[{"type":"thinking","thinking":"DL-4's kernel lemma requires `n ≤ t²` — but for I_I, t ranges over [−T, T] including t near 0! The condition t² ≥ n FAILS for |t| < √n.\n\nHmm. So for I_I (the right vertical of the MAIN rectangle, from −T to T at re = 2√n), the kernel bound e^{√n/2}+1 needs to hold for ALL t ∈ [−T, T], including t = 0.\n\nLet me redo the math for t = 0: |1 + 1/(w−1)| with w = 2√n: |1+1/(2√n−1)| = 1 + 1/(2√n−1). (1+1/(2√n−1))ⁿ ≤ e^{n/(2√n−1)}. n/(2√n−1) vs √n/2: n/(2√n−1) = √n·√n/(2√n−1). For √n = x: x²/(2x−1) vs x/2 ⟺ 2x² vs x(2x−1) = 2x²−x ⟺ 0 vs −x: so x²/(2x−1) > x/2 — the exponent is BIGGER than √n/2. By how much: x²/(2x−1) = x/2 · 1/(1−1/(2x)) ≈ x/2·(1+1/(2x)) = x/2 + 1/4. So e^{n/(2√n−1)} ≤ e^{√n/2+1/4+o(1)}.\n\nHmm — so at t = 0 the honest bound is e^{√n/2+~1/4}, NOT e^{√n/2}. IIMaj charges (e^{√n/2}+1). Factor e^{1/4} ≈ 1.284 over the charge. Does the fold survive? IIMaj n = 2(√n+1)(e^{√n/2}+1)·0.57·2^{−(2√n−2)}. The true bound would be 2T·(e^{√n/2+~1/4}+1)·0.57·2^{−(2√n−2)} ≈ e^{1/4}·IIMaj ≈ 1.284·IIMaj. The margin has IIMaj with coefficient 1.5 total (1 + 2×(1/4)) — the Bn charge is IIMaj·1 and DogLegRight includes IIMaj/4 each side. Hmm wait — but the convexity used exactly 1.5·IIMaj with the 1/(3n) bracket. If I_I's true bound is c·IIMaj with c ≈ 1.284, the total IIMaj coefficient becomes c + 0.5 ≈ 1.784, and the convexity needs coef·(1/(3n)) ≤ 2? The marginTilde_convex's IIM row: the budget was vIIM·(IIMaj″) with IIMaj″ ≤ 1/(3n) and the row sum needing ≤ 2 total... Actually the convexity inequality: margin″ ≥ 0 needs ψ(F)−blocks (negative) + Bn″ ≤ something. Let me not spiral — the coefficient 1.5 was chosen because DogLegRight = ... + IIMaj/4 per side ×2 = IIMaj/2, plus I_I's 1× IIMaj = 1.5 total. If I_I actually costs 1.284×IIMaj, total = 1.784, and 1.784·(1/(3n)) vs the budget 1/(2n)·...: 1.784/3 = 0.5947 vs 1/2 = 0.5 — EXCEEDS the budget! That would break the convexity re-proof (the hbridge : 1.5·(1/(3n)) = 1/(2n)).\n\nHold on — wait. Let me recompute more carefully. Is the true kernel bound at t=0 really e^{√n/2+1/4}?\n\n|w/(w−1)| at w = 2√n (t=0): = 2√n/(2√n−1) = 1/(1−1/(2√n)). Raised to n: exp(n·log(1/(1−1/(2√n)))) = exp(n·(1/(2√n) + 1/(2(2√n)²) + ...)) = exp(√n/2 + 1/8 + O(1/√n)). So indeed ≈ e^{√n/2+1/8}.\n\nHmm wait: n·log(1/(1−u)) with u = 1/(2√n): log(1/(1−u)) = u + u²/2 + u³/3 + ... = 1/(2√n) + 1/(8n) + ... times n = √n/2 + 1/8 + O(1/√n). Yes — e^{√n/2+1/8}.\n\nBut WAIT — the DL-4 lemma says the bound e^{√n/2}+1 holds for n ≤ t². And DL-6's vIntegral_Jlag_right_le_IIMaj_quarter proves V_right ≤ IIMaj/4 where V_right spans t ∈ [T, U] — t ≥ T ≥ √n so t² ≥ n ✓ the DL-4 lemma applies. But the MAIN rectangle's right vertical I_I spans t ∈ [−T, T] — DL-4 does NOT apply near t = 0.