Proof. We distinguish throughout the target perturbation $\delta>0$ tolerated by the finite packet calculation from the source conductor exponent $\epsilon_*>0$ in Pintz’s density theorems. They are not the same parameter.

The relative-conductor hypothesis $M\leq C$ is used to place one class in Zhao’s $M=O(1)$ alternative. In that alternative, Lemma 3.1 uses $$k=2(\phi+3x+y+z),\qquad \phi=\frac13,$$(2.5) and the fixed-class form of Lemma 3.3 is $$(\Delta^2-\varepsilon_z)N+2\Delta D\leq1.$$(2.6) After this alternative has been selected, neither (2.5) and (2.6), nor Zhao’s displayed constant $\mathcal C(x,y,z,\Lambda,\lambda_0)$ contains the numerical value of $M$. The same is true of Zhao’s fixed-class integer consequences, Abel summation, the $R$-decomposition, and the two fixed-class $D$-alternatives. Thus $C$ is absent from every limiting finite inequality.

By hypothesis, the exact certificate consists of finitely many rational inequalities, finitely many continuous operations whose denominators are strictly separated from zero, and tail inequalities which are uniform in $H$. Its common target tolerance $\delta=\delta(A,c)>0$ is such that perturbing all source inequalities by at most $\delta$ changes the final packet ceiling by less than $\kappa_0/2$. This choice precedes $H$ and $C$.

Now fix $H\geq H_c$. At the source of Zhao’s fixed-class estimates, Pintz’s Theorems D and I are applied under $$\operatorname{cond}(\chi\overline{\chi'})\leq Q^{\epsilon_*}.$$(2.7) Pintz’s Theorem D, specifically his (4.3), and Theorem I, specifically his (4.34), enter Zhao’s Lemmas 3.1 and 3.3 with errors of the form $$O_H(\epsilon_*)+o_{H,\epsilon_*}(1) \qquad(Q\longrightarrow\infty);$$(2.8) in particular, the multiplier $1+C_D(H)\epsilon_*$ in Theorem D is height-dependent. Choose $\epsilon_*=\epsilon_*(A,c,H)>0$ sufficiently small that every $O_H(\epsilon_*)$-term, after its finitely many uses in the packet calculation, consumes less than $\delta/2$.

Only now fix $C$. If $$Q\geq C^{1/\epsilon_*},$$(2.9) then $M\leq C$ implies (2.7). Enlarge $Q_0$ until all the $o_{H,\epsilon_*}(1)$-terms in (2.8), together with the fixed-height zero-labeling and Deuring–Heilbronn errors, consume less than $\delta/2$. The resulting finite packet ceiling is at most $1-\kappa_0/2$.

For fixed $A$, changing the exponential coefficient affects only fixed numerical weights and detector parameters. None of the limiting conductor reductions introduces the numerical value of $C$. This proves the asserted order (2.4) and the quantifiers in Definition 4. ◻

Proof. If $L(\rho_j,\chi_j^*)=0$, then $L(\overline{\rho_j},\overline{\chi_j^*})=0$. The involution $$(\rho_j,\chi_j^*)\longmapsto (\overline{\rho_j},\overline{\chi_j^*})$$(2.14) preserves $1-\beta_j$, height, conductor, multiplicity, and the condition $r_j\mid q$. It converts (2.10) bijectively into (2.11). This is the reindexing between Pintz’s (2.13) and (2.36)–(2.37).

Since $q\leq Y$, one has $$q\mid Q,\qquad Y/2<Q\leq Y.$$(2.15) If $\chi^*$ has conductor dividing $Q$, its induction $\chi_Q$ satisfies $$L(s,\chi_Q)=L(s,\chi^*) \prod_{\substack{p\mid Q\\p\nmid\operatorname{cond}\chi^*}} \left(1-\chi^*(p)p^{-s}\right).$$(2.16) The primitive $L$-function inducing $\chi_Q$ is $L(s,\chi^*)$, so the primitive packet zero multiset is unchanged by definition. The additional Euler-factor zeros of the imprimitive function lie on $\Re s=0$ and do not enter that multiset. Moreover, $$\operatorname{cond}\bigl((\chi_Q\overline{\chi'_Q})^*\bigr) = \operatorname{cond}\bigl((\chi^*\overline{\chi'^*})^*\bigr).$$(2.17)

