[{"type":"thinking","thinking":"User chose: Repair proof only. Proceed with 6-R-1: D-073 §§3–7:\n- §3 Dependency audit (D2-i): every ingredient of Thm 6.1's proof listed and marked T-generic / T-dependent.\n- §4 The corrected stepping lemma (D2-ii): per-step product ratio ≤ q := e^{1/c²}/2 < 1, geometric series from σ = 3, explicit right-tail bound ∫₃^{σ₁}, total horizontal bound |I_II| + |I_IV| with EXPLICIT constants (no untraced O's — use D-071's machinery: kernel charge e^{4/c²}+1, |ζ′/ζ(2)|, the zero-avoiding (log T)² bound).\n- §5 The choice of c (D2-iii): recommended c for the main line, quantified (2 margin / 1.5 range).\n- §6 Consequence audit (D2-iv): Thm 6.1 conclusion + Thm 1.1 height.\n- §7 Cross-validation (D2-v, E-033): numerical cross-validation of all new constants.\n\nThe ζ-specialized setting: Lagarias's Thm 6.1 is for general automorphic L; our use is ζ only (D-036/D-037/R-004 are ζ-specific). The repair should be written for ζ (with the general-π case noted where relevant for the letter). Constants from D-071: kernel charge e^{4/c²}+1, |ζ′/ζ(2)| = 0.5699..., the tight R-bound (10.5 log + 48 — wait, that's for our Lean assembly; Lagarias's framework uses the Lemma 6.1 near-sum + O((log T)²)). \n\nHmm — what level of explicitness? The mission says \"explicit right-tail bound ∫₃^{σ₁}, total horizontal bound |I_II| + |I_IV| with EXPLICIT constants (no untraced O's — use D-071's machinery: kernel charge e^{4/c²}+1, |ζ′/ζ(2)|, the zero-avoiding (log T)² bound)\".\n\nSo the structure of the repaired proof (ζ case):\n- The horizontal (II): σ from σ₀ = −5/2 to σ₁ = 2√n at im = T (T = c√n + ε zero-avoiding).\n- Split: [−5/2, 3] (mid part) + [3, 2√n] (right tail).\n- Mid part: |k_n| ≤ e^{4/c²}+1 (kernel charge at T = c√n: |k_n(s)| ≤ (1 + 4/(16+T²))ⁿ + 1 ≤ e^{4n/T²}+1 = e^{4/c²}+1 — from (6.11)-type bound); |ζ′/ζ| ≤ (the zero-avoiding bound) — Lagarias uses O((log T)²) from Lemma 6.1; explicitly we have the D-056/D-069 machinery: near sum + 10.5 log(T+2) + 48, near-sum integral NI(T) ≤ Z2(T)·2·log(16.5/c_clear) etc. (D-071 §4.4).\n- Right tail: the corrected stepping: at σ ≥ 3, |ζ′/ζ(σ+iT)| ≤ 2^{−(σ−2)}·|ζ′/ζ(2)| ((6.10) with explicit |ζ′/ζ(2)| = 0.569960...) and the kernel ratio |k_n(s+1)+1|/|k_n(s)+1| ≤ (1+1/T²)ⁿ ≤ e^{n/T²} = e^{1/c²} ((6.12)); product ratio per unit step ≤ e^{1/c²}/2 =: q < 1 ⟺ c > 1/√(log 2).\n- So the product |k_n(σ+iT) + 1|·|ζ′/ζ(σ+1+iT)|... wait — need care about which s the kernel and zeta are evaluated at. In Lagarias's stepping, the product at step j (σ = 3+j): |k_n(s)|·|ζ′/ζ(s+1)|? Let me re-read the (6.12) passage: \"the estimate (6.10) gains a factor of 2 going one unit to the right. In comparison |k_n(s+1)+1|/|k_n(s)+1| ≤ e.\" and \"We obtain an upper bound for |k_n(s)L′/L(s,π)| that decreases geometrically\". Hmm — actually in Lagarias the integrand on (II) is k_n(s)·(−L′/L(s+1))? or k_n(s)·(−L′/L(s))? From our Lean work (rectInt'_Jlag), the integrand is Jlag n w = k_n(w−1)·(−ζ′/ζ(w)) with w = s+1 where... hmm no. Let me fix the convention once: integrand f(w) = k_n(w−1)·(−ζ′/ζ(w)), w = σ+iT, σ ∈ [σ₀, σ₁]. At the right tail, σ ≥ 3, |ζ′/ζ(σ+iT)| ≤ 2^{−(σ−2)}|ζ′/ζ(2)| by (6.10) (valid for σ > 2 ✓). Kernel: k_n(w−1), w−1 = (σ−1)+iT.\n\nThe stepping: from σ to σ+1: zeta factor halves (≤ 2^{−(σ−2)}|ζ′/ζ(2)| → 2^{−(σ−1)}|ζ′/ζ(2)|). Kernel: k_n((σ+1)−1+iT)/k_n(σ−1+iT): |k_n(s+1)+1|/|k_n(s)+1| where s = σ−1+iT: ratio = |1+1/(s+1)|ⁿ/|1+1/s|ⁿ ≤ (1 + 1/(|s||s+1|))ⁿ ≤ (1+1/T²)ⁿ ≤ e^{n/T²} = e^{1/c²}.