[{"type":"thinking","thinking":"Let me re-establish where I am and what's next.\n\n**Current state:** ii′-6-a2 committed (0dfbd6e). I started ii′-6-c (the four side bounds of the L0 assembly). While checking signatures, I discovered **Finding F-E**: the I_I right-vertical kernel bound `e^{√n/2}+1` is off by one — at `w = 2√n + it` with `t = 0`, the true kernel is `(2√n/(2√n−1))ⁿ ≈ e^{√n/2+1/8} > e^{√n/2}`, so the charged `IIMaj` (constant 0.57) is genuinely exceeded. The repair plan: fatten IIMaj's constant 0.57 → 0.8 (since `n/(2√n−1) ≤ √n/2 + 1/3` for `n ≥ 4`, and `0.57·e^{1/3} ≤ 0.57·1.4 = 0.798 ≤ 0.8`).\n\nI was checking how load-bearing the literal `0.57`/`4.56` is in `iimaj_bracket_bound` — and I just read the proof. Critical observation from what I just read:\n\n**`iimaj_bracket_bound` is NOT a ratio statement.** Its statement (LiSkeleton:1511–1517) is literally:\n```\n4.56 * (2 * (-iimA eA - iimB eB) + (s+1)(iimA² eA + iimB² eB)) -\n  (4.56 * ((eA + eB) + (s+1)(-iimA eA - iimB eB))) / s < 4/3\n```\nThe 4.56 (= 2·2·0.57·4) appears as a LINEAR factor in the statement itself, and the certified product `4.56 × 1.2237 × 0.2124 × 1.1099 = 1.3155 < 4/3` has 4.56 as one of the four factors. So the bracket bound is ψ̃″ − ψ̃′/s < 4/3 where ψ̃ is the FULL majorant including the 0.57 constant — NOT a ratio.\n\nHmm wait, but that's ψ̃″(s) − ψ̃′(s)/s < 4/3 where ψ̃(s) = 4.56·(s+1)·(e^{−as} + e^{−bs}). So the statement is \"the psiTilde bracket (second derivative minus first derivative over s) is < 4/3\" — an ABSOLUTE bound, tiny numbers (~1.3155 at s = √20, decaying). The bound 4/3 is absolute.\n\nIf I change IIMaj's constant 0.57 → 0.8, then 4.56 = 2·0.57·4 → 2·0.8·4 = 6.4, and the certified product becomes 6.4 × 1.2237 × 0.2124 × 1.1099 = 1.8470 > 4/3 = 1.3333! **BREAKS.**\n\nSo the 4/3-bracket does NOT survive fattening the constant to 0.8. The margin: 1.3155 vs 4/3: slack factor 1.0127. So the constant can only grow by ≤ 1.2%: 0.57 → ≤ 0.5772. But the I_I repair needs ×1.404 (0.8). DEAD END for option (D′′).\n\nHmm. So fattening IIMaj's constant breaks the convexity bracket. Options reconsidered:\n\n**Option (B): charge Bn's I_I term as `e^{1/3}·IIMaj n` (a new explicit factor in Bn, NOT touching IIMaj's definition).**\n\nThen:\n- I_I ≤ e^{1/3}·IIMaj: kernel ≤ e^{√n/2+1/3}+1 ≤ e^{1/3}(e^{√n/2}+1), length 2T ≤ 2(√n+1), zeta 0.57·2^{−(2√n−2)} ⟹ I_I ≤ 2(√n+1)·e^{1/3}·(e^{√n/2}+1)·0.57·2^{−(2√n−2)} = e^{1/3}·IIMaj n ✓.\n- Bn's I_I line: `Real.exp (1/3) * IIMaj n` instead of `IIMaj n`.\n- DogLegRight: UNCHANGED (IIMaj/4 — V_right's true bound uses DL-4's kernel at t² ≥ n which is VALID, and the 0.57-constant IIMaj accounting stays exact) ✓✓ no LiDogLeg changes!\n- iimaj_bracket_bound/lt_three: UNCHANGED (IIMaj untouched) ✓.\n- Base points: the IIMaj term in Bn is now e^{1/3}·IIMaj — the base-point proofs bound this piece: old hII : IIMaj 80000 ≤ 2590.08; new need e^{1/3}·IIMaj ≤ 1.4·2590.08 = 3626.2. The Bn upper bounds shift: Bn(8e4) ≤ 332000 + 0.4·2590.08·... wait: the I_I term increase: (e^{1/3}−1)·IIMaj ≤ 0.4·2590.08 = 1036. Bn(8e4) ≤ 333036; margin(8e4) ≥ 4001 − 1036 = 2965 > 0 ✓. Bn(10⁵): increase 0.4·2900.16 = 1160: ≤ 353160; margin(10⁵) ≥ 80501 − 1160 = 79341 > 66001 ✓ hbase2 survives.\n- **Convexity (the crux)**: the marginTilde IIM-coefficient becomes −(e^{1/3} + 1/2)·IIMaj (I_I's e^{1/3} + DogLegRight's 2×1/4 = 1/2) = −(e^{1/3}+0.5) ≈ −1.8956·IIMaj. The row needs: coef·IIMaj″(n) ≤ coef·(1/(3n)) vs the budget 1/(2n). 1.8956/3 = 0.6319 > 0.5 ✗ **BREAKS the exact bridge** (1.5·(1/(3n)) = 1/(2n) saturated exactly).\n\nDamn. The 1/(2n) budget is exactly saturated by 1.5. Any increase of the total IIMaj coefficient beyond 1.5 breaks it — UNLESS the budget itself has more room. Where did 1/(2n) come from? From the convexity proof's structure: the total budget for ALL Bn-blocks' second derivatives is bounded by the sInftyLower's (1/2)n·log n second derivative contribution ~ (1/2)·log n + ..., and the ψ-table distributes it. The IIM row's allocation was 1/(2n). Is there slack in other rows? From the summary: \"vId/vIIM literals updated; hnn4 new coefficient with Real.exp_pos (5/2) nlinarith; final linarith with hIIM (lt_three) + hbridge : (1.5)·(1/(3n)) = 1/(2n)\". \n\nThe dog-leg addition vId row was just added in a2 — the DogLegRight block (√n+1)(e^{5/2}+1)/2 + 16.76 contributes to mId, and IIMaj/2 to mIIM. The final linarith: the total must be ≥ 0. The sInftyLower gives (1/2)·log n + c — GROWING, while all Bn″-blocks decay (1/(4n)·stuff, (log T)²-stuff/n^{3/2}...). So at the BASE point n = 8×10⁴ the inequality is tightest?? Actually the proof is ψ-functional: margiñ″(n) = ψ[F](T)/(4n) + ... hmm, chain_rule_sqrt_shift: margiñ(n) = φ(T(n)) with T(n) = √n+1, φ″(T) computed via the ψ-table, then margiñ″(n) = φ″(T)/(4n) + φ′(T)·(−1/(4n^{3/2}))-ish. The blocks: each Bn-block b(T) contributes −[b″(T)/(4n) − b′(T)/(4n(T−1))] = −ψ[b](T)/(4n). The sInftyLower's leading (1/2)n log n contributes (1/2)(log n + 1) − ... direct. \n\nThe IIM row: IIMaj is a function of n directly (not through T(n)... wait — IIMaj n = 2(√n+1)(e^{√n/2}+1)·0.57·2^{−(2√n−2)} — it IS a function of √n: IIMaj(n) = ψ̃(√n)·... the psiTilde machine: ψ̃(s) = 4.56(s+1)(e^{−as}+e^{−bs}) with s = √n. And the bracket ψ̃″(s) − ψ̃′(s)/s < 4/3 relates to the n-second-derivative via the chain rule: d²/dn² [ψ̃(√n)] = ψ̃″(√n)/(4n) − ψ̃′(√n)/(4n^{3/2}) = [ψ̃″(s) − ψ̃′(s)/s]/(4n) < (4/3)/(4n) = 1/(3n). THAT's where 1/(3n) comes from — the ABSOLUTE bound (4/3)/(4n) using the bracket < 4/3 as an absolute ψ̃-bracket.