Conductor submultiplicativity along a simple path of at most $K-1$ edges gives (2.13). For $\delta_\ell=1-\beta_\ell$ and $B=1/\theta>A$, (2.15) gives $$X^{-(\delta_i+\delta_j)} \leq Q^{-B(\delta_i+\delta_j)} =\mathrm e^{-B(\lambda_i+\lambda_j)} \leq\mathrm e^{-A(\lambda_i+\lambda_j)}.$$(2.18) Every retained ordered occurrence is therefore bounded by the corresponding ordered occurrence in its component square. The Cartesian-product multiplicity hypothesis ensures that no occurrence is used more often than it appears in that square. Enlarging the selected zeros to the full multisets $\mathcal Z_\nu(U)$ only adds nonnegative terms, which proves (2.12). ◻

Proof. For even $m$, the Euler product in Pintz’s (2.9) gives $\mathfrak S(m)\geq\mathfrak s_0$. The pole–pole term in Pintz’s Theorem A is exactly $\mathfrak S(m)m$. For every selected non-pole–pole singularity pair, Pintz’s beta-integral estimate (2.20) gives, uniformly over the fixed rectangle, $$\left| \frac{\Gamma(\rho_i)\Gamma(\rho_j)} {\Gamma(\rho_i+\rho_j)} \right| m^{-(\delta_i+\delta_j)} \leq (1+\omega_X)X^{-(\delta_i+\delta_j)}.$$(2.23) Indeed, if $s=\delta_i+\delta_j\leq2H_0/\log X$, then $$m^{-s}\leq2^sX^{-s} =\left(1+O_{H_0}\!\left(\frac1{\log X}\right)\right)X^{-s},$$ which is the precise use of $m\geq X/2$.

By hypothesis the retained non-pole terms are exactly the zero–zero terms counted by $\mathcal C_X(m)$. There are at most $(K+1)^2$ non-pole–pole pairs. For every pair failing one of Pintz’s three retention conditions, his (2.12) gives $\lvert\mathfrak S(\chi_i,\chi_j,m)\rvert\leq\eta$; after increasing $X_0$, the beta factor in (2.23) is at most two. Their normalized total is therefore at most the second error term in (2.20). Dividing the remainder in Pintz’s Theorem A by $\mathfrak S(m)m\geq\mathfrak s_0X/2$ gives the final line of (2.20).

Choose $H_0,T_0$ so that the first two explicit-formula terms contribute at most $\kappa/16$ each. The log-free density bound then fixes $K=O_\tau(\mathrm e^{2H_0})$. Choose $\eta$ so that its term is at most $\kappa/16$. Finally enlarge $X_0$ until the $X^{-\tau}$-term is at most $\kappa/16$ and $\omega_X\leq\kappa/4$. Since $$(1+\kappa/4)(1-\kappa)\leq1-\frac{3\kappa}{4},$$ (2.20) yields (2.22). ◻

Proof. It is enough to count exceptions in $X/2<m\leq X$. Fix, independently of $X$, $$0<\tau< \min\left\{\tau_0,\frac{2/5-\theta}{2},\frac{2}{45}\right\},$$(2.27) where $\tau_0$ is the small absolute upper bound in Pintz’s Theorem A. Set Pintz’s preliminary short-interval parameter $\varepsilon_0=2\tau$, and apply the explicit formula with $$\vartheta_{\mathrm P}=\theta+2\tau, \qquad \varepsilon_{\mathrm P}=\tau.$$(2.28) Its hypotheses hold because $$\tau<\vartheta_{\mathrm P}<\frac25<\frac49-\tau,$$ and it supplies $$X^{\theta+\tau}\leq P\leq X^{\theta+2\tau}.$$(2.29) The fact that this one $P$ works simultaneously for every $m\leq X$, rather than being selected separately for each target, is the uniformity assertion in Pintz’s published Theorem 1 (Pintz 2023). Pintz’s minor-arc estimate (2.6) therefore leaves $$O(XP^{-1}\log^{10}X) =o(X^{1-\theta})$$(2.30) targets.