\n\nSo the product P(σ) := |k_n(σ−1+iT)+1|·|ζ′/ζ(σ+iT)| satisfies P(σ+1)/P(σ) ≤ e^{1/c²}/2 = q. At σ = 3: P(3) ≤ (e^{4/c²}+2)·|ζ′/ζ(2)|?? hmm: |k_n(2+iT)+1| ≤ |k_n(2+iT)|+1 ≤ e^{4/c²}+1+1. Wait the kernel charge on [−5/2, 3]: |k_n| ≤ e^{4/c²}+1, so |k_n+1| ≤ e^{4/c²}+2. Hmm, but the stepping bounds the ratio of |k_n+1| — the \"+1\" form is because k_n(s)+1 = (1+1/s)ⁿ exactly. So the natural product is P(σ) = (1+1/(σ−1+iT))ⁿ magnitude times zeta magnitude. And the integrand is k_n(w−1) = (1+1/(w−1))ⁿ − 1 — the −1. For the right tail bound: |k_n(w−1)| ≤ |(1+1/(w−1))ⁿ| + 1. The stepping gives the geometric decay of |(1+1/(w−1))ⁿ|·|ζ′/ζ|; the \"+1\" leftover: ∫₃^{σ₁} 1·|ζ′/ζ(σ+iT)| dσ ≤ |ζ′/ζ(2)|·Σ_{j≥0} 2^{−(1+j)}... = |ζ′/ζ(2)|·(2^{−1}+2^{−2}+...) wait: ∫₃^∞ 2^{−(σ−2)} dσ ·|ζ′/ζ(2)| = (1/(2 log 2))·|ζ′/ζ(2)|. OK.\n\nLet me structure the corrected stepping lemma cleanly:\n\n**Lemma (corrected stepping).** Let T ≥ c√n with c > 1/√(log 2). For σ ≥ 3, define P(σ) = |1+1/(σ−1+iT)|ⁿ · |ζ′/ζ(σ+iT)|·... hmm, actually cleaner: bound the product at integers and use monotonicity within unit intervals? The (6.10) bound is monotone in σ for all real σ > 2. The kernel |1+1/(σ−1+iT)|ⁿ is NOT monotone in σ necessarily, but the per-step ratio bound applies between any σ and σ+1 (the ratio (1+1/T²)ⁿ is uniform in σ!). Indeed: |1+1/(s+1)|/|1+1/s| ≤ 1 + |1/(s+1) − 1/s|·|s/(s+1)|^{-1}... the (6.12) chain: |1+1/(σ+1+iT)|ⁿ/|1+1/(σ+iT)|ⁿ = |1 + (1/(s+1) − 1/s)/(1+1/s)|ⁿ ≤ (1 + 1/(|s+1|·|s|))ⁿ... the printed chain gives ≤ (1+1/T²)ⁿ. This is uniform for ALL σ (any real part, as long as |s|, |s+1| ≥ T in the im part — yes |σ+iT| ≥ T). So the ratio bound holds for every unit step anywhere on the line. And (6.10) holds for every real σ > 2 pointwise: |ζ′/ζ(σ+iT)| ≤ 2^{−(σ−2)}·|ζ′/ζ(2)| — actually careful: (6.10) is |L′/L(s)| ≤ 2^{−(σ−2)}|L′/L(2)| for σ > 2, from the Dirichlet series: |Σ b(m) m^{−s}| ≤ Σ |b(m)| m^{−σ} ≤ 2^{−(σ−2)} Σ|b(m)| m^{−2}. For ζ: b(m) = Λ(m), |ζ′/ζ(2)| = Σ Λ(m)/m² = 0.56996099306... ✓.\n\nSo for ALL σ ≥ 3 (not just integers): P(σ+j)/P(σ) ≤ q^j for integer j, with P(σ) := |1+1/(σ−1+iT)|ⁿ·|ζ′/ζ(σ+iT)|... wait, need the kernel argument shift care: stepping in σ moves both. Let me define Q(σ) = |1+1/(σ−1+iT)|ⁿ · 2^{−(σ−2)} — then Q(σ+1)/Q(σ) = (kernel ratio ≤ e^{1/c²})·(1/2) ≤ q for all σ > 3... and the product |kernel(σ)|·|ζ(σ)| ≤ Q(σ)·|ζ′/ζ(2)|. Hence:\n\n∫₃^{σ₁} |kernel·ζ′/ζ| dσ ≤ |ζ′/ζ(2)|·∫₃^{σ₁} Q(σ) dσ + |ζ′/ζ(2)|·∫₃^{σ₁} 2^{−(σ−2)} dσ\n(the second from the −1 in k_n = (1+1/s)ⁿ − 1).\n\nFor the first: write σ = 3 + j + r, r ∈ [0,1): Q(σ) ≤ Q(3+r)·q^j. ∫₃^{σ₁} Q(σ) dσ ≤ (Σ_{j≥0} q^j)·∫₀¹ Q(3+r) dr = (1/(1−q))·∫₀¹ Q(3+r) dr. And Q(3+r) = |1+1/(2+r+iT)|ⁿ·2^{−(1+r)} ≤ (e^{4/c²}+... hmm — |1+1/(2+r+iT)|ⁿ ≤ (1 + (3+r)/((2+r)²+T²))ⁿ via the (6.11) bound ≤ (1+4/(16+T²))ⁿ ≤ e^{4/c²} for r ≤ 1 (the (6.11) chain: |1+1/s| ≤ 1 + (σ+1)/(σ²+T²); at σ = 2+r ≤ 3, T² ≥ n: (σ+1)/(σ²+T²) ≤ 4/(16+T²)... wait that's the printed −3≤σ≤3 bound with the kernel at s = 2+r+iT, i.e. Re s = 2+r ∈ [2,3] ⊂ [−3,3] ✓ so |1+1/s|ⁿ ≤ (1+4/(16+T²))ⁿ ≤ e^{4n/T²} = e^{4/c²} ✓).\n\nSo ∫₀¹ Q(3+r) dr ≤ e^{4/c²}·∫₀¹ 2^{−(1+r)} dr = e^{4/c²}·(1/(2 log 2) − 1/(4 log 2))... ∫₀¹ 2^{−1−r} dr = ½·∫₀¹ 2^{−r} dr = ½·(1−1/2)/log 2 = 1/(4 log 2). \n\nFirst integral ≤ e^{4/c²}/(4 log 2·(1−q))·|ζ′/ζ(2)|. Second ≤ |ζ′/ζ(2)|/(2 log 2).