\n\nHmm interesting — so lt_three: deriv(deriv (IIMaj)) n < 1/(3n) uses ψ̃″−ψ̃′/s < 4/3 ABSOLUTE (≈1.3155·10^{−100}-ish actual value? no wait — the bracket ψ̃″−ψ̃′/s at s=√20 ≈ 1.3155, and at s = √10⁵ ≈ 316 it's ~e^{−a·316}·poly ~ 10^{-121}). The 4/3 is an absolute uniform bound on the bracket.\n\nOK so the budget: margiñ″(n) = [positive stuff from sInftyLower] − Σ_blocks [block-brackets]/(4n)... and the IIM row contributes coef·[ψ̃″−ψ̃′/s]/(4n) ≤ coef·(4/3)/(4n) = coef/(3n). With coef = 1.5: 1.5/(3n) = 1/(2n). For this to fit, the positive side must cover Σ all negative rows — and the IIM row's allocated share was 1/(2n). \n\nCan the IIM row's allocation grow to 1.8956/(3n) = 0.6319/n? That requires finding an extra 0.1319/n of budget somewhere. Without reading the full convexity proof, I can't tell if other rows have slack. From the a2 summary: \"hnn4 new coefficient with Real.exp_pos (5/2) nlinarith; final linarith with hIIM (lt_three) + hbridge\" — the final linarith presumably sums: positive sInftyLower terms ≥ Σ |negative rows| with each row bounded by its literal. If any row was estimated with slack, the total has slack.\n\nHmm. This is getting deep. Let me step back and think about what's cleanest.\n\n**Option (C): split I_I and absorb the overrun into the horizontal slack (cross-subsidy).** The pinch inequality currently: NI + 5.5(10.5 log+48) ≤ 7Ĥ with ratio ≤ 0.40 measured (E-030 V3: ≤ 0.40·(7Ĥ+3.3) — huge slack). The I_I overrun: (e^{1/3}−1)·IIMaj ≈ 0.3956·IIMaj — at n = 10⁵: IIMaj(10⁵) ~ 10^{-120}·... vs 7Ĥ(T) ~ 21·Ĉ₀² ~ 21·(0.433·5.77+...)² ~ 21·16 ~ 340. The overrun is 10^{-118} vs slack ~200 — absurdly overkill. But formalizing cross-subsidy means the pinch lemma becomes: NI + 5.5(...) + (e^{1/3}−1)·IIMaj/(2(e⁴+1)) ≤ 7Ĥ — the extra term is so tiny that the pinch proof (a log/loglog comparison with certified endpoints) would absorb it trivially... but it COUPLES two unrelated pieces and complicates the pinch statement. Ugly but Bn-convexity-untouched.\n\n**Option (D): sharpen the I_I bound below IIMaj by a smarter kernel estimate.** The kernel on I_I: |w/(w−1)|ⁿ with w = 2√n+it. Note |w|²/|w−1|² = ((2√n)²+t²)/((2√n−1)²+t²) = 1 + (4√n−1)/((2√n−1)²+t²). The integral ∫_{-T}^{T} (1 + (4√n−1)/((2√n−1)²+t²))^{n/2} dt. Laplace: the peak at t=0 has width ~ 2√n/√(n/√n)·... ~ n^{3/4}-ish?? Let me redo: exponent ≈ (n/2)·(4√n−1)/((2√n−1)²+t²) ≈ (2n√n)/(4n+t²). At t = 0: 2n√n/(4n) = √n/2 ✓. Decays when t² ~ 4n·(√n/2)/(1) hmm: the exponent drops by O(1) when t²·(2n√n)/(4n)² ~ 1 ⟹ t² ~ 16n²/(2n√n) = 8√n ⟹ t ~ √8·n^{1/4}. So the integral ~ e^{√n/2}·∫exp(−c·t²/(8√n))dt ~ e^{√n/2}·√(8π√n) ~ e^{√n/2}·5·n^{1/4}. vs IIMaj's charge 2(√n+1)·e^{√n/2}·(0.57·2^{−...}). The zeta factor 0.57·2^{−(2√n−2)} multiplies both. So TRUE I_I ~ (5·n^{1/4}/(2√n))·IIMaj ~ 2.5·n^{−1/4}·IIMaj ≪ IIMaj. So a Laplace-type bound gives I_I ≤ IIMaj EASILY (factor n^{1/4} headroom). But formalizing Laplace on this integral: split at |t| = δₙ := n^{1/4}: \n- Inner: (e^{√n/2+1/3}+1)·0.57·2^{−...}·2δₙ... hmm inner length 2n^{1/4}, bound ≤ 2n^{1/4}·e^{1/3}(e^{√n/2}+1)·0.57·2^{−...} = e^{1/3}·IIMaj·n^{1/4}/(√n+1) ≤ 1.4·IIMaj·n^{1/4}/√n. For n ≥ 10⁵: n^{1/4}/√n = n^{−1/4} ≤ 10^{-5/4} ≈ 0.0562. So inner ≤ 1.4·0.0562·IIMaj = 0.0787·IIMaj ✓ tiny!\n- Outer (n^{1/4} ≤ |t| ≤ T): kernel ≤ (1+(4√n−1)/((2√n−1)²+√n))^{n}... exponent: n·(4√n)/(4n) · 4n/(4n+√n)-ish = √n/2·(1/(1+1/(4√n))) ≈ √n/2 − 1/8. So kernel ≤ e^{√n/2−1/8+o(1)}+1. Length 2T ≤ 2(√n+1). Outer ≤ 2(√n+1)·(e^{√n/2−1/8}+1)·0.57·2^{−...} = IIMaj·(e^{√n/2−1/8}+1)/(e^{√n/2}+1) ≤ IIMaj·1 ✓ (e^{√n/2−1/8}+1 ≤ e^{√n/2}+1 ✓ trivially!). \nTotal: ≤ (0.0787 + 1)·IIMaj = 1.0787·IIMaj. STILL over IIMaj! Because the outer estimate's kernel at t = n^{1/4} is e^{√n/2−1/8} which vs e^{√n/2}: ratio e^{−1/8} = 0.882 — so outer ≤ 0.882·IIMaj·(1+small) hmm: (e^{√n/2−1/8}+1)/(e^{√n/2}+1) ≤ (0.882·e^{√n/2}+1)/(e^{√n/2}+1) ≤ 0.882 + 0.118·(1+1)/(e^{√n/2}+1)... for √n ≥ 316: e^{√n/2} huge so ≈ 0.882. So outer ≤ 0.883·IIMaj, total ≤ 0.883 + 0.0787 = 0.96·IIMaj < IIMaj ✓✓!! \n\nWait let me redo this more carefully — this could give I_I ≤ IIMaj with NO Bn change at all!!\n\nSplit at δ = n^{1/4} (or any convenient split — maybe δ = √n/4? Let me optimize). We need:\nI_I ≤ [2δ·e^{1/3} + 2T·K(δ)]·(e... hmm let me parametrize. For |t| ≥ δ: kernel² = (1+(4√n−1)/((2√n−1)²+t²))ⁿ ≤ exp(n·(4√n−1)/((2√n−1)²+δ²)) (using 1+x ≤ e^x... wait: (1+u)^n ≤ e^{nu} ✓ with u = (4√n−1)/((2√n−1)²+δ²)·... hmm careful: |w/(w−1)| = sqrt(1 + (4√n−1)/((2√n−1)²+t²)) so |w/(w−1)|ⁿ = (1+…)^{n/2} ≤ e^{(n/2)·u}, u = (4√n−1)/((2√n−1)²+δ²). (n/2)·u = (n/2)(4√n−1)/((2√n−1)²+δ²). At δ² = √n (i.e. δ = n^{1/4}): (2√n−1)² + √n = 4n − 4√n + 1 + √n = 4n − 3√n + 1. (n/2)(4√n−1)/(4n−3√n+1) = (2n√n − n/2)/(4n−3√n+1). For large n: ≈ (2n√n)/(4n)·1/(1 − 3/(4√n)) = (√n/2)(1 + 3/(4√n) + …) = √n/2 + 3/8 + … So kernel ≤ e^{√n/2+3/8} — WORSE than e^{√n/2+1/3}?! \n\nWait that can't be right — at t = δ = n^{1/4} the kernel should be SMALLER than at t=0 (e^{√n/2+1/8-ish}). Let me recompute the t=0 case: u₀ = (4√n−1)/(2√n−1)² = (4√n−1)/(4n−4√n+1). (n/2)·u₀ = (n/2)(4√n−1)/(4n−4√n+1) = (2n√n−n/2)/(4n−4√n+1). Divide num and denom by 4n: = (√n/2 − 1/8)/(1 − 1/√n + 1/(4n)) ≈ (√n/2)(1 − 1/(4√n))(1 + 1/√n + …) ≈ √n/2·(1 + 3/(4√n)) = √n/2 + 3/8. Hmm so even at t=0, (n/2)·u₀ ≈ √n/2 + 3/8?? But earlier I computed n·log(2√n/(2√n−1)) = √n/2 + 1/8. Discrepancy: because |w/(w−1)|ⁿ = (|w|²/|w−1|²)^{n/2} and (1+u)^{n/2} ≤ e^{(n/2)u} LOSES vs the true n·log|w/(w−1)| = (n/2)·log(1+u₀) ≈ (n/2)(u₀ − u₀²/2). The 1+x ≤ e^x route loses the −u²/2 term which contributes −(n/2)(u₀²/2) = −n·u₀²/4 ≈ −n·(1/√n)²/4 = −1/4. So true ≈ √n/2 + 3/8 − 1/4 = √n/2 + 1/8 ✓ consistent with the earlier sharp computation. So the 1+x ≤ e^x route gives e^{√n/2+3/8} UNIFORMLY (even at t=0 — the t² denominator only helps at larger t).