\n\nHow did the ORIGINAL D-036/D-071 design handle this? D-071 §4.4 I_I: \"kernel ≤ e^{√n/2}+1 (kFun_shift_norm_le_exp_sqrt_half_add_one ✓)\". The ii′-4 lemma kFun_shift_norm_le_exp_sqrt_half_add_one is stated at (s+1) with re s ≥ 2√n — i.e., at w = s+1 with re w ≥ 2√n+1. At the vertical re w = 2√n that lemma does NOT apply either (off by one: re w = 2√n means re s = 2√n−1 < 2√n).\n\nHmm, so was the original I_I design ALSO broken at the boundary? Let's check: ii′-4's lemma at w with re w ≥ 2√n+1: |w/(w−1)| ≤ 1+1/|w−1|, |w−1| ≥ re w − 1 ≥ 2√n ⟹ ≤ (1+1/(2√n))ⁿ ≤ e^{√n/2} ✓. So the ii′-4 lemma is for re w ≥ 2√n+1, but the rectangle's right edge is at re w = 2√n. So even the ORIGINAL design had an off-by-one here?? \n\nLet me recheck: is the rectangle's right edge at 2√n or 2√n+1? From LiAssembly: rectW n T = 2√n + iT. And from D-036 variant B: w = 2√n + iT? The original D-036 contour: the rectangle is [−5/2, σ₁] × [−T, T] with σ₁ = 2√n... From D-071: \"I_I (right vertical σ₁ = 2√n)\". And the ii′-4 lemma proved for re s ≥ 2√n at the SHIFTED argument s+1 — so the lemma applies at points w = s+1 with re w ≥ 2√n + 1.\n\nSo there's a genuine off-by-one: the design cites kFun_shift_norm_le_exp_sqrt_half_add_one for the vertical at re = 2√n, but that lemma needs re ≥ 2√n+1 (after the shift), and DL-4's direct variant needs t² ≥ n.\n\nAt t = 0 the true bound is e^{√n/2+1/8}·(1+o(1)) — so the honest charge is e^{1/8}·IIMaj-ish. OR — redo IIMaj's definition to absorb it? NO — IIMaj is load-bearing in the convexity (coefficient 1.5 with the exact 1/(2n) bridge). Changing IIMaj's definition would re-open ii′-6-a2.\n\nOptions:\n(a) Bound the kernel on I_I by e^{√n/2+1/4}+1 honestly: (1+1/(2√n−1))ⁿ ≤ e^{n/(2√n−1)} ≤ e^{√n/2+1/4} for n ≥ ?: n/(2√n−1) ≤ √n/2 + 1/4 ⟺ n ≤ (√n/2+1/4)(2√n−1) = n − √n/2 + √n/2 − 1/4 = n − 1/4. FALSE — n > n − 1/4. So e^{n/(2√n−1)} > e^{√n/2+1/4}?? wait: (√n/2+1/4)(2√n−1) = n − √n/2 + (2√n)/4 − 1/4 = n − √n/2 + √n/2 − 1/4 = n − 1/4 < n. So n/(2√n−1) > √n/2 + 1/4. Hmm so e^{√n/2+1/4} is NOT an upper bound either! Need +1/8+log-correct... n·log(1/(1−u)) ≤ n·(u + u²) for u ≤ 1/2 (log(1/(1−u)) ≤ u + u² for u ∈ [0, 1/2]: since log(1/(1−u)) = u + u²/2 + u³/3+... ≤ u + u²(1/2+1/4+...) = u + u² ✓ for u ≤ 1/2 hmm: u³/3 + u⁴/4 + ... ≤ u³/3·(1+u+u²+...) = u³/(3(1−u)) ≤ 2u³/3 ≤ u²/2·(4u/3) ≤ u²/2 when u ≤ 3/8... roughly). n·u = √n/2, n·u² = 1/4. So n·log ≤ √n/2 + 1/4 for u ≤ 1/2-ish ⟹ the e^{√n/2+1/4} bound via the LOG form (not the linear 1+x ≤ e^x form). The linear form gives n/(2√n−1) which is too big; the log form gives n·log(1+1/(2√n−1)) = n·log(2√n/(2√n−1)) = n·log(1/(1−1/(2√n))) ≤ n·(u+u²) = √n/2 + 1/4 ✓ with u = 1/(2√n) ≤ 1/4 for n ≥ 4... need log(1/(1−u)) ≤ u + u² valid for u ≤ some u₀. Let me verify: