several primes, gives

$$\sum_{\substack{d\le x^\vartheta\\(d,u)>1}} \frac{\mu^2(d)}{\varphi(d)} \le\left(\sum_{l\mid u}\frac1{l-1}\right) \sum_{d'\le x}\frac{\mu^2(d')}{\varphi(d')} \ll L^{-18}.$$

Multiplication by the nonnegative candidate weights proves the $X_AL^{-18}$ error. Together these computations establish the stated Type I bound with the order $c'$, then $D$, then $C'$, then $K$. No proxy precision is involved. Constants or eventual thresholds may depend on fixed $\vartheta$; no uniform threshold as $\vartheta\uparrow1/2$ is needed or claimed.

Balanced-box application and the full choice order

This check concerns the actual consumers of E1, E3, E4, and E5. It does not use or verify the truth of P.

The assignment retaining every micro-band entry and one mark from each P-group has $D_1\le x^{1/20}$. The Q contribution is at most $x^{1/25}$; the $O(T)$ marks contribute $\exp(O(TL^{2/5}))=x^{o(1)}$. The harmonic sum of each normalized mark is 1, so $\sum\lambda/D_1=H_Q^{\rm mic}$ and $\sum\lambda\le x^{1/20}H_Q^{\rm mic}$. The exceptional integer is retained whole; its factors are not mistakenly treated as belonging to prime bands.

Expanding the marked weight gives exactly one mark from each group. Independent permissions with probability $q$ for the remaining P-primes reproduce $q^{\omega(u)-s}$, because $\omega$ counts distinct primes. The normalized number of omitted pure-band lists for a fixed residual product is $\prod1/a_p!\le1$. Dropping those restrictions gives the stated pointwise majorant. This is an exact expectation of a nonnegative majorant, not an averaging heuristic about primes.

For $D_1\mid u$ and $mn=2u-1$, one has $mn\equiv-1\bmod D_1$. Even when $D_1$ has prime powers, there are exactly $\varphi(D_1)$ pairs of unit residue classes. At an odd sieve prime $l\nmid D_1$, $mn=0$ has $2l-1$ pairs and $mn=-1$ has $l-1$, and these sets are disjoint. CRT thus gives $\varphi(D_1)\nu(d)$ pairs modulo $D_1d$. The count per pair of residue classes has error $O(1+(M_1+M_2)/(D_1d))$. Using the E5 divisor sums yields total error

$$O\left(((M_1+M_2)D+D_1D^2)(1+\log D)^2\right), \qquad D=x^{10\theta}\le x^{1/20}.$$

Dropping the parity of the actual odd $m,n$ only enlarges the box count. The omission of the prime 2 changes Euler-product comparisons by a fixed factor.

For a banned prime below $W$, the avoidance factor is $(1-1/l)(1-2/l)$; above $W$ it is $1-1/l+1/l^2$. Compare these with $(1-1/l)^{1+2\mathbf1_{l\le W}}$, paying a convergent $1+O(l^{-2})$ factor where needed. An allowed P-prime has factor $(1-1/l)^2$ and needs one compensating inverse factor. Primes removed because they divide $D_1$ need at most three. Their compensation is bounded, since each is at least $L^{20}$ and $D_1\le x^{1/20}$. The mask expectation of the remaining compensation is at most $\prod_{p\text{ P-prime}}(1+q/(p-1))\le H_P$.

Summing main terms gives $O(xV(W)^2V(x^\theta)H_Q^{\rm mic}H_P)$. E4, with constants independent of $j_{**}$, and $V(x^\theta)\ll_\theta1/L$, make this $O(X_AV(W)^2)$. Balanced dyads satisfy $M_i\ll x^{0.54}$, uniformly for $0<\kappa<1/50$. Summing their lattice errors over assignments gives at most $x^{0.64}L^{O(1)}H_Q^{\rm mic}$. Compared to $X_AV(W)^2\gg xH_QH_P/L^2$, this tends to zero using $H_Q^{\rm mic}\ll H_Q$ and $H_P\ge1$.

There are $O(\kappa L+1)$ possible dyads for one balanced factor, and $O(1)$ for the other once its partner and product scale are fixed. The two prime proxies contribute at most $9/(L^2V(W)^2)$. Hence the balanced removal is $C_{\rm bal}(\kappa+L^{-1})X_A/L+o(X_A/L)$, with a constant depending only on the early geometry, $j',\theta,\Psi$, and the E4 comparison constants. No later $\kappa,b_1,j_{**},S$ enters that constant.

The full order is therefore consistent: early geometry and prefix; $j',\theta,j_0$; then $\kappa$ using this constant; then sufficiently small $b_1$, $b_*=b_1/3$, fixed $j_{**}$, and a fixed mesh; then Type II saving and its internal $K_0,D_0,B$; then $S>2B+3$ and PNT accuracies; then $x$. The Type I accuracy $D>C_A+10$, its conductor split, and its SW accuracy can be fixed after the early exponent $C_A$ is known and before taking $x$ large. They do not constrain $S$.

The conditional prime mass and conversion to a distinct-prime count use positivity, $X_A\gg xL^{-C_A}$, the pointwise $\exp(C_w\sqrt L)$ bound, and injectivity of $u\mapsto2u-1$. The prime-factor bound transfers to $p+1=2u$ for large $x$. The resulting lower bound exceeds $x^{1-\eta}$ for any fixed $\eta>0$, since $\eta L$ dominates both $\log L$ and $\sqrt L$. These are valid consequences if the stated Type II hypothesis P and the consumer derivations hold. They are not unconditional consequences of examining E.