\n\nTotal right-tail: |I_right| ≤ |ζ′/ζ(2)|·[e^{4/c²}/(4(1−q)log 2) + 1/(2 log 2)].\n\nAt c = 2: q = e^{1/4}/2 = 1.2840/2 = 0.6420; 1−q = 0.358; e^{4/4} = e; first term: e/(4·0.358·0.6931) = e/0.9925 = 2.739; second: 0.7213. Sum = 3.461; ×|ζ′/ζ(2)| = 0.56996 → 1.973. So |I_right| ≤ 1.98 at c = 2. Compare E-032 measured |I_right| = 0.354 at n=10⁵, envelope 0.355 — the bound 1.98 is honest and tight-ish (within 6×). \n\nAt c = 1.5: q = e^{4/9}/2 = 1.5596/2 = 0.7798; 1−q = 0.2202; e^{4/2.25} = e^{1.7778} = 5.922; first: 5.922/(4·0.2202·0.6931) = 5.922/0.6104 = 9.702; +0.7213 = 10.42; ×0.56996 = 5.94. |I_right| ≤ 5.94 at c = 1.5. E-032 measured 1.00 ✓.\n\nNice — clean explicit bounds. But wait — should the repaired proof be ζ-specific or general-π? For the Lagarias letter, the general-π statement is what repairs his Thm 6.1. The general case: |L′/L(2,π)| explicit only under Ramanujan; the kernel charge same; q same. The letter paragraph can note the general form with L′/L(2,π) in place of 0.57. The main-line use is ζ — write the proof for ζ with the general form noted in one line.\n\nAlso the mid part at T = c√n: |k_n| ≤ e^{4/c²}+1 on [−5/2, 3] (kernel charge), and |ζ′/ζ| ≤ zero-avoiding bound: Lagarias's (6.9) O((log T)²) — for explicitness use the D-071 machinery: |ζ′/ζ| ≤ near sum + 10.5 log(T+2) + 48 with the near-sum integral NI(T) ≤ Z2(T)·2·log(16.5/c_clear), c_clear = 1/(9·C0hat(a+2)) the wide-pigeonhole clearance. Hmm — but the wide pigeonhole was designed for T ∈ [√n, √n+1]; at T = c√n the window is [c√n, c√n+1] — carries verbatim (D-072 §6: \"The zero-avoiding T pigeonhole and the wide pigeonhole carry over verbatim (window [c√n, c√n+1])\"). ✓.\n\nSo the mid part: ∫_{−5/2}^{3} |k_n|·|ζ′/ζ| dσ ≤ (e^{4/c²}+1)·[NI + 5.5·(10.5 log(T+2)+48)] — the D-071 §4.4 segment-integral machine. That's the mid horizontal bound, explicit. (I_II total = mid + right.)\n\nThen the horizontal total (both (II) and (IV)): ×2.\n\nThe consequence audit (§6): Thm 6.1's conclusion Sf = λ_n(√n) + O(√n log n): with T = c√n + ε:\n- The bridge |λ_n(T) − λ_n(√n)| = O(log n): zeros in [√n, c√n]: count O(c√n log n)·... each zero contributes ≤ 2 (the |1+1/n|^{n/2} ≤ 2 argument needs |γ|² ≥ ... hmm: for √n ≤ |γ| ≤ c√n: |(ρ−1)/ρ|ⁿ ≤ (1+1/|ρ|²)^{n/2} ≤ (1+1/n)^{n/2}·... |ρ|² ≥ γ² ≥ n → ≤ e^{1/2}·... the printed argument: |(1−β−iγ)/(β²+γ²)|ⁿ ≤ (1+1/n)^{n/2} ≤ 2 — wait that used |γ| ≥ √n ⟹ γ² ≥ n ⟹ (β²+γ²) ≥ n... (1−β)/(β²+γ²) ≤ 1/(β²+γ²)·... the term (1−β−iγ)/(β²+γ²) = 1/ρ̄... whatever: |ρ/(ρ−1)|ⁿ... The printed bound: |((ρ−1)/ρ)ⁿ| = |ρ/... each zero term ≤ 2 for |γ| ∈ [√n, T]. Count: N(c√n) − N(√n) = O((c−1)√n log n). So |λ_n(T) − λ_n(√n)| = O((c−1)√n log n) — wait, that's bigger than the printed O(log n)! The printed claim: \"|λn(√n, π∨) − λn(T, π∨)| = O(log n)\" because |T − √n| < 1 and there are O(log n) zeros per unit interval. With T = c√n, the window is (c−1)√n wide with O((c−1)√n log n) zeros, each ≤ 2 → O((c−1)√n log n). That's the same order as the final O(√n log n) remainder — so Thm 6.1's conclusion Sf(n) = λ_n(√n) + O(√n log n) SURVIVES with a bigger (c-dependent) constant! The theorem statement is unchanged; the constant in the O changes. \n\nHmm wait — actually in our R-004 assembly, we don't go through λ_n(√n) at all: we bridge directly |λ_n(T)| ≤ 4·FAlt(T) (hRH). The D-072 §6 repair costs already account: bridge 4·FAlt(c√n) grows by ≈ c×; I_III length doubles; etc. The consequence audit (D2-iv) needs: (a) Thm 6.1 conclusion form unchanged, constant c-dependent; (b) Thm 1.1 height: Lagarias Thm 1.1 says λ_n > 0 for n ≤ (T₀−1)²... hmm — actually what exactly does Thm 1.1 say? From R-004: N₁ = (T₀−1)² with T₀ = 3×10¹² — the bridge \"if RH holds to height T₀ then λ_n > 0 for n ≤ (T₀−1)²\" — the mechanism: Sf(n) = λ_n(T) + O(√n log n), and λ_n(T) ≤ 2·N(T)·... under RH-to-T₀, and T ≈ √n must satisfy T ≤ T₀ → n ≤ (T₀/...)