f(u) = u + u² + log(1−u), f(0) = 0, f′(u) = 1 + 2u − 1/(1−u) = ((1+2u)(1−u) − 1)/(1−u) = (1 + u − 2u² − 1)/(1−u) = u(1−2u)/(1−u) ≥ 0 for u ∈ [0, 1/2]. ✓ So log(1/(1−u)) ≤ u + u² for u ∈ [0, 1/2]. So kernel ≤ e^{√n/2+1/4} + 1 for n ≥ 1 (u = 1/(2√n) ≤ 1/2 ✓). \n\n(b) Then I_I ≤ 2T·(e^{√n/2+1/4}+1)·0.57·2^{−(2√n−2)} = e^{1/4}·IIMaj·[T/(√n+1)]·[(e^{√n/2+1/4}+1)/(e^{√n/2}+1)] ≤ e^{1/4}·IIMaj. So the honest I_I charge is e^{1/4}·IIMaj ≈ 1.284·IIMaj.\n\nThen the total IIMaj coefficient in Bn: e^{1/4} + 2·(1/4) = e^{1/4} + 1/2 ≈ 1.784. The convexity bridge needs coef·(1/(3n)) ≤ 1/(2n)·... the budget: marginTilde_convex's IIM row contributes −coef·IIMaj″ ≥ −coef·(1/(3n))... hmm wait actually let me look at how the convexity used the 1.5: hbridge : (1.5)·(1/(3n)) = 1/(2n). The total budget for all blocks was 2 (the sInftyLower leading term gives n·log n whose second derivative contributes...). The ψ-table sums negative contributions and the IIM row's ψ[φ](T)/(4n) + 11-block table... I don't remember the exact structure. The summary says \"vId/vIIM literals updated; hnn4 new coefficient; final linarith with hIIM (lt_three) + hbridge : (1.5)·(1/(3n)) = 1/(2n)\". So the budget for the IIM row was exactly 1/(2n) and 1.5·(1/(3n)) = 1/(2n) EXACTLY saturates it. With coefficient 1.784: 1.784/(3n) = 0.5947/n > 0.5/n — BREAKS.\n\nHmm. But wait — maybe there's headroom elsewhere (the vId row got a new coefficient with slack?). This is getting deep into the convexity proof. Alternative:\n\n(c) Split I_I into |t| ≤ √n and √n ≤ |t| ≤ T:\n  - For √n ≤ |t| ≤ T: DL-4's kernel bound e^{√n/2}+1 applies (t² ≥ n ✓). Zeta: 0.57·2^{−(2√n−2)}. Length 2(T−√n) ≤ 2.\n  - For |t| ≤ √n: kernel ≤ e^{√n/2+1/4}+1 (the t=0 worst case, by the log-form bound which is uniform in t! |1+1/(w−1)| = sqrt(1 + 2·re(w)/(normSq(w−1))−...) hmm — actually |1+1/(w−1)|² = 1 + (2·re(w−1)+1)/|w−1|² = 1 + (2re(w)−1)/((2√n−1)²+t²). For fixed re, this DECREASES in t². So the worst case is t = 0: |w/(w−1)| = 2√n/(2√n−1) exactly. So the uniform bound (2√n/(2√n−1))ⁿ ≤ e^{√n/2+1/4} holds for ALL t ✓ — the t=0 case is the max.\n  \n  So actually the honest uniform kernel bound on the whole I_I is e^{√n/2+1/4}+1, no split needed. The charge is e^{1/4}·IIMaj·(T/(√n+1)) ≤ e^{1/4}·IIMaj.\n\n(d) OR: sharpen by noting the I_I integral with the ACTUAL kernel: ∫_{-T}^{T} (2√n/√((2√n−1)²+t²))ⁿ dt — the t=0 peak is narrow (width ~√(2√n)/√n...) — the integral might be much smaller than length×max. But that's a refinement; the crude charge is e^{1/4}IIMaj.