\n\nOK so with the crude route: kernel(t) ≤ e^{(n/2)·u(t)}, u(t) = (4√n−1)/((2√n−1)²+t²). For the outer region |t| ≥ δ: (n/2)u(δ) = (2n√n−n/2)/(4n−4√n+1+δ²). For this to be ≤ √n/2 − c for some c > 0, need: (2n√n−n/2) ≤ (√n/2−c)(4n−4√n+1+δ²) ⟹ LHS ≈ 2n√n, RHS ≈ (√n/2)(4n+δ²) − c(4n) = 2n√n + (√n/2)δ² − 4cn ⟹ need (√n/2)δ² ≥ 4cn + lower ⟹ δ² ≥ 8c√n·... (√n/2)δ² ≥ 4cn ⟺ δ² ≥ 8c√n. For c = 1/8: δ² ≥ √n ✓ EXACTLY the δ = n^{1/4} choice: (n/2)u(n^{1/4}) = (2n√n−n/2)/(4n−3√n+1): is this ≤ √n/2 − 1/8? ⟺ 2n√n−n/2 ≤ (√n/2−1/8)(4n−3√n+1) = 2n√n − (3/2)n + √n/2 − n/2 + (3/8)√n − 1/8 = 2n√n − 2n + (7/8)√n − 1/8. ⟺ −n/2 ≤ −2n + (7/8)√n − 1/8 ⟺ (3/2)n ≤ (7/8)√n − 1/8 — FALSE for large n!! Hmm so (n/2)u(n^{1/4}) ≈ √n/2 + 3/8 − δ²-correction... let me just numerically: n = 10⁵, √n = 316.23: (n/2)u(δ) at δ² = √n = 316: denom = 4n − 3√n + 1 = 400000 − 948.7 + 1 = 399052; numer = 2n√n − n/2 = 2·10⁵·316.23 − 50000 = 6.3246×10⁷ − 5×10⁴ = 6.3196×10⁷. Ratio: 6.3196×10⁷/399052 = 158.37. And √n/2 = 158.11. So (n/2)u ≈ 158.37 = √n/2 + 0.26 — BARELY above √n/2, not below! So at δ = n^{1/4}, the crude-bound exponent is √n/2 + 0.26 > √n/2. The t² correction is too weak at δ = n^{1/4} because the crude bound overshoots by 3/8−1/4 = 1/8 at t=0 and the t² term only subtracts ~ (n/2)·u₀·δ²/(4n) ≈ (√n/2)·δ²/(4n)·... ≈ δ²/(8√n): at δ² = √n: 1/8. So net: √n/2 + 1/8 (true t=0 value, crude gives +3/8) − 1/8 (t² correction) = crude at δ: +3/8−1/8 = +1/4 ≈ +0.26 ✓ matches. So crude route can never beat √n/2 + (3/8 − δ²/(8√n))... need δ²/(8√n) ≥ 3/8 ⟹ δ² ≥ 3√n ⟹ δ = √(3)·n^{1/4}. At δ² = 3√n: crude exponent ≈ √n/2 + 3/8 − 3/8 = √n/2 ✓ borderline. For a strict margin: δ² = 4√n: exponent ≈ √n/2 − 1/8. And the inner piece: length 2δ = 8·n^{1/4}, bound 8n^{1/4}·e^{√n/2+1/3}·0.57·2^{−...} = e^{1/3}·IIMaj·4n^{1/4}/(√n+1)·... hmm: IIMaj = 2(√n+1)(e^{√n/2}+1)·0.57·2^{−...}: inner ≤ 2δ·e^{1/3}(e^{√n/2}+1)·0.57·2^{−...} = e^{1/3}·IIMaj·δ/(√n+1) = e^{1/3}·IIMaj·2√2·n^{1/4}/(√n+1) (δ = 2n^{1/4}) ≈ 1.3957·2.828·IIMaj·n^{−1/4}·(√n/(√n+1)) ≈ 3.95·IIMaj·n^{−1/4}. At n = 10⁵: n^{1/4} = 17.78: 3.95/17.78 = 0.222. Inner ≤ 0.222·IIMaj. And DECREASING (n^{−1/4}) ✓.\n\nOuter (|t| ≥ 2n^{1/4}): kernel ≤ e^{(n/2)u(2n^{1/4})}+1 where (n/2)u = (2n√n−n/2)/(4n−4√n+1+4√n) = (2n√n−n/2)/(4n+1) = (2√n−1/2)·n/(4n+1)·... = n(2√n−1/2)/(4n+1). Divide: = (2√n−1/2)/(4+1/n) = (√n/2−1/8)/(1+1/(4n)) ≤ √n/2 − 1/8 + tiny... actually (√n/2−1/8)/(1+1/(4n)) ≤ √n/2−1/8 ⟺ true (denominator > 1, numerator > 0) ✓ EXACTLY. So kernel ≤ e^{√n/2−1/8}+1. Outer ≤ 2T·(e^{√n/2−1/8}+1)·0.57·2^{−...} ≤ 2(√n+1)·(e^{√n/2−1/8}+1)·0.57·2^{−...} = IIMaj·(e^{√n/2−1/8}+1)/(e^{√n/2}+1). Now (e^{√n/2−1/8}+1)/(e^{√n/2}+1) ≤ ? Write = (e^{−1/8}·e^{√n/2}+1)/(e^{√n/2}+1) ≤ e^{−1/8} + 1/(e^{√n/2}+1) ≤ e^{−1/8} + e^{−√n/2}. For √n ≥ √10⁵ = 316: e^{−√n/2} ≤ e^{−158} — negligible. So outer ≤ (0.8825 + 10^{−69})·IIMaj ≤ 0.8825·IIMaj.\n\nTotal: ≤ (0.222 + 0.8825)·IIMaj = 1.105·IIMaj. STILL OVER. Hmm — because at n = 10⁵ the inner ratio 3.95·n^{−1/4} = 0.222 is too big, and it decays only like n^{−1/4}. At n = N₁ = 9×10²⁴: n^{1/4} = (9×10²⁴)^{1/4} ≈ 1.73×10⁶: 3.95/1.73×10⁶ ≈ 2.3×10^{−6} ✓ tiny there. The problem is only the base region. At n = 10⁵: need inner-ratio + outer-ratio ≤ 1: 3.95·n^{−1/4} + 0.8825 ≤ 1 ⟺ 3.95·n^{−1/4} ≤ 0.1175 ⟺ n^{1/4} ≥ 33.6 ⟺ n ≥ 1.28×10⁶. Just misses at 10⁵!\n\nTune the split: δ² = c√n. Inner ratio: e^{1/3}·2√c·n^{1/4}/(√n+1) ≈ 2.795·√c·n^{−1/4}. Outer exponent: (n/2)u = (2n√n−n/2)/(4n−4√n+1+c√n) ≈ √n/2 + 3/8 − c/8 − (small). For c = 6: exponent ≈ √n/2 − 3/8 ⟹ outer ratio ≈ e^{−3/8} = 0.687. Inner ratio: 2.795·√6·n^{−1/4} = 6.846·n^{−1/4}: at 10⁵: 0.385. Total: 1.072. Still over! c = 9: outer e^{−9/8+3/8} = e^{−3/4} = 0.472; inner 2.795·3·n^{−1/4} = 8.385/17.78 = 0.472. Total 0.944 < 1 ✓! c = 9, δ = 3n^{1/4}: total ≈ 0.944. Tight-ish at base, improving as n^{−1/4} + const. Margin 5.6% at the worst point. Hmm — workable but requires the (n/2)u(δ) ≤ √n/2 − 3/4-ish inequality cleanly: (2n√n−n/2)/(4n+(c−4)√n+1) ≤ √n/2 − α: cross-multiply: 2n√n−n/2 ≤ (√n/2−α)(4n+(c−4)√n+1) = 2n√n + ((c−4)/2)n + √n/2 − 4αn − α(c−4)√n − α ⟹ 0 ≤ ((c−4)/2 − 4α + 1/2)n + (1/2 − α(c−4))√n − α ⟹ need (c−4)/2 + 1/2 ≥ 4α (leading) ⟺ α ≤ (c−3)/8. c = 9: α ≤ 6/8 = 3/4 ✓ (and the √n term: 1/2 − (3/4)(5) = −3.25 < 0 — uh oh: the √n-coefficient is negative: 0 ≤ ((9−4)/2−3+1/2)n + (1/2−15/4)√n − 3/4 = (0)n·... wait: (c−4)/2 + 1/2 − 4α with c=9, α=3/4: (5/2)+(1/2)−3 = 0 ✓; √n term: 1/2 − (3/4)(9−4) = 1/2 − 15/4 = −13/4 <0; constant −3/4 < 0. So the inequality is 0 ≤ 0·n − (13/4)√n − 3/4 — FALSE!! Let me recompute. Inequality: 2n√n − n/2 ≤ (√n/2 − α)(4n + (c−4)√n + 1). Expand RHS: (√n/2)(4n) + (√n/2)(c−4)√n + (√n/2) − α(4n) − α(c−4)√n − α = 2n√n + ((c−4)/2)n + √n/2 − 4αn − α(c−4)√n − α. RHS − LHS = 2n√n + ((c−4)/2)n + √n/2 − 4αn − α(c−4)√n − α − 2n√n + n/2 = [((c−4)/2) + 1/2 − 4α]n + [1/2 − α(c−4)]√n − α. c = 9, α = 3/4: [(5/2)+(1/2)−3]n + [1/2−(15/4)]√n − 3/4 = 0·n − (13/4)√n − 3/4 < 0 ✗. So α = 3/4 is NOT achieved; the leading-n coefficient must be ≥ 0 AND the √n one too (for large n the n-term dominates but here it's exactly 0, so the negative √n-term kills it). Need both: ((c−4)/2)+1/2−4α ≥ 0 AND 1/2 − α(c−4) ≥ 0: α ≤ 1/(2(c−4)). c = 9: α ≤ 1/10. So with c = 9: α = 1/10: outer ratio e^{−0.1} = 0.905; inner 0.472; total 1.377 ✗ worse. Hmm so the α is limited by the √n-coefficient: α ≤ 1/(2(c−4)) which for large c shrinks — counterproductive. Balance: α(c) = min((c−3)/8, 1/(2(c−4))). c = 5: min(1/4, 1/2) = 1/4; inner ratio 2.795·√5·n^{−1/4} = 6.25/17.78 = 0.352; outer e^{−1/4} = 0.779; total 1.13 ✗. c = 6: min(3/8, 1/4) = 1/4; inner 6.846/17.78 = 0.385; total 1.164 ✗. c = 4.5: min(3/16, 1) = 3/16 = 0.1875; inner 2.795·2.121/17.78 = 0.333; outer e^{−0.1875} = 0.829; total 1.162 ✗. c = 3: min(0, 1/−2) — degenerate. Hmm the c-range (4,5): α = 1/(2(c−4)) for c < 5-ish: c = 4.2: α = 2.5?? no: 1/(2·0.2) = 2.5, min((1.2)/8, 2.5) = 0.15; inner 2.795·2.049/17.78 = 0.322; outer e^{−0.15} = 0.861; total 1.183 ✗. Hmm all ≥ 1.1 at n = 10⁵. The inner ratio ~ 2.8√c·n^{−1/4} is just too big at 10⁵ unless c tiny, but then α → small and outer → 1. Fundamental tension: total ≈ e^{−α} + 2.8√c·n^{−1/4} with α ≈ (c−3)/8: minimize over c: at n=10⁵: try c = 8: α = 5/8 = 0.625 BUT √n-coefficient check: 1/2 − (5/8)(4) = −2 < 0 ✗ α invalid. The √n-coefficient constraint α ≤ 1/(2(c−4)) binds for c > 4.5: c = 4.6: α ≤ 1/1.2 = 0.833, min((1.6)/8, 0.833) = 0.2; inner 2.795·2.1448/17.78 = 0.337; outer e^{−0.2} = 0.819; total 1.156. STILL over.\n\nHmm. So the 1+x ≤ e^x crude route can't beat IIMaj at n = 10⁵ with ANY split. The overshoot +3/8 vs the true +1/8 is the killer (needs the −u²/2 log term). With the SHARP kernel bound (log form, −log(1−v) ≤ v+v² machinery): true exponent at t: (n/2)·log(1+u(t)) ≤ (n/2)(u − u²/2 + u³/3). This gives t=0: √n/2+1/8; the split-analysis redone with sharp bounds would give outer ratios e^{−c/8}-ish with the same √n-coefficient constraint... this is getting REALLY heavy for Lean.\n\nSTOP. Rethink strategy. The overrun factor is e^{1/8} ≈ 1.133 (sharp) — the charged IIMaj is too small by ≤ 13.3% (at t=0, the max). The convexity budget is EXACTLY saturated at coefficient 1.5. But WAIT — maybe I should look at the actual convexity proof budget before assuming 1/(2n) is a hard wall. The final linarith: positive-side ≥ Σ negative rows. If the positive side has ANY slack ≥ 0.2·(1/(3n)) ≈ 0.0667/n... Actually hold on — maybe the cleaner question: does the convexity proof's IIM row use lt_three (IIMaj″ < 1/(3n)) — an ABSOLUTE bound — while the ACTUAL IIMaj″ at n ≥ 8×10⁴ is ~10^{-120}? The budget allocation 1/(2n) for the IIM row is enormously wasteful — the row's true value is 10^{-120}. So the PROOF STRUCTURE wastes it, not the math. If the coefficient grows to 1.8956, the row bound 1.8956·(1/(3n)) = 0.632/n exceeds the 0.5/n allocation by 0.132/n — and that excess must come from somewhere. The sInftyLower side: (1/2)n·log n's second derivative = (1/2)(log n + 1)... vs the sum of ALL row allocations — I need to read the proof to see the total accounting. If the proof ends with `linarith` over literals like h1 : X ≥ (1/2)log n + ..., hIIM : row ≤ 1/(2n), hId : row ≤ ..., and the final inequality has slack, then bumping the IIM literal from 1/(2n) to 0.632/n MIGHT still close — depends on the slack.\n\nLet me read the marginTilde_convex proof's final assembly to see the actual slack.\n\nALTERNATIVELY — much simpler idea!! **Charge I_I as e^{1/3}·IIMaj in Bn but keep the mIIM convexity row at −1.5·IIMaj + make up the difference in the mK CONSTANT row?** No — the convexity is about second derivatives; a constant has zero second derivative and doesn't help.\n\nANOTHER idea: **the mId row!** The DogLegRight block (√n+1)(e^{5/2}+1)/2 + 16.76 + IIMaj/4 per side ×2 = (√n+1)(e^{5/2}+1)/2 + 16.76 + IIMaj/2 — wait no: 2·DogLegRight = (√n+1)/2·(e^{5/2}+1) + 16.76 + IIMaj/2. Hmm from marginExpanded: mIdb := fun m => (−(((8/3)*2.7 + 192) + (e^{5/2}+1)/2))·(√m+1) — the (√m+1)-coefficient includes (e^{5/2}+1)/2 (the DogLeg V_left piece) and mIIM := (−1.5)·IIMaj. The mId row's second derivative: (√n+1)-type terms have ψ/derivatives ~ 1/(4n^{3/2})·coef — the row was just re-done in a2 with \"hnn4 new coefficient with Real.exp_pos (5/2) nlinarith\". Maybe the mId row has slack. But shuffling slack between rows = re-proving chunks of the convexity. \n\nHmm, actually — WAIT. Simplest of all: **can I prove I_I ≤ IIMaj + (something already overcharged)?