². With T = c√n: need c√n ≤ T₀ → n ≤ (T₀/c)². So Thm 1.1's height becomes (T₀/c)² — as D-072 §6 states (N₁ → (T₀/c)²). ✓ That's the Thm 1.1 consequence.\n\nAlso S_∞(n) > B(n) margin: the margin crossover — D-072 §6 says the margin survives comfortably with crossover of the same order (n ≳ 10⁴–10⁵). The consequence audit should quantify the new B(n) with c and re-verify the fold — but that's the MAIN LINE's job (the 8-step protocol re-run at the chosen c = ii′-6-a redo). The audit here just lists what changes and confirms viability (E-033 cross-validates the new constants numerically).\n\nThe choice of c (§5): quantify margin vs range:\n- c = 2: q = 0.642, right-tail ≤ 1.98, kernel charge e¹+1 = 3.72, N₁ = 2.25×10²⁴, bridge 4F(2√n), I_III 4√n[...].\n- c = 1.5: q = 0.780, right-tail ≤ 5.94, kernel charge e^{16/9}+1 = 6.92, N₁ = 4×10²⁴.\n- c = 1.25: q = e^{0.64}/2 = 1.8965/2 = 0.948 — margin thin: 1−q = 0.052, right-tail: e^{2.56}/(4·0.052·0.693) = 12.94/0.1441 = 89.8·... ×0.57 = 53.6 — and the stepping's margin q→1 makes the geometric sum fragile. Recommend c = 2 for proof comfort (main line), c = 1.5 if the N₁ range matters for the headline. D-072 said \"c = 2 preferred for margin comfort, c = 1.5 for range\". The mission: \"(iii) recommended c for the main line, quantified (2 margin / 1.5 range)\". So: recommend c = 2 for the Lean main line (max slack in the fold; N₁ = 2.25×10²⁴ still criterion-3), with c = 1.5 documented as the range option (N₁ = 4×10²⁴) at a cost quantified: right-tail 1.98→5.94, kernel charge 3.72→6.92, bridge ×1.5 vs ×2 — actually the bridge cost is SMALLER at c = 1.5 (4·FAlt(1.5√n) < 4·FAlt(2√n)). The trade: bigger N₁ vs tighter fold. The margin at N₁: margin(N₁) = 2.458e26 (unchanged since it depends on T₀ endpoint... wait no — margin(N₁) with N₁ = (T₀/c)²: S_∞(N₁) = ½N₁ log N₁ vs B(N₁) ~ 21c·√N₁ log N₁ = 21c·(T₀/c)·log = 21·T₀·log — the margin RATIO S_∞/B at N₁ scales like √N₁/c·... hmm: S_∞/B ~ [½N₁ log N₁]/[4c√N₁·polylog] ~ √N₁/(8c)·... = T₀/(8c²)·polylog. So smaller c gives MORE asymptotic margin ratio but the absolute margin at N₁: both grow; fine — the E-030-style fold check needs re-running per c. The recommendation: c = 2 because the fold slack is what the Lean proof needs (the margin at N₁ in absolute terms is enormous either way: S_∞(N₁) ~ ½·2.25e24·56 ~ 6e25 vs B ~ 21·2·1.5e12·55·... ~ 3.5e15·... whatever, E-033 quantifies). Hmm wait — that doesn't sound right: margin(N₁) = 2.458e26 at the OLD N₁ = 9e24. At N₁ = 2.25e24: S_∞ ~ ½·2.25e24·log(2.25e24) ≈ 1.125e24·56.1 ≈ 6.3e25; B(N₁) with c=2: ~4·FAlt(T₀) + 4T₀[(2/3)Z2+10.5·56+48] ≈ [4·0.159·T₀ log T₀-ish...]. OK this is E-030/E-033's job to compute exactly. For §5 I'll quantify the c-trade via the per-piece table and let E-033 verify margins at both c values for n = 10⁵ and at the new N₁'s.