Invoke Pintz’s Theorem B with $\varepsilon'=\tau$. If its exceptional real zero exists, then $$E(X)\ll X^{3/5+\tau}=O(X^{1-\theta}),$$ because $\theta+2\tau<2/5$. We may henceforth work in the complementary alternative. The classical primitive zero-free region (Davenport 2000, Ch. 14), together with the absence of that single possible exceptional real zero, supplies a fixed $h=h(\tau)>0$ such that every relevant bounded-height zero satisfies $$1-\beta\geq\frac h{\log X}.$$(2.31) The constant $h$ is chosen from the absolute zero-free-region constant and Pintz’s Theorem B before any truncation height below is chosen. For every later fixed height $U$, $$\log(r(U+2))\leq(2/5)\log X+O_U(1).$$ Thus taking $U=H_Z$ below affects only $X_0$, not $h$; there is no circular dependence.

Set $c=h\theta/2$, and let $\kappa=\kappa(A,c)>0$ be furnished by $\mathsf Z(A)$, decreased if necessary so that $\kappa\leq1/2$. Choose $H_0,T_0,K,\eta$ in the order (2.21), decreasing $\eta$ further so that $\eta\leq1$. Write $$C_0=\eta^{-3}, \qquad C_{\mathrm{rel}}=C_0^{K-1}.$$(2.32) Here Pintz’s selected labelled-zero multiset is conjugation-stable, and the number of distinct primitive characters is no larger than the number of labelled zeros. Thus the log-free density estimate supplies one $K\geq1$ which bounds both quantities (even if the selected set is empty). These are fixed numbers. If one writes $\eta=\varepsilon_{\mathrm{cut}}/K^2$, then $\eta^{-3}=K^6/\varepsilon_{\mathrm{cut}}^3$; this is the reciprocal forced by Pintz’s (2.13) and (2.24), rather than the opposite parenthetical quantity printed in his (2.34).

Pintz partitions the targets into at most $2^K$ subsets according to which selected primitive conductors divide $C_1(\eta)m$; see his (2.26). For one subset let $q$ be their least common multiple, with the convention $\operatorname{lcm}(\varnothing)=1$, and put $Y=\lfloor X^\theta\rfloor$. For the empty subset the retained zero sum is empty. If $q>Y$, Pintz’s divisibility count (2.29) gives $O_{\eta,K}(X^{1-\theta})$ targets in that subset. We may therefore suppose $q\leq Y$ and form the modulus $Q$ in Lemma 7.

The retained critical sum contains no pole–zero pair once $X$ is large. Indeed, such a pair would force the zero character to have primitive conductor $<C_0$. There are only finitely many such primitive $L$-functions, and none vanishes at $1$; their bounded-height zeros eventually lie outside $\beta\geq1-H_0/\log X$. We may therefore apply Lemma 7 to the retained zero–zero pairs. These pairs form a submultiset of the natural ordered zero product, with the multiplicities required there.

Pintz’s horizontal truncation and (2.15) give $$(1-\beta)\log Q\leq\theta H_0, \qquad |\gamma|\leq T_0.$$ Thus the selected zeros lie in a fixed Zhao rectangle with $$H_Z:=\max\left\{ \theta H_0,\ T_0,\ 5.68,\ 1.09\log\frac1c \right\}.$$(2.33) Moreover, $\log Q\geq(\theta/2)\log X$ for all sufficiently large $X$, and (2.31) gives $$\lambda_\rho=(1-\beta)\log Q\geq c.$$(2.34) The same bound holds for every additional zero of the selected primitive characters in this full Zhao rectangle. In particular, $H_Z\geq H_c$, so the moving cutoff required in the near-Siegel branch is contained in the rectangle before $\mathsf Z(A)$ is invoked.

All parameters in $$Q_0(A,c,H_Z,C_{\mathrm{rel}})$$ are now fixed, and $Q>Y/2\to\infty$. Hence, for sufficiently large $X$, $\mathsf Z(A)$ and (2.12) give $$\mathcal C_X(m) \leq \sum_\nu \left(\sum_{\rho\in\mathcal Z_\nu(H_Z)} \mathrm e^{-A\lambda_\rho}\right)^2 \leq1-\kappa.$$(2.35) Lemma 8 now yields $R_1(m)>0$ outside the minor-arc exceptions and the large-$q$ subsets. Summing their bounds over the at most $2^K$ subsets proves the desired count: indeed, on every remaining target, $$R_1(m)\geq\frac{\kappa}{2}\mathfrak S(m)m \geq\frac{\kappa\mathfrak s_0}{4}X,$$ which dominates Pintz’s minor-arc bound $\lvert R_2(m)\rvert\leq X/\sqrt{\log X}$. Hence $R(m)=R_1(m)+R_2(m)>0$ for sufficiently large $X$. This proves (2.25) on $[X/2,X]$, and dyadic summation proves it up to $X$.