** The horizontals: E-030 measured ratio ≤ 0.40 — the pinch 7Ĥ has 60% slack ~ 200+ at n=10⁵ (in absolute terms 0.6·2(e⁴+1)·7Ĥ ~ 66·Ĥ ~ 66·340 ~ 22440 at n=10⁵, growing like L²). The I_I overrun: (e^{1/3}−1)·IIMaj(10⁵) ~ 0.4·10^{-119}. So cross-subsidy: the fold 2(e⁴+1)(NI+5.5(...)) + e^{1/3}IIMaj ≤ 2(e⁴+1)·7Ĥ + IIMaj ⟺ 2(e⁴+1)(NI + 5.5(...) − 7Ĥ) + (e^{1/3}−1)IIMaj ≤ 0 ⟸ NI + 5.5(...) ≤ 7Ĥ − (e^{1/3}−1)·IIMaj/(2(e⁴+1)). The RHS subtraction is ~10^{-122} — and the pinch's OWN proof margin (ratio 0.40 ⟹ absolute slack ~0.6·7Ĥ ~ 200) — so the strengthened pinch NI + 5.5(...) ≤ 7Ĥ − tiny is THE SAME PROOF with a microscopically different endpoint (the certified endpoint evaluation changes in the 10^{-122}-th digit — the rational interval bounds don't even see it!). So the pinch lemma's statement becomes NI-total + e^{1/3}·IIMaj ≤ 7Ĥ-charge + IIMaj... hmm — but formally the cleanest packaging: keep Bn EXACTLY as-is; prove the fold directly: sides-total ≤ Bn-charges by ONE combined inequality: [horizontal-integral ≤ (e⁴+1)(NI+5.5(...))] + [I_I ≤ e^{1/3}IIMaj] and then NI + 5.5(...) + (e^{1/3}−1)IIMaj/(e⁴+1) ≤ 7Ĥ. The pinch lemma gets a \"+ tiny\" on the LHS. The pinch proof = certified endpoint + growth direction: the tiny term is ≤ 10^{-100} everywhere on [10⁵, ∞)·... — the growth-direction argument: is (e^{1/3}−1)IIMaj/(2(e⁴+1)) decreasing? IIMaj is decreasing for n ≥ ... (bracket/antitone machinery exists — iim_u2_antitone etc.). 7Ĥ − NI − 5.5(...) is increasing (that's the pinch's growth argument) minus a decreasing tiny ⟹ still increasing. So the strengthened pinch: at base 10⁵: LHS−RHS ≤ −200 + 10^{-119} < 0 ✓ certified; growth: d/dn[RHS−LHS] ≥ 0 − (tiny)' ≥ 0 − 0... The proof structure IDENTICAL, with one more negligible term tracked. This keeps Bn, marginTilde, marginExpanded, convexity, base points, DogLegRight, IIMaj — EVERYTHING — untouched!! The only changes: the pinch lemma (6-d territory anyway) and the I_I bound proof (e^{1/3} factor). \n\nActually hmm, wait — but is that even needed? Let me re-examine: maybe just charge I_I as e^{1/3}·IIMaj directly in the fold with the pinch absorbing. YES that's exactly option (C) above but now I see it's clean: the pinch lemma is in 6-d anyway; Bn untouched; convexity untouched.\n\nHmm, but actually — hold on, let me reconsider option (E): **sharpen the zeta side on I_I**. norm_logDeriv_zeta_le_iimZeta ≤ 0.57·2^{−(2√n−2)}. How sharp is the 0.57? From DL-6: dirichletCX ≤ log2·2^{−(2√n−2)} + 3^{−(2√n−2)}·dirichletC2 + ... ≈ (0.6931 + 9·0.164)·2^{−...} = (0.6931+1.476)·... hmm the 0.57: log_nine_quarters_ge + three_pow...: 0.57 = log 2·(something)? From the banked: \"0.6931471808 + 9·0.164 ≤ 2.28\" and then ≤ 0.57·4 = 2.28 ✓ — so 0.57 = (log 2 + 9·0.164)/4 = 2.169/4·... log2 + 1.476 = 2.169; /4 = 0.5423 ≤ 0.57 (slack 5%). So 0.57 has ~5% slack — not enough to absorb e^{1/8} = 13%.\n\nOption (F): **sharpen the kernel bound via the log-form** −log(1−v) ≤ v+v²: kernel ≤ e^{√n/2+1/4} EXACTLY (computed earlier: n·log(1/(1−u)) ≤ n(u+u²) = √n/2+1/4 for u = 1/(2√n) ≤ 1/2 ✓ n ≥ 1). Then I_I ≤ e^{1/4}·IIMaj — overrun e^{1/4}−1 = 28.4%... wait that's WORSE than e^{1/8}?? No: e^{√n/2+1/4} vs the charged e^{√n/2}: ratio e^{1/4} = 1.284. The sharp true value is e^{√n/2+1/8} (ratio 1.133) but the v+v² log bound gives +1/4 (loses another 1/8 vs truth). So option F: overrun 28.4% — worse than the 39.6% of e^{1/3} but still needs subsidy. The v+v² route (via exp-partial-sum polynomial argument) is more Lean work than add_one_le_exp (one line). So e^{1/3} it is: I_I ≤ e^{1/3}·IIMaj, overrun 39.56% of IIMaj — immaterial in absolute terms (10^{-119}), absorbed by the pinch.\n\nDECISION: **Option (C): Bn/convexity/IIMaj untouched. I_I bound = e^{1/3}·IIMaj (new kernel lemma uniform in t). The 6-d pinch lemma absorbs the (e^{1/3}−1)·IIMaj overrun** (strengthened statement NI-total + e^{1/3}·IIMaj ≤ 7Ĥ-charge + IIMaj, same proof, microscopically different certified constants — verify at pinch-build time that the endpoint evaluation's rational interval bounds don't feel it).\n\nWait — one more consideration: is there maybe an even cleaner home for the overrun? The I_III charge: 4T((2/3)Z2hat + 10.5 log + 48) — the near-sum bound (2/3)·Z2hat used ‖s−ρ‖ ≥ 3/2, but ACTUALLY ‖s−ρ‖ ≥ |re| = 5/2 + β ≥ 5/2 ⟹ 2/5 ≤ 2/3 — slack factor 5/3 in the near-sum piece: (2/3−2/5)·Z2hat·4T = (4/15)·Z2hat·4T ~ (4/15)·3.3·4·317 ~ 1115 at n=10⁵ — also huge slack vs 10^{-119}. But touching I_III's bound = touching its charge bookkeeping. The horizontal pinch is the most natural subsidy home (it's a single real-analytic inequality in 6-d anyway).\n\nHmm — actually hold on, even simpler: why not absorb it into the \"+1\"?? Bn has a trailing +1 (the triv term). The triv term's true bound: |1−(2/3)ⁿ| ≤ 1 — the charge is EXACTLY 1 and true value 1−(2/3)ⁿ — slack (2/3)ⁿ ~ 10^{-17617} at n=10⁵ — plenty! But the +1 is charged by the BRIDGE (zeroSum ≤ 4FAlt + 1), a different additive piece of the fold: |Sf| ≤ ‖RectInt'‖ + ‖zeroSum‖ where ‖RectInt'‖ ≤ (1/2π)·(sides) ≤ sides and ‖zeroSum‖ ≤ 4F+1. The I_I overrun is inside ‖RectInt'‖'s (1/2π)!! **THE 2π FACTOR!!** ‖RectangleIntegral'‖ = (1/(2π))·‖RectangleIntegral‖ ≤ (1/2π)·(sides-total). The sides-total gets DIVIDED BY 2π in the fold!! D-071 §4.5: \"RectangleIntegral' is 2π-normalized so ‖RectInt‖ ≤ Σ sides/(2π) ≤ Σ sides ✓ the fold is immediate per-piece\" — the design DROPS the 2π (charges sides/(2π) ≤ sides). So the actual fold has a FREE factor 2π ≈ 6.28 of slack on ALL side pieces!!! The I_I overrun e^{1/3}·IIMaj/(2π) ≤ IIMaj·(1.3957/6.2832) = 0.2221·IIMaj ≤ IIMaj ✓✓✓ — the 2π factor alone absorbs the overrun with factor 4.5 to spare!!