\n\nE-033 (cross-validation, D2-v): numerical script verifying:\n1. The corrected stepping bound vs the true integral at c = 1.5, 2 (E-032 already gives |I_right| = 1.00/0.354 at n=10⁵ — check ≤ 5.94/1.98 ✓ — E-033 can just verify the constants: |ζ′/ζ(2)| = 0.569960..., q values, the bound values vs E-032 measured).\n2. The mid-part bound (e^{4/c²}+1)·[NI + 5.5·(10.5 log+48)] vs the measured env_mid (E-032: 14.74 at n=10⁵ — the charged bound is ~3.9e4?? hmm that's the OLD charge with e⁴+1 = 55.6. At c = 2: e¹+1 = 3.72 × [NI + 5.5·(10.5·log 635+48)] = 3.72×[NI + 5.5×115.8] = 3.72×[NI+637]; NI at T=633: Z2(633)·2·log(16.5/c_clear): Z2 = 0.6·log 635+0.886·loglog 635+2.7 = 3.87+1.65+2.7 = 8.2; c_clear = 1/(9·C0hat(635)): C0hat = 0.433·log 637 + 0.886·loglog 637 + 2.9 = 2.79+1.65+2.9 = 7.34; c_clear = 0.0151; log(16.5/0.0151) = 7.0; NI = 8.2·2·7.0 = 115. So mid ≤ 3.72×(115+637) = 2798. True env_mid ~14.7 — slack 190×. Fine — honest, provable, and dwarfed by nothing. The fold's horizontal share 7Ĥ+3.3 ~ 7·Ĉ₀(T)²·... ~ 7·54 = 378?? Hmm wait — 7Ĥ(T) where Ĥ ~ Ĉ₀² ~ 54 at T=633 → 378; but the honest charge is 2798+right 2·... the horizontal per-side is charged (e⁴+1)[NI+5.5(...)+3.3] — at c=2: 3.72·(115+637+2) = 2802 vs share 7Ĥ+3.3 ~ 7·Ĉ₀(633)² + 3.3: Ĉ₀(633) = 0.433·6.45+0.886·1.86+2.9 = 7.34 → 7·53.9+3.3 = 380. 2802 > 380!! The horizontal share FAILS at c = 2 with the current Bn form!\n\nHold on. Let me recompute. In D-071's Bn, the horizontal share is 2(e⁴+1)·[7Ĥ(T)+3.3] with T = √n+1 and Ĥ(T) := Ĉ₀(T)²·... what IS Ĥ? From D-071 §2: \"horizontals | 2(e⁴+1)·[NI + 5.5(10.5 log + 48) + 3.3] | 2(e⁴+1)[7Ĥ+3.3] | V3: ≤ 0.40\". And \"the Ĉ₀² term dwarfs everything: NI ~ 1.2·L·log L vs 21Ĉ₀² ~ 3.94·L²\". So 7Ĥ ~ 21·Ĉ₀²·(1/3)... Ĥ(T) is presumably Ĉ₀-based: H(T) from D-069 Finding 2 — \"H(T)'s near sum has a clearance-coverage gap (honest ≈ 7·Ĉ₀² vs baked 3·Ĉ₀²)\". So Ĥ(T) ≈ 3·Ĉ₀(T)² and 7Ĥ ≈ 21·Ĉ₀². At T = 633: Ĉ₀ = 7.34 → 21·53.9 = 1132. The honest mid charge at c=2: NI + 5.5·(10.5·log 635 + 48) = 115 + 637 = 752 ≤ 1132 ✓ (fits, ratio 0.66 — hmm, above E-030 V3's 0.40 but that was at T = √n+1 = 317 where Ĉ₀² is similar but NI is smaller... let me recompute at c=2 honestly: the fit 752 ≤ 1132 ✓ works).\n\nBut NOTE: the kernel factor (e⁴+1) multiplies BOTH sides identically in D-071's fold (the share is 2(e⁴+1)[7Ĥ+3.3] and the charge 2(e⁴+1)[NI+5.5(...)+3.3]). With the repair, the kernel factor changes to (e^{4/c²}+1) on BOTH sides — it cancels. The right-tail 3.3 becomes the explicit right bound (1.98 at c=2 — SMALLER than 3.3 ✓ improves). So the fold fit at c=2: NI(2√n) + 5.5(10.5 log(2√n+2)+48) + 1.98 ≤ 7Ĥ(2√n)+3.3? At n=10⁵, T=633: 752+1.98 = 754 ≤ 1132+3.3 ✓ ratio 0.66. At larger n: NI ~ (2/3-ish)·L·logL·2·... vs 21Ĉ₀² ~ 3.94L² — the L² beats L·logL asymptotically ✓ but the 5.5·10.5·L term is common... the fit should hold on [10⁵, N₁] — E-030 V3-style check at the new c is needed (that's part of the main-line redo; E-033 cross-validates it).\n\nOK — I have all the pieces. Now, the repaired proof §4 needs to be written carefully with exact constants:\n\nConstants:\n- |ζ′/ζ(2)| = Σ Λ(m)/m² = 0.5699609930696... (E-033 certifies).\n- q(c) = e^{1/c²}/2.\n- Kernel charge K(c) = e^{4/c²}+1 on [−5/2, 3] (T² ≥ c²n; (6.11) bound (1+4/(16+T²))ⁿ ≤ e^{4n/T²} = e^{4/c²}).\n- Right-tail bound R(c) = |ζ′/ζ(2)|·[e^{4/c²}/(4(1−q)log 2) + 1/(2 log 2)].\n- Wide-pigeonhole clearance c_clear = 1/(9·C0hat(a+2)), a = ⌊c√n⌋... window [c√n, c√n+1].