\n\n(e) OR: redefine IIMaj to use the honest kernel constant? IIMaj is used in: Bn (coefficient 1), DogLegRight (IIMaj/4), the convexity (coefficient 1.5), base points, E-034. If IIMaj's 0.57 became 0.57·e^{1/4} ≈ 0.732: then V_right ≤ IIMaj/4 still holds (DL-6's bound has slack — DL-6 used dirichletCX ≤ 0.57 with the direct bound... the zeta side is unchanged; the kernel side for V_right uses DL-4 (t² ≥ n ✓ valid) — so V_right's true bound is unchanged, and IIMaj/4 with the fatter IIMaj still bounds it ✓). I_I ≤ IIMaj ✓ by construction. Convexity: coefficient stays 1.5 ✓✓✓. Base points: IIMaj is tiny (2^{−2√n}·e^{√n/2} — at n=8e4: 2√n ≈ 565.7, 2^{−563.7} — ASTRONOMICALLY small; e^{1/4} factor irrelevant at the certified precision? The base-point proofs bound IIMaj ≤ some tiny rational — e.g. IIMaj/4 ≤ 2590.08/4?? wait that was DogLegRight's IIMaj/4 ≤ 2590/4?? That seems big... hmm, from the summary: \"hDL : 2·DogLegRight 80000 ≤ 3184 with certified DogLegRight computation (div_le_div_of_nonneg_right outer /2, mul_le_mul with hT2 and exp_five_halves_le, IIMaj/4 ≤ 2590.08/4)\". Hmm, IIMaj(8e4)/4 ≤ 2590/4? IIMaj(8e4) = 2(√8e4+1)(e^{√8e4/2}+1)·0.57·2^{−(2√8e4−2)}: √8e4 ≈ 282.84; e^{141.4} ≈ 4×10⁶¹; 2^{−563.7} ≈ 10^{−169.7}. So IIMaj ≈ 2·283.8·4×10⁶¹·0.57·10^{−169.7} ≈ 3.7×10^{−106}. So IIMaj/4 ≤ 2590/4 is a VERY loose bound (probably they bounded (e^{√n/2}+1)·2^{−(2√n−2)} ≤ something loose like 1 and then 2(√n+1)·0.57 ≤ 2·283.8·0.57... no that's 323.6. Hmm, 2590.08: 2·(√8e4+1)·(e^{√8e4/2}+1)·0.57·2^{−(2√8e4−2)}/4 — no. Whatever — the point: if IIMaj's definition changes 0.57 → 0.732, the base-point bounds' certified numerics must be re-checked (the loose bounds almost certainly still hold — the slack is 10¹⁰⁰×). And the iimaj_bracket_bound < 4/3 (the factor-monotone product 1.3155) — IIMaj's bracket: IIMaj(n+1)/IIMaj(n)·... the bracket bound compares IIMaj at consecutive... Actually iimaj_bracket_bound: IIMaj(n)·(4n)...? It's the certified product 4.56×1.2237×0.2124×1.1099 = 1.3155 < 4/3 where 0.2124 ≈ ... The factor 0.57 appears inside one of those factors? If IIMaj grows by e^{1/4}, one factor grows by ≤ e^{1/4}·(something) — the product 1.3155·e^{1/4} = 1.69 > 4/3 = 1.333 — BREAKS the bracket!\n\nHmm wait, no: iimaj_bracket_bound is about the RATIO structure IIMaj(n)/(n+1)... I need to look. Actually the bracket: IIMaj_deriv2 ≤ (4/3)·IIMaj/(4n)... the 4/3 came from IIMaj(n)·... Let me look at the actual statement: `iimaj_bracket_bound ... < 4/3` and `IIMaj_deriv2_lt_three : deriv (deriv IIMaj) n < 1/(3n)`. The deriv2 of IIMaj ~ IIMaj·(stuff)/(4n)-ish; the bracket < 4/3 relates deriv2·(4n)/IIMaj < 4/3. If IIMaj's constant 0.57 → 0.732, IIMaj and its derivatives BOTH scale by the same factor ⟹ the RATIO (deriv2·4n/IIMaj) is UNCHANGED ⟹ the bracket bound is invariant!! Because the bracket is a ratio of IIMaj-derivatives to IIMaj. YES — constant factors cancel in the logarithmic-derivative-type bracket. Let me verify: IIMaj(n) = 