\n\nBut careful: if the fold uses sides/(2π) ≤ sides for the OTHER pieces (charging I_III ≤ its full bound etc.), then each piece is charged at its full bound — and I_I needs e^{1/3}·IIMaj ≤ IIMaj·2π·... the per-piece fit for I_I: [I_I-true ≤ e^{1/3}·IIMaj] needs [e^{1/3}·IIMaj ≤ 2π·IIMaj-charge] — but the charge in Bn is IIMaj WITHOUT a 2π... The fold: |Sf| ≤ ‖RectInt'‖ + ‖zeroSum‖ ≤ (1/2π)·(I_III + Horiz + I_I) + (4F+1) where I_III ≤ A, Horiz ≤ B, I_I ≤ e^{1/3}·IIMaj. Need (1/2π)(A + B + e^{1/3}IIMaj) + 4F + 1 ≤ Bn = A' + B' + IIMaj + 4F + 1 where A' = A's charge = A exactly (I_III's charge = the bound), B' = B's charge... if A ≤ A' and B ≤ B' (per-piece fits), then need (1/2π)(A' + B' + e^{1/3}IIMaj) ≤ A' + B' + IIMaj ⟸ A' + B' + e^{1/3}IIMaj ≤ 2π(A' + B' + IIMaj) ⟸ e^{1/3}IIMaj ≤ (2π−1)(A'+B') + 2π·IIMaj ⟸ e^{1/3} ≤ 2π ✓ (1.3957 ≤ 6.2832) — TRIVIAL linarith, no pinch coupling, no strengthened pinch!! The 2π factor gives it for free with the PER-PIECE fits UNCHANGED. \n\nSo: **the fold just needs ONE extra linarith step: e^{1/3}·IIMaj ≤ 2π·IIMaj (via e^{1/3} ≤ 1.4 ≤ 2π·1 — wait need e^{1/3} ≤ 2π: 1.4 ≤ 6.28 ✓)**. NO Bn change, NO convexity change, NO pinch change, NO IIMaj change. The only new content: the I_I kernel bound lemma (uniform in t: kernel ≤ e^{√n/2+1/3}+1) and the I_I side bound statement: ‖VIntegral (Jlag n) (2√n) (−T) T‖ ≤ e^{1/3}·IIMaj n.\n\nHmm wait, actually — hmm, but hold on: is the fold really done with the 2π slack available, or does the design deliberately charge sides/(2π)?? D-071 §4.5: \"Bn charges the raw side bounds (the paper's convention; RectangleIntegral' is 2π-normalized so ‖RectInt‖ ≤ Σ sides/(2π) ≤ Σ sides ✓ the fold is immediate per-piece)\". So Bn charges the RAW side bounds (no 2π division) — i.e., the fold drops a free 2π. The overrun e^{1/3} fits inside the dropped 2π with the per-piece logic: (1/2π)(A'+B'+e^{1/3}IIMaj) ≤ A'+B'+IIMaj needs e^{1/3}IIMaj ≤ (2π−1)(A'+B') + 2π·IIMaj — since IIMaj ≥ 0 and A',B' ≥ 0: suffices e^{1/3}·IIMaj ≤ 2π·IIMaj ✓ e^{1/3} ≤ 2π ✓. But actually cleaner: (1/2π)(A'+B'+e^{1/3}IIMaj) ≤ (1/2π)(A'+B'+2π·IIMaj)... no wait that's ≤ A'+B'+... hmm (1/2π)(A'+B') + IIMaj ≤ A'+B'+IIMaj ✓ (A'+B' ≥ 0, 1/2π ≤ 1). So: (1/2π)(A+B+e^{1/3}IIMaj) ≤ (1/2π)(A'+B') + (1/2π)e^{1/3}IIMaj ≤ (A'+B') + IIMaj ✓✓ PERFECT — two linarith steps with e^{1/3} ≤ 2π (certified: e^{1/3} ≤ 1.4, 1.4 ≤ 2π via π ≥ 3.14: 2π ≥ 6.28 ✓).\n\nBut hmm — one subtlety: this requires A', B' ≥ 0 (for dropping the 1/2π on them) — the charges ARE nonnegative ✓ (positivity).\n\nAlso — the HORIZONTAL per-piece fit: horiz ≤ B' uses the pinch NI+5.5(...) ≤ 7Ĥ — unchanged ✓.\n\nGREAT. So F-E's repair is ENTIRELY local to 6-c-iii (the I_I bound, charged as e^{1/3}·IIMaj) + one linarith step in 6-d's fold (e^{1/3} ≤ 2π). The Finding F-E note goes into D-074 (or a small erratum note) + the ii′-7 paper-correction queue (the paper's B(n) is still VALID — the true I_I ≤ e^{1/3}·IIMaj ≤ 2π·IIMaj and the paper's fold also drops the 2π — so the paper's Theorem statement survives EXACTLY, only the I_I piece-bound justification in the derivation needs the kernel footnote). Actually — for the paper: B(n) as stated remains a valid bound; the derivation D-036's I_I kernel claim \"e^{√n/2}+1\" gets the erratum \"e^{√n/2+1/3}+1, absorbed by the 2π normalization slack\". \n\nLet me now also double check the kernel bound e^{√n/2+1/3}: |w/(w−1)|ⁿ, w = 2√n+it: |w/(w−1)| = sqrt(1+(4√n−1)/((2√n−1)²+t²)) ≤ sqrt(1+(4√n−1)/(2√n−1)²) = sqrt(((2√n−1)²+4√n−1)/(2√n−1)²) = sqrt((4n)/(2√n−1)²) = 2√n/(2√n−1) = 1+1/(2√n−1). So |w/(w−1)|ⁿ ≤ (1+1/(2√n−1))ⁿ ≤ e^{n/(2√n−1)} (add_one_le_exp ^ n: (1+x)ⁿ ≤ (e^x)ⁿ = e^{nx} ✓). n/(2√n−1) ≤ √n/2 + 1/3 for n ≥ 4: ⟺ 3n ≤ (√n/2+1/3)·3·(2√n−1)... redo: n/(2√n−1) ≤ √n/2 + 1/3 ⟺ n ≤ (√n/2 + 1/3)(2√n−1) = n − √n/2 + (2/3)√n − 1/3 = n + √n/6 − 1/3 ⟺ 0 ≤ √n/6 − 1/3 ⟺ √n ≥ 2 ⟺ n ≥ 4 ✓ CLEAN. So kernel ≤ e^{√n/2+1/3} for n ≥ 4, all t ✓. Then +1: ‖kFun‖ = ‖(w/(w−1))ⁿ − 1‖ ≤ |w/(w−1)|ⁿ + 1 ≤ e^{√n/2+1/3} + 1.\n\nI_I bound: ‖VIntegral (Jlag n) (2√n) (−T) T‖ = ‖I • ∫_{-T}^{T} Jlag‖ = ‖∫_{-T}^{T} Jlag‖ ≤ ∫_{-T}^{T} ‖Jlag‖ (norm_integral_le_of_norm_le_const with constant majorant — need integrability of Jlag on the segment: Jlag continuous (ζ ≠ 0 on re = 2√n > 1 ✓ riemannZeta_ne_zero_of_one_le_re... re = 2√n ≥ 2 > 1 ✓; s ≠ 1 ✓) ⟹ ContinuousOn ⟹ IntervalIntegrable ✓; constant majorant integrable ✓).\n‖Jlag n w‖ = ‖kFun‖·‖logDeriv ζ w‖ ≤ (e^{√n/2+1/3}+1)·0.57·2^{−(2√n−2)} (norm_logDeriv_zeta_le_iimZeta, n ≥ 5 ✓ — note Jlag = kFun·(−(deriv ζ/ζ)); logDeriv f = f'/f so −(deriv ζ/ζ) = −logDeriv ζ; ‖−x‖ = ‖x‖; norm_logDeriv_zeta_le_iimZeta is stated for logDeriv ✓).\nSo ‖I_I‖ ≤ 2T·(e^{√n/2+1/3}+1)·0.57·2^{−(2√n−2)} ≤ 2(√n+1)·e^{1/3}·(e^{√n/2}+1)·0.57·2^{−(2√n−2)} = e^{1/3}·IIMaj n (T ≤ √n+1 ✓; e^{√n/2+1/3}+1 = e^{1/3}e^{√n/2}+1 ≤ e^{1/3}(e^{√n/2}+1) ✓ e^{1/3} ≥ 1).\n\nThe equality e^{1/3}·IIMaj n = 2(√n+1)·e^{1/3}·(e^{√n/2}+1)·0.57·2^{−(2√n−2)}: unfold IIMaj + ring ✓.\n\nIIMaj n for n : ℕ — IIMaj takes ℝ: IIMaj (n : ℝ) ✓.\n\nSo 6-c-iii is now well-defined with the F-E repair built in. \n\nNow let me also make sure about the LEFT vertical's zero-freeness for integrability: re = −5/2: ζ(−5/2+it) ≠ 0 — zeta_zero_classification: zeros = nontrivial ∪ {0} ∪ {−2n}: re −5/2 excludes nontrivial (re ∈ (0,1)) and 0 and −2n (even integers) ✓. And s ≠ 1 ✓. So Jlag continuous on the left vertical ✓.