Finally put $\alpha=\min\{2/5,1/A\}$. Given $\varepsilon>0$, choose once and for all a fixed $\theta<\alpha$ with $\alpha-\theta<\varepsilon$. Then $$X^{1-\theta}\leq X^{1-\alpha+\varepsilon},$$ which proves (2.26). No parameter is allowed to vary with $X$ in this endpoint passage. ◻

Proof. For every $\lambda$, $$\mathrm e^{-A\max(\lambda,\Lambda)} = \mathrm e^{-(A-k)\max(\lambda,\Lambda)} \mathrm e^{-k\max(\lambda,\Lambda)} \leq \mathrm e^{-(A-k)\Lambda} \mathrm e^{-k\max(\lambda,\Lambda)}.$$ Sum and apply (3.3). ◻

Proof. The kernel support is $0\leq u\leq2/x$. For $d=\Lambda-\lambda_j\geq0$, Zhao compares $\mathrm e^{ud}-1$ with $\mathrm e^{Ad}-1$. For $0\leq u\leq A$, $$d\longmapsto\frac{\mathrm e^{ud}-1}{\mathrm e^{Ad}-1}$$ is nonincreasing. Thus $2/x\leq A$ supplies the ratio monotonicity, while Zhao’s density inequality independently requires $x\geq4\lambda_0/5$. Every certified detector is chosen strictly above both thresholds, which is also the strict support hypothesis in Pintz’s Theorem K. ◻

Proof. Every certified zero contributes at least $\psi(v_j)-\psi(u)$ because $\psi$ is decreasing. If $N_i(u)\geq B+1$, the left side of (4.4) exceeds one by (4.6) for all sufficiently small $\varepsilon_z>0$, a contradiction. ◻

Proof. If $\mu_i\in[a,b]$, every zero in the class is at least $a$, and the class itself supplies a zero no later than $b$. Thus Lemma 13 applies with the positive charge $\psi(b)-\psi(u)$. Abel summation of the resulting certified integer envelope gives $c_A(\Lambda;a,b)$. If $\mu_i>\Lambda$, the ordinary cap at lower defect $\Lambda$ applies. The cases in (4.10), together with this late case, exhaust every possible $\mu_i$. ◻

Proof. We first prove the bilinear statement $$\sum_i r_it_i\leq\sum_i g_i^Rg_i^T.$$(5.4) Decreasing rearrangement preserves feasibility. Indeed, if the $k$-th largest coordinate of $r$ exceeded $a_k$, then at least $k$ original coordinates would exceed $a_k$, while only the first $k-1$ cap positions can do so. The same holds for $t$. The rearrangement inequality gives $$\sum_i r_it_i\leq\sum_i r_i^\downarrow t_i^\downarrow.$$ For every $k$, $$\sum_{i\leq k}r_i^\downarrow \leq\min\left\{U,\sum_{i\leq k}a_i\right\} =\sum_{i\leq k}g_i^R.$$(5.5) Abel summation against the decreasing nonnegative sequence $t^\downarrow$ gives $$\sum_i r_i^\downarrow t_i^\downarrow \leq\sum_i g_i^Rt_i^\downarrow.$$ Apply the corresponding $b,V$ prefix majorization against the decreasing sequence $g^R$ to obtain (5.4).

Apply (5.4) to two copies of $r$, two copies of $t$, and to $r,t$. Adding the resulting bounds proves (5.3). The greedy pair itself is feasible, so the bound is attained. The countable case follows by truncation, using the finite mass budgets. ◻

Proof. The zero alternatives are precisely Zhao’s Lemma 2.4, including its three unconditional final assertions. If $c>H$, the packet is empty and there is nothing to prove. Otherwise, intersect the displayed closed intervals with $[c,H]$ and discard empty intersections; the resulting rows cover $[c,H]$, with harmless overlaps at their endpoints. In the active interval, (6.5) exhausts the possible placements of the at most two low zeros. In every secondary interval the verifier checks each admissible allocation $(m,k)$, where $m$ is the number of known low zeros and $k$ the number of classes which contain them. The remaining cases are the four scalar rows and the near-Siegel majorant (7.18). ◻