2(√n+1)(e^{√n/2}+1)·C·2^{−(2√n−2)} with C = 0.57. deriv2 scales linearly in C; the bracket divides by IIMaj ⟹ C cancels ✓. So changing 0.57 → 0.57·e^{1/4} leaves iimaj_bracket_bound and IIMaj_deriv2_lt_three PROOF-UNCHANGED (they're ratio statements... unless the proof literally computes with 0.57 — the certified product 4.56×1.2237×0.2124×1.1099: is 0.57 in there? If the proof's algebra cancels C symbolically, no change needed. If C appears in the factors, it must cancel there too (since the bound is C-free: 4/3). 0.2124... hmm 0.57·e^{−2log2·...}? Whatever — if C canceled in the original proof it cancels now.)\n\nHmm, but actually — wait. Changing IIMaj's definition also affects: DogLegRight (IIMaj/4 — the DL-6-ii theorem `vIntegral_Jlag_right_le_IIMaj_quarter` proves V_right ≤ IIMaj/4 with the CURRENT IIMaj. If IIMaj fattens, the theorem still holds (V_right ≤ old-IIMaj/4 ≤ new-IIMaj/4 — but the PROOF's final accounting `e4 : IIMaj/4 = ((√n+1)/2)(e+1)·0.57·2^{−(2√n−2)}` by unfold/ring would BREAK — the equality is definitional!). So the DL-6-ii proof's e4 would need updating (0.57 → new constant). Mechanical but touches LiDogLeg.\n\nAlternatively (f): DON'T touch IIMaj. Instead bound I_I ≤ e^{1/4}·IIMaj and charge Bn's I_I line as e^{1/4}·IIMaj instead of IIMaj. Then Bn changes: `... + 2·DogLegRight n + (e^{1/4})·IIMaj n + 1` — wait but then convexity coefficient: 1.5 → e^{1/4}+1/2 ≈ 1.784, breaking the exact 1/(2n) bridge. Unless the bridge has slack... it doesn't (exact equality).\n\n(g) Sharpen the I_I bound below IIMaj by using the actual structure: is e^{1/4} avoidable? The zeta side: norm_logDeriv_zeta_le_iimZeta ≤ 0.57·2^{−(2√n−2)}. The 0.57 has slack? DL-6's 0.57: dirichletCX ≤ 3^{−(2√n−2)}·dirichletC2 + log-terms... the 0.57 = 9·0.164·... + log 2 ≈ ... it's honest-ish. Kernel honest max: (2√n/(2√n−1))ⁿ = e^{√n/2+1/8+O(1/√n)}. So the product honest bound: e^{1/8}·[2(√n+1)(e^{√n/2}+1)·0.57·2^{−(2√n−2)}]·(1+o(1)) = e^{1/8}·IIMaj·(1+o(1)). e^{1/8} ≈ 1.133. So even the SHARP uniform bound is ≈1.13·IIMaj > IIMaj. The charged IIMaj is simply too small for the t=0 kernel. This was a latent defect in the ORIGINAL D-036 charge too (the original B(n) charged IIMaj = ... 0.57 ... with the same e^{√n/2}+1 kernel claim citing the off-by-one lemma!).\n\nHmm hold on — let me double-check the original D-036/D-037 charge for I_I. Bn's line: `IIMaj n` with IIMaj = 2(√n+1)(e^{√n/2}+1)·0.57·2^{−(2√n−2)}. And D-071 §4.4 I_I cites \"kernel ≤ e^{√n/2}+1 (kFun_shift_norm_le_exp_sqrt_half_add_one ✓)\". So the defect predates D-074 — it's in the original design chain (E-023/E-028 validated the MARGIN numerically, not the I_I kernel lemma applicability). Actually wait — did E-028/E-030 validate the I_I piece against the true integral? E-027/E-028 were about margins; the I_I piece bound was \"validated\" only through the lemma citation. And E-031 (the F-D diagnostic) was about the horizontal. Hmm — actually E-032 measured the true horizontal integral; was there any measurement of I_I? Probably not (it's obviously tiny).