\n\nAnd the horizontal integrability: T zero-avoiding-wide ⟹ no zeros on the horizontal path (clearance > 0) + s ≠ 1 (im = T ≥ 316) ✓.\n\nNOW — let me also settle the remaining signatures for 6-c-i scaffolding:\n1. HIntegral split: HIntegral f a c y = HIntegral f a b y + HIntegral f b c y — unfold HIntegral + intervalIntegral.integral_add_adjacent_intervals (needs IntervalIntegrable on both) ✓.\n2. VIntegral norm: VIntegral f x y1 y2 = I • ∫ — norm_smul + Complex.norm_I: ‖I • v‖ = ‖I‖·‖v‖ = ‖v‖ ✓.\n3. Jlag = kFun·(−logDeriv): Jlag def: kFun n s * (−(deriv riemannZeta s / riemannZeta s)). logDeriv riemannZeta s = deriv riemannZeta s / riemannZeta s (logDeriv_def — rfl? logDeriv f x = deriv f x / f x by def ✓). So Jlag n s = kFun n s * (−logDeriv riemannZeta s) — by rfl (unfold logDeriv). ✓ norm: ‖kFun‖·‖−logDeriv‖ = ‖kFun‖·‖logDeriv‖ (norm_mul + norm_neg).\n4. The even-symmetry for the left vertical's negative tail: g(t) = ‖Jlag n (−5/2 + tI)‖; g(−t) = g(t) via Jlag_conj (needs s ≠ 1: re = −5/2 ✓ never 1): Jlag n (conj s) = conj (Jlag n s)?? Jlag_conj (LiDogLeg:105): Jlag n (starRingEnd s) = starRingEnd (Jlag n s) for s ≠ 1. conj(−5/2 + tI) = −5/2 − tI ✓. So g(−t) = ‖Jlag(−5/2−tI)‖ = ‖conj(Jlag(−5/2+tI))‖ = ‖Jlag(−5/2+tI)‖ = g(t) (Complex.norm_conj) ✓.\n5. intervalIntegral.integral_comp_neg: ∫ x in a..b, f(−x) = ∫ x in −b..−a, f x — Mathlib name: `intervalIntegral.integral_comp_neg` ✓ (exists).\n\nNow — before writing any Lean, per the crash-safe protocol I should:\n1. Log Finding F-E in D-074 (progress log + the repair decision) — and note it for the ii′-7 paper queue.\n2. Update STATE.md with the ii′-6-c sub-stage map + F-E.\n3. Then start 6-c-i.\n\nActually — the todo list is set. Let me also reconsider the overall 6-c file plan: new file `LiSideBounds.lean`. Contents:\n- §0: scaffolding — rectangle decomposition (RectangleIntegral_Jlag_eq_sides), HIntegral split, VIntegral/HIntegral norm lemmas, Jlag-norm factorization, integrability on the three paths, Z2hat_mono, even-symmetry helper.\n- §1: I_III (6-c-ii).\n- §2: I_I (6-c-iii, F-E form).\n- §3: mid-horizontal machine (6-c-iv): per-zero segment integral (log-FTC split), near-sum swap, R-part, conj for the bottom.\n- §4: horizontal assembly (6-c-v): mid + dog-leg tails, both sides ⟹ ≤ 2(e⁴+1)(NI-charge) + 2·DogLegRight... wait the pinch is 6-d; 6-c-v outputs ≤ (e⁴+1)·(NI + 5.5(...))·2 + 2·DogLegRight-form. Hmm — actually the horizontal bound per side: ‖HIntegral (−5/2) (2√n) (±T)‖ ≤ (e⁴+1)·(NI + 5.5(...)) + DogLegRight(n). Both sides summed: ≤ 2(e⁴+1)(NI+5.5(...)) + 2·DogLegRight. Then 6-d: pinch NI+5.5(...) ≤ 7Ĥ ⟹ ≤ Bn's horizontal charge.\n\nWhere NI(T) := 2·Z2hat(T)·Real.log(16.5·9·C0hat(√n+2)) — hmm: 16.5/c with c = 1/(9·C0hat(√n+2)) ⟹ 16.5/c = 16.5·9·C0hat(√n+2) = 148.5·C0hat(√n+2). Note: NI depends on n (through the clearance) AND T.\n\nWait — the clearance c_clear = 1/(9·C0hat(a+2)) with a = √n (the wide pigeonhole's window [√n, √n+1] — exists_zero_avoiding_T_wide (ha : 3 ≤ a): T ∈ [a, a+1], clearance vs zeros with γ ∈ (a−1, a+2)). For the assembly: a := √n ≥ √(10⁵) ≥ 316 ≥ 3 ✓.\n\nThe near zeros for the horizontal at height T: nearZerosSub T = |γ − T| < 1 ⟹ γ ∈ (T−1, T+1) ⊆ (√n−1, √n+2) ✓ (T ∈ [√n, √n+1]) — clearance applies ✓ δ ≥ 1/(9·C0hat(√n+2)) ✓.\n\nOK — the per-zero integral: ∫_{-5/2}^{3} dσ/‖σ+iT−ρ‖ where ‖σ+iT−ρ‖ = sqrt((σ−β)²+(T−γ)²) ≥ ... with δ := |T−γ|. Plan: split the interval at β (β ∈ (0,1) ⊂ [−5/2,3] ✓ interior):\n∫_{-5/2}^{β−δ'} + ∫_{β−δ'}^{β+δ'} + ∫_{β+δ'}^{3} where δ' := min(δ, ...) hmm — actually the crude split: inner [β−δ, β+δ] ∩ [−5/2,3]: integrand ≤ 1/sqrt((σ−β)²+δ²) ≤ 1/δ: integral ≤ 2δ/δ = 2 ✓ (interval length 2δ; but if β−δ < −5/2 the inner piece is shorter — still ≤ 2). Outer: σ ∈ [−5/2, β−δ] ∪ [β+δ, 3] (if nonempty): 1/sqrt((σ−β)²+δ²) ≤ 1/|σ−β|: ∫_{β+δ}^{3} dσ/(σ−β) = log(3−β) − log δ (integral_one_div? or FTC: HasDerivAt log); ∫_{-5/2}^{β−δ} dσ/(β−σ) = log(β+5/2) − log δ. Total outer ≤ log((3−β)(β+5/2)) − 2 log δ ≤ log((3)(7/2)) + 2·log(1/δ) — (3−β) ≤ 3, (β+5/2) ≤ 7/2: log((3−β)(β+5/2)) ≤ log(10.5): total ≤ 2 + log 10.5 + 2·log(1/δ) ≤ 2·log(16.5/δ)?? 2·log(16.5/δ) = 2·log 16.5 + 2·log(1/δ) = 5.606 + 2 log(1/δ). Need 2 + log 10.5 + 2 log(1/δ) ≤ 5.606 + 2 log(1/δ) ⟺ 2 + 2.351 = 4.351 ≤ 5.606 ✓ (log 10.5 ≤ 2.352: 10.5 ≤ e^{2.352} = 10.51 ✓ certified-ish). So per-zero: ∫ ≤ 2·log(16.5/δ) ✓ with slack 1.25.\n\nHmm wait — D-071's design said ≤ 2·asinh(5.5/δ) ≤ 2·log(16.5/c) — asinh(5.5/δ) ≤ log(2·5.5/δ + 1) = log(11/δ+1) ≤ log(16.5/δ) using 1 ≤ 5.5/δ ⟺ δ ≤ 5.5 ✓. Either route lands at log(16.5/δ); my crude-split route: 2 + log 10.5 + 2 log(1/δ) ≤ 2 log(16.5/δ) — fine.\n\nEdge cases in Lean: the inner/outer split needs case analysis (β−δ vs −5/2, β+δ vs 3). To AVOID case analysis: bound the outer integrals over the FULL side intervals: ∫_{β+δ}^{3} ≤ ∫_{β+δ}^{3} dσ/(σ−β) = log((3−β)/δ) — need β+δ ≤ 3 for the integral to make sense; if β+δ > 3 the outer-right is empty (≤ 0 ✓ fine, but the FTC form breaks). Cleaner: use interval intersection and total ≤ with intervalIntegral over the full [−5/2,3] of the majorant min(1/δ, 1/|σ−β|)... the min function's integral is annoying too.\n\nALTERNATIVE cleaner route — bound the integral by the asinh-free inequality via a SINGLE majorant integral: for all σ: 1/sqrt((σ−β)²+δ²) ≤ 1/δ · χ_{|σ−β|≤δ} + 1/|σ−β| · χ_{|σ−β|>δ}. The RHS integrable piecewise... still case analysis.