\n- NI(T) = Z2(T)·2·log(16.5/c_clear) — wait, the asinh bound: ∫_{-5/2}^{3} dx/|w−ρ| ≤ 2 asinh(5.5/δ) ≤ 2 log(16.5/δ)... asinh(x) ≤ log(2x+1)... 2 asinh(5.5/δ) ≤ 2 log(11/δ + 1) ≤ 2 log(16.5/δ) for δ ≤ 1 ✓ (δ = c_clear ≤ 1/(9·2.9) < 0.04 ✓). Hmm, the D-071 form: NI(T) := Z2(T)·2·log(16.5/c). ✓.\n- The mid charge: (e^{4/c²}+1)·[NI(T) + 5.5·(10.5 log(T+2) + 48)].\n- The horizontal total: |I_II| + |I_IV| ≤ 2·{(e^{4/c²}+1)·[NI + 5.5·(10.5 log(T+2)+48)] + R(c)}.\n\nWait — one nuance: the near-sum integral over [−5/2, 3] uses the tight R-bound's near sum; the tight R-bound (10.5 log + 48) covers [−5/2, 3] ✓ (σ ∈ [−5/2, 7/2] in the 5f capstone... actually 5c/5d cover σ ∈ [−5/2, 1/2] and [1/2, 7/2] — combined [−5/2, 7/2] ✓, and the tight version was proved for the strip... The 5f capstone logDeriv_riemannZeta_sub_nearZeros_le_tight: I believe σ range [−5/2, 7/2] or similar — the D-071 §4.4 uses it on [−5/2, 3] ✓).\n\nAlso the mid part's right endpoint: [−5/2, 3] vs the right tail [3, 2√n] — at σ ∈ [3, 7/2] both the R-bound and the Dirichlet stepping apply; we use Dirichlet from 3 (better). ✓ consistent with D-071.\n\nAlso the kernel charge (e^{4/c²}+1) on [−5/2, 3]: the (6.11) derivation needs σ ∈ [−3, 3]: at σ = −5/2 fine. The printed: |1+1/s| ≤ 1+(σ+1)/(σ²+T²) ≤ 1+4/(16+T²) for −3 ≤ σ ≤ 3: at σ = −3: (σ+1) = −2 <0... the bound (σ+1)/(σ²+T²) max on [−3,3]: at σ=3: 4/(9+T²); printed uses 4/(16+T²) — for σ² ≤ 9 < 16... hmm printed says \"for −3 ≤ σ ≤ 3: |k_n(s)| ≤ (1+4/(16+n))ⁿ+1 ≤ e⁴+1\" — with σ²+T² ≥ T² ≥ n and σ+1 ≤ 4: (σ+1)/(σ²+T²) ≤ 4/T² ≤ 4/n?? Then (1+4/n)ⁿ ≤ e⁴. The \"16+n\" form: σ²+T² with σ² ≤ 9... 4/(9+T²) vs 4/(16+T²) — at σ = −3, σ+1 = −2: negative, bound trivial. Anyway (1+4/(16+T²))ⁿ is a valid uniform bound (since (σ+1)/(σ²+T²) ≤ 4/(16+T²) on [−3,3] — check worst case σ = 3: 4/(9+T²) ≤ 4/(16+T²)? No — 4/(9+T²) > 4/(16+T²)!! The printed bound (1+4/(16+n))ⁿ with T² = n: at σ=3 it's (1+4/(9+n))ⁿ which is LARGER than (1+4/(16+n))ⁿ. Hmm — so is the printed e⁴ charge slightly wrong too?? Let me check (6.11): |1+1/s| ≤ 1 + (σ+1)/(σ²+T²). At σ = 3, T² = n: (σ+1)/(σ²+T²) = 4/(9+n). So |k_n| ≤ (1+4/(9+n))ⁿ + 1 ≤ e^{4n/(9+n)}+1 ≤ e^4+1 — the exponent 4n/(9+n) < 4 ✓ so e⁴+1 still bounds it! The intermediate \"(1+4/(16+n))ⁿ\" is wrong-ish for σ=3 (should be 9 not 16 — hmm, 16 = 4² maybe they used σ² ≤ ... no wait: (σ+1)/(σ²+T²): maybe they bounded σ²+T² ≥ ... by (|σ|+... hmm 16+T² with σ² ≤ 9: σ²+T² ≤ 16+T² gives (σ+1)/(σ²+T²) ≥ (σ+1)/(16+T²) — wrong direction. But (σ+1)/(σ²+T²) ≤ 4/(σ²+T²)... ugh. The safe form: max over [−3,3] of (σ+1)/(σ²+T²). Derivative: [(σ²+T²) − (σ+1)2σ]/(...)² = [T² − σ² − 2σ]/(...)² — zero at σ = −1+√(1+T²) ≈ T−1: max value ≈ (T)/( (T−1)² + T² ) ≈ T/(2T²) = 1/(2T) — that's much SMALLER than 4/T² for large T. So the true max of (σ+1)/(σ²+T²) over [−3,3] is ~1/(2T), and (1+1/(2T))ⁿ = (1+1/(2c√n))ⁿ → e^{√n/(2c)} — WAIT that grows! Hmm no: (σ+1)/(σ²+T²) at σ ≈ T−1?? But σ ∈ [−3,3] — the critical point σ = −1+√(1+T²) ≈ T−1 is OUTSIDE [−3,3] for T > 4. On [−3,3] the function is... derivative at σ ∈ [−3,3]: T²−σ²−2σ > 0 ⟺ σ²+2σ < T² ✓ always for T ≥ 4 — so the function is INCREASING on [−3,3], max at σ = 3: 4/(9+T²). So |k_n| ≤ (1+4/(9+T²))ⁿ+1 ≤ e^{4n/(9+T²)}+1. At T² = c²n: e^{4n/(9+c²n)}+1 ≤ e^{4/c²}+1 ✓ (since 4n/(9+c²n) < 4/c²). So the charge e^{4/c²}+1 is VALID (the printed \"(1+4/(16+n))ⁿ\" intermediate form is a harmless looseness at σ < 3 — it under-bounds at σ = 3 exactly but e⁴+1 remains a valid bound — no NEW bug; the e⁴+1 charge stands. Our Lean kFun_shift_norm_le_exp_four_add_one was proved independently ✓).