\n\nSo this is Finding F-E (a minor sibling of F-A/F-B): the I_I kernel bound e^{√n/2}+1 does NOT apply at the vertical re = 2√n near t = 0 — the honest uniform bound is (2√n/(2√n−1))ⁿ ≤ e^{√n/2+1/4} (log form, u = 1/(2√n) ≤ u+u² for n ≥ 1... wait need u ≤ 1/2: 1/(2√n) ≤ 1/2 ⟺ n ≥ 1 ✓). The repair: charge I_I honestly.\n\nThe cleanest repair that does NOT touch IIMaj/convexity/DL-6: prove I_I ≤ e^{1/4}·IIMaj, then change Bn's I_I term from `IIMaj n` to `Real.exp (1/4) * IIMaj n`. But the convexity coefficient becomes e^{1/4}+0.5...\n\nLet me look at the actual convexity budget more carefully — maybe there IS slack. The mIIM row: mIIM := fun m => (−1.5)·IIMaj m. In the ψ-table the row needs: the second derivative contribution −1.5·IIMaj″(n) ≥ ... the chain rule: margiñ″(n) = [stuff] − 1.5·IIMaj″(n)·... hmm actually marginTilde is a function of n directly (not through T for the IIMaj piece — IIMaj n is a function of n). The convexity proof: marginTilde″ ≥ 0. The IIMaj piece contributes −1.5·IIMaj″(n). IIMaj″ ≤ 1/(3n) (the lt_three theorem — WAIT: IIMaj_deriv2_lt_three : deriv (deriv IIMaj) n < 1/(3n) — the bound doesn't even have IIMaj on the RHS! It's an ABSOLUTE bound: IIMaj″(n) < 1/(3n). Since IIMaj is astronomically small, IIMaj″ is too; 1/(3n) is a loose absolute bound. The 4/3-bracket: deriv2 IIMaj n < (4/3)·IIMaj(n)·(1/(4n))?? and then IIMaj(n) ≤ 1 gives < 1/(3n)?? Hmm: (4/3)/(4n) = 1/(3n) ✓ — so the chain is deriv2 IIMaj < (4/3)·IIMaj(n)/(4n) ≤ (4/3)·1/(4n) = 1/(3n), using IIMaj(n) ≤ 1!! So the bracket bound < 4/3 IS a ratio (C-invariant) but the lt_three step uses IIMaj(n) ≤ 1.\n\nSo: if IIMaj fattens by e^{1/4} (0.57 → 0.732), then: (i) bracket < 4/3 UNCHANGED (ratio) ✓; (ii) lt_three needs IIMaj(n) ≤ 1 — at the fatter constant: IIMaj(8e4)-ish ≤ 1? IIMaj(n) for n ≥ 8e4 is ~10^{-106} ✓✓ but IIMaj is used for n ≥ 10⁵ in the margin... and the lt_three theorem's domain: probably n ≥ 8×10⁴ or n ≥ 1? If the proof shows IIMaj(n) ≤ 1 for n in some range, the e^{1/4} factor keeps it ≤ 1 trivially (10^{-106}·1.284). But IIMaj(small n) — e.g. IIMaj(16) = 2·5·(e²+1)·0.732·2^{−6} = 10·8.39·0.732/64 ≈ 0.96 — borderline but the margin domain is n ≥ 8e4 ✓.\n\nHmm OK. So option (e'): fatten IIMaj's constant 0.57 → 0.57·e^{1/4} — wait, NO. Let me reconsider. The cleaner option: keep IIMaj as-is; charge I_I ≤ e^{1/4}·IIMaj; change the Bn I_I term to e^{1/4}·IIMaj; and handle the convexity coefficient 1.5 → e^{1/4}+1/2.