\n\nSIMPLEST IN LEAN — actually use arcsinh?? Mathlib has Real.arcsinh with `Real.hasDerivAt_arcsinh`?? Hmm — does Mathlib have arcsinh? I recall `Real.arcsinh : ℝ → ℝ := log (x + sqrt (x ^ 2 + 1))` EXISTS (Mathlib.Analysis.SpecialFunctions.Arsinh). And its derivative: hasDerivAt arcsinh (1/√(x²+1)) — `Real.hasDerivAt_arcsinh` ✓ I believe exists. Then ∫_{-5/2}^{3} dσ/√((σ−β)²+δ²) = ∫ dx/ (δ·√(((σ−β)/δ)²+1)) = arcsinh((σ−β)/δ) evaluated: ∫_{-5/2}^{3} = δ·... substitution u = (σ−β)/δ: ∫ du/√(u²+1) = arcsinh(u): so the integral = arcsinh((3−β)/δ) − arcsinh((−5/2−β)/δ) = arcsinh((3−β)/δ) + arcsinh((β+5/2)/δ) ≤ 2·arcsinh(5.5/δ) (each arg ≤ 5.5/δ: (3−β) ≤ 3 ≤ 5.5 ✓, β+5/2 ≤ 7/2 ≤ 5.5 ✓; arcsinh monotone). arcsinh(x) = log(x+√(x²+1)) ≤ log(x + (x+1)) = log(2x+1) (√(x²+1) ≤ x+1 for x ≥ 0 ✓). So ≤ 2·log(11/δ + 1) ≤ 2·log(16.5/δ) (11/δ + 1 ≤ 11/δ + 5.5/δ = 16.5/δ for δ ≤ 5.5 — δ ≤ 1/2?? δ = |T−γ| < 1 < 5.5 ✓ actually the near zeros have |γ−T| < 1 so δ < 1 ✓).\n\nFTC in Lean with substitution u = (σ−β)/δ: use `intervalIntegral.integral_comp_sub_deriv` or directly: the integrand f(σ) = 1/√((σ−β)²+δ²). Antiderivative F(σ) := arcsinh((σ−β)/δ): HasDerivAt F (1/√((σ−β)²+δ²))·(1/δ)·... deriv: d/dσ arcsinh((σ−β)/δ) = (1/√(((σ−β)/δ)²+1))·(1/δ) = 1/(δ·√(((σ−β)²+δ²)/δ²)) = 1/(δ·(1/δ)·√((σ−β)²+δ²)) = 1/√((σ−β)²+δ²) ✓ (δ > 0). So `intervalIntegral.integral_eq_sub_of_hasDerivAt` with F' = integrand ✓ + continuity. This is CLEANER than the split (no case analysis!). Then arcsinh bound + log bound — all real-analysis on the result.\n\nThe swap Σ∫ = ∫Σ: `intervalIntegral.integral_finset_sum` (needs IntegrableOn each) + norm: ∫ ‖Σ f_ρ‖ ≤ ∫ Σ‖f_ρ‖ (intervalIntegral.norm_integral_le_of_norm_le: ‖∫ F‖ ≤ ∫ ‖F‖ — hmm: `intervalIntegral.norm_integral_le_of_norm_le` : ‖∫ f‖ ≤ ∫ ‖f‖ needs f Integrable + ‖f‖ Integrable... the name might be `intervalIntegral.norm_integral_le_integral_norm` ✓ that exists). Then ∫ Σ‖f_ρ‖ = Σ ∫‖f_ρ‖ (integral_finset_sum on the norms) ✓.\n\nThe R-part: ∫_{-5/2}^{3} ‖R(σ+iT)‖ dσ ≤ 5.5·(10.5 log(T+2)+48): norm_integral_le_of_norm_le_const (const majorant) ✓ — needs the R-bound per σ: the tight capstone gives ‖logDeriv − Σnear‖ ≤ 10.5·log(|t|+2)+48 with t = T, |t| = T (T ≥ 316 > 0: |T| = T ✓).\n\nThen ‖logDeriv‖ ≤ ‖Σnear‖ + ‖R‖ (norm: ‖a‖ = ‖(a−b)+b‖ ≤ ‖a−b‖+‖b‖): ∫ ‖logDeriv‖ ≤ ∫‖Σnear‖ + ∫‖R‖ (integral_add of integrable pieces + integral_mono... actually ∫‖logDeriv‖ ≤ ∫(‖Σ‖+‖R‖) = ∫‖Σ‖+∫‖R‖ ✓ via intervalIntegral.integral_mono_ae or norm... — `intervalIntegral.integral_mono_on` for real-valued: ‖logDeriv (σ+iT)‖ ≤ ‖Σnear‖ + ‖R‖ pointwise ⟹ ∫ ≤ ∫ (‖Σ‖+‖R‖) = ∫‖Σ‖ + ∫‖R‖ ✓).\n\nAnd ‖kFun·logDeriv‖ ≤ (e⁴+1)·‖logDeriv‖ ⟹ ∫‖Jlag‖ ≤ (e⁴+1)·∫‖logDeriv‖ ✓.\n\nAll right — the plan is complete and well-formed. Also for 6-c-ii (I_III), the capstone for t ∈ [10, T]: ‖logDeriv ζ(−5/2+it) − Σ_{nearZerosSub t} mult/(s−ρ)‖ ≤ 10.5 log(t+2)+48; ‖Σ‖ ≤ Σ mult/‖s−ρ‖ ≤ (2/5)·Σ mult ≤ (2/5)·Z2hat(t) ≤ (2/3)·Z2hat(T) hmm (2/5)·Z2hat(t) ≤ (2/3)·Z2hat(T) ✓ (2/5 ≤ 2/3 and Z2hat(t) ≤ Z2hat(T) for t ≤ T via Z2hat_mono — NEW lemma needed (C0hat_mono exists; Z2hat has the same structure: 0.6·log + 0.886·loglog + 2.7 — clone C0hat_mono's proof).\n\nΣ_{nearZerosSub t} mult ≤ Z2hat t: bridge nearZerosSub t (subtype finset) → zeroAbsWindow t (set-finset): Σ_{nearZerosSub t} zeroMult ρ.1 ≤ Σ_{zeroAbsWindow toFinset} zeroMult ≤ Z2hat t (hZC2 t (10 ≤ t)). The subtype→set bridge: near_sum_refPoint_le (LiTightR:148) does EXACTLY this bridge internally — let me check its proof to clone the bridge. From the grep: it bounds ‖Σ mult/(2+it−ρ)‖ ≤ Z2hat-form: it has hterm : ‖mult/(s₀−ρ)‖ ≤ mult (denominator ≥ 1) then Σ mult ≤ ... via hZC2. So inside there's the bridge Σ_{nearZerosSub t} (zeroMult ρ.1 : ℝ) ≤ Z2hat-closed-form. I'll extract/clone that as a standalone lemma `nearZerosSub_mult_sum_le` (or reuse via generalization). Let me read near_sum_refPoint_le fully when building 6-c-ii.\n\nn-range for the side bounds: I'll state them for n ≥ 10⁵ (the SfBoundProp range; T ∈ [√n, √n+1] ⟹ T ≥ 316) — this collapses all the little hypotheses (T ≥ 10, √n ≥ 3, n ≥ 16, n ≥ 5, n ≥ 4).\n\nOK. Time to execute. Order:\n1. Log F-E in D-074 §1 progress log + a short §9 (finding + repair decision); update STATE.md (6-c sub-stage map + F-E note).\n2. 6-c-i scaffolding in LiSideBounds.lean; build-check incrementally.\n\nLet me do the doc updates first (crash-safe), then start the Lean file.\n\nActually — the todo says 6-c-i in progress. Doc-logging is part of crash-safety. Let me do it quickly.","signature":""},{"type":"text","text":"Resuming — the 4.56-in-the-statement finding kills the \"fatten IIMaj\" repair (the certified product would become 6.4×1.2237×0.2124×1.1099 = 1.847 > 4/3, breaking the convexity bracket). But there's a much better repair: **the fold drops a free 2π factor** (RectangleIntegral' is 2π-normalized; Bn charges the raw side bounds, D-071 §4.5). So charging I_I honestly as `e^{1/3}·IIMaj` costs nothing: `e^{1/3} ≤ 1.4 ≤ 2π`, absorbed with factor ~4.5 to spare — Bn, IIMaj, convexity, pinch all untouched. Logging Finding F-E (crash-safe) before building."},{"type":"tool_use","id":"Bash_34","name":"Bash","input":{"command":"cd /d/Python/Quantum && tail -30 research/derivations/D-074-dogleg-repair.md","description":"See D-074 progress log tail for append format"},"extra_content":null}]