\n\nOK. Also for the letter paragraph (D3): the repair is one line: \"choose T = c√n + ε_n with any fixed c > 1/√log 2; then (6.12) gives ratio e^{1/c²} < 2 and the stepping decreases geometrically as printed; the height in Thm 1.1 becomes (T₀/c)².\" Plus the E-032 note: the T = √n per-side statement appears true (hidden cancellation measured) but unprovable by the stepping.\n\nNow — what about §3 dependency audit. List Thm 6.1's ingredients and T-dependence:\n1. Contour rectangle σ₀ ∈ (−3,−2), σ₁ = 2√n, im = ±T, T = √n+ε — T-generic (any T ≥ something); σ₁ = 2√n unchanged in the repair.\n2. Residue identity (6.2)/(6.7): I_2 = Sf − λ_n(T) + O(1) — T-generic (residue theorem).\n3. Trivial zero contribution O(1) — T-independent.\n4. The symmetry ρ → 1−ρ̄ rewrite (6.5)–(6.7) — T-independent.\n5. |λ_n(√n) − λ_n(T)| = O(log n) — T-DEPENDENT: printed for |T−√n| < 1; at T = c√n becomes O((c−1)√n log n) — still ≤ O(√n log n), absorbed (Thm 6.1 conclusion unchanged, constant c-dependent).\n6. Lemma 6.1 (6.8) — T-generic (general s).\n7. (6.9) |L′/L(s+1)| = O((log T)²) on the zero-avoiding horizontal — T-generic (needs T zero-avoiding; the pigeonhole works in any unit window [c√n, c√n+1] via (2.11)'s O(log n) per unit count — count is O(log T) per unit window at height T, window count 3·O(log T) — pigeonhole clearance O(1/log T) ✓ verbatim).\n8. (6.10) Dirichlet halving — T-generic (pointwise in σ > 2).\n9. (6.11) kernel bound + e⁴+1 charge on [−3,3] — T-DEPENDENT (improves: e⁴+1 → e^{4/c²}+1).\n10. I_I (right vertical) o(1) — T-generic-ish: |k_n| ≤ e^{√n/2}/... printed: |(1−1/s)ⁿ−1| ≤ (1+1/σ₁)ⁿ+1 ≤ e^{√n/2} < 2^{√n}... with σ₁ = 2√n unchanged, and |L′/L| ≤ C₀2^{−2(√n+2)}: the product ≤ 2^{√n}·C₀2^{−2√n−4}·length(2T = 2c√n) → o(1) ✓ verbatim (length grows c× — still o(1)).\n11. (6.12) stepping — **T-DEPENDENT, BROKEN at T = √n**; repaired at T = c√n, c > 1/√log 2.\n12. (6.13) I_III = O(√n log n) — T-generic (length 2T = 2c√n: constant c×).\n13. (6.14) λ_n(T) = O(T log T) under RH — T-generic; the Thm 1.1 height T ≤ T₀ becomes c√n ≤ T₀ ⟹ n ≤ (T₀/c)² — T-DEPENDENT (the range cost).\n14. Zero-avoiding choice σ₀ (trivial zeros, spacing 2) — T-independent.\n\nThat's the audit. §6 consequence audit: Thm 6.1 conclusion (Sf = λ_n(√n) + O(√n log n)) UNCHANGED as a statement, constant now c-dependent (items 5, 12, 13); Thm 1.1 height: (T₀−1)² → (T₀/c)² (item 13); everything else unchanged. Plus our R-004-specific consequences (already mapped in D-072 §6: N₁ → (T₀/c)², bridge ×c, I_III ×c, kernel charge drops, I_I shrinks, margin survives — the 8-step protocol re-run).\n\nE-033 script (§7 cross-validation) — verify:\n1. |ζ′/ζ(2)| = 0.569960993... (certified: Dirichlet partial + tail bound).\n2. q(c) < 1 for c = 1.25, 1.5, 2; ≥ 1 for c = 1 (q(1) = e/2 = 1.359 ✓ the bug).\n3. The corrected right-tail bound R(c) vs E-032's measured |I_right| (n = 10³, 10⁴, 10⁵): R(1.5) = 5.94 ≥ 1.12/1.14/1.00 ✓; R(2) = 1.98 ≥ 0.62/0.31/0.35 ✓.\n4. The kernel charge e^{4/c²}+1 valid: max over σ ∈ [−3,3] of |k_n| at T = c√n — numerical max vs bound (sample).