\n\nActually — hold on. Let me reconsider whether we even need the uniform-in-t bound. The I_I integral: ∫_{-T}^{T} ‖kFun‖·‖logDeriv‖ dt. The kernel's t-dependence: |w/(w−1)|ⁿ with w = 2√n+it: |w|²/|w−1|² = ((2√n)²+t²)/((2√n−1)²+t²). At t=0: (2√n/(2√n−1))². As |t| grows the ratio → 1. The exponent: n·log(|w|/|w−1|)... the integral ∫ exp(n·log(...)) dt — Laplace-type: peak at t=0 of width ~? n·log(|w|/|w−1|) ≈ n·(re-diff)/(...) ≈ n·(2√n·(4√n−2).../... let me just: log(|w|²/|w−1|²)/2 = (1/2)log(1 + (4√n−1)/((2√n−1)²+t²)) ≈ (4√n−1)/(2((2√n−1)²+t²)). At t=0: ≈ (4√n)/(2·4n) = 1/(2√n)·... times n = √n/2 ✓. Width: the t² denominator — (2√n−1)² + t², so width ~ 2√n−1 ~ 2√n. The Gaussian-ish integral: e^{√n/2}·∫ exp(−n·(4√n)t²/(2(2√n)⁴))dt ~ e^{√n/2}·√(π/(n·(4√n)/(2·16n²))) = e^{√n/2}·√(π·32n²/(4n√n)) = e^{√n/2}·√(8πn/√n)... ~ e^{√n/2}·√(8π)·n^{1/4}. So the integral ~ e^{√n/2}·n^{1/4}·const vs the crude charge 2T·e^{√n/2} ~ 2√n·e^{√n/2}. So the TRUE integral is smaller than IIMaj by a factor ~n^{1/4}/(2√n) → 0! So a Laplace-analysis would give I_I ≤ IIMaj easily (even ≪). But formalizing Laplace is heavy.\n\nSimplest honest fix preserving everything: **charge I_I ≤ e^{1/4}·IIMaj and set the Bn I_I term to e^{1/4}·IIMaj**, then re-do the convexity row with coefficient e^{1/4}+1/2. Budget check: the IIM row needs coef·IIMaj″ ≤ coef·(1/(3n)) vs the allocated 1/(2n). coef = e^{1/4}+1/2 ≈ 1.784: 1.784/(3n) = 0.5947/n > 0.5/n ✗ BREAKS as computed.\n\nHmm. What IS the actual budget structure? Let me look at the marginTilde_convex proof to see where the 1/(2n) came from and whether other rows have slack. Actually — the total budget: margiñ″(n) = (1/2)·log n + (1/2) + (c₁) − Bn″(n) and the ψ-table shows Bn″(n) ≤ [something]/(4n)-ish... I really need to read the proof. Let me look at marginTilde_convex in LiSkeleton.\n\nActually, let me step back. Alternatives:\n\n(A) **Split I_I at |t| = √n**: \n  - Outer (√n ≤ |t| ≤ T): DL-4 kernel e^{√n/2}+1 ✓, zeta 0.57·2^{−(2√n−2)}, length 2(T−√n) ≤ 2 ⟹ ≤ 2·(e^{√n/2}+1)·0.57·2^{−(2√n−2)} = IIMaj/(√n+1) ≤ IIMaj/... — IIMaj = 2(√n+1)(...) so 2·(...) = IIMaj/(√n+1). \n  - Inner (|t| ≤ √n): kernel ≤ e^{√n/2+1/4}+1 ≤ 2·e^{√n/2+1/4}... zeta same, length 2√n ⟹ ≤ 2√n·(e^{√n/2+1/4}+1)·0.57·2^{−(2√n−2)} = e^{1/4}·[2√n(e^{√n/2}+1)·0.57·2^{−(2√n−2)}] = e^{1/4}·IIMaj·√n/(√n+1) ≤ e^{1/4}·IIMaj.\n  Total: ≤ IIMaj·(e^{1/4}√n + 1)/(√n+1) ≤ e^{1/4}·IIMaj. Same. No gain.\n\n(B) **Use a better zeta bound on the inner piece**: norm_logDeriv_zeta_le_iimZeta is uniform; no t-improvement available (Dirichlet series at σ = 2√n is t-independent-ish).\n\n(C) **Absorb the e^{1/4} into the 0.57**: I_I ≤ 2T·(e^{√n/2+1/4}+1)·0.57·2^{−(2√n−2)}. Note e^{√n/2+1/4}+1 ≤ e^{1/4}(e^{√n/2}+1) ✓ (e^{1/4} ≥ 1). So I_I ≤ e^{1/4}·IIMaj. Equivalently 0.57·e^{1/4} = 0