\n5. The horizontal fold fit: NI + 5.5(10.5 log+48) + R(c) ≤ 7Ĥ + 3.3 at c ∈ {1.5, 2} on the n-grid [10⁵, N₁(c)] (log-space grid) — the E-030 V3 analogue at the new c. Hmm — wait: what's Ĥ exactly? D-071: \"NI + 5.5(10.5 log + 48) + 3.3 ≤ 7Ĥ + 3.3\" and \"21Ĉ₀² ~ 3.94·L²\". Let me look at what 7Ĥ is precisely — from D-069 Finding 2: \"H(T)'s near sum has a clearance-coverage gap (honest ≈ 7·Ĉ₀² vs baked 3·Ĉ₀²)\". So H(T) = 3·Ĉ₀(T)² baked; honest 7·Ĉ₀²; and Bn's horizontal share is 7Ĥ = 7·Ĥ where Ĥ = 3·Ĉ₀²?? Then 7Ĥ = 21·Ĉ₀² — the honest near-sum coverage. I need the exact def of Ĥ from LiSkeleton... Actually for E-033 I should pull the exact definitions from the Lean file (Hhat). Let me grep for Ĥ/Hhat in LiSkeleton.lean later when writing E-033.\n\nAlso 6: the margin check: margin(n) > 0 on [10⁵, N₁(c)] for c ∈ {1.5, 2} with the repaired Bn (bridge 4·FAlt(c√n+1), I_III 4c√n[(2/3)Z2+10.5 log+48], horizontals 2(e^{4/c²}+1)[7Ĥ+R(c)], I_I IIMaj — hmm IIMaj changes with c? I_I: kernel ≤ e^{√n/(2c)}+1... wait kFun_shift_norm_le_exp_sqrt_half_add_one at σ₁ = 2√n: |k_n| ≤ e^{√n/2}+1 — that's T-independent (σ₁ = 2√n fixed): |1+1/s|ⁿ at Re s = 2√n: (1+1/(2√n))ⁿ ≤ e^{√n/2} ✓ independent of c. |ζ′/ζ| ≤ 2^{−(2√n−2)}·0.57; length 2T = 2c√n (grows c×) → IIMaj n gets a factor c. Tiny anyway (o(1)). \nAnd triv ≤ 1, bridge 4·FAlt(T), T = c√n+1.\n\nThis is the main-line redo preview — E-033 certifies the fold works at the recommended c BEFORE the main line touches Lean. \n\nThe margin function: margin(n) = S_∞(n) − B(n)·FAlt(n)?? no wait — Bn IS the bound: |Sf n| ≤ Bn FAlt n where the notation is Bn(F, n) — from D-071 §2 the formula Bn(F,n) := 4·F(T) + 4T[...] + ... with F = FAlt T. Hmm, the name \"Bn FAlt n\" in SfBoundProp is the function Bn applied to FAlt... Let me not guess — for E-033 I'll replicate E-030's fold computation (E-030 script exists — the honest-Bn fold validation). Let me look at E-030's script when writing E-033 and clone its Bn/margin computation with the c-parameterization.\n\nOK. Plan: write D-073 §§3–6 now (the analytic content), then E-033 script, run, then §7 + §8 drafts, commit 6-R-1 and 6-R-2. Let me check E-030 first for the fold formulas, and grep the Lean Hhat def.","signature":""},{"type":"text","text":"Direction set: repair proof only. Now building 6-R-1. First, pulling the exact Bn/Ĥ definitions from E-030 and the Lean sources so the repaired constants slot into the existing fold machinery."},{"type":"tool_use","id":"TodoWrite_28","name":"TodoWrite","input":{"todos":[{"content":"Read all mission inputs (D-073 §0, D-072, Lagarias pp. 27–29, D-071, E-031 script)","status":"completed","activeForm":"Reading mission inputs"},{"content":"6-R-0: E-032 diagnostic — DONE, verdict HIDDEN CANCELLATION, committed ded9603","status":"completed","activeForm":"Running E-032 diagnostic"},{"content":"6-R-1: D-073 §§3–6 (dependency audit, corrected stepping lemma explicit constants, choice of c, consequence audit)","status":"in_progress","activeForm":"Writing the repaired proof"},{"content":"6-R-1: E-033 cross-validation script + run + db registration; D-073 §7","status":"pending","activeForm":"Cross-validating constants (E-033)"},{"content":"6-R-2: D-073 §8 drafts (letter paragraph + per-side verdict paragraph); close thread; commit","status":"pending","activeForm":"Drafting letter/finding-note paragraphs"}]},"extra_content":null},{"type":"tool_use","id":"Glob_29","name":"Glob","input":{"pattern":"research/numerics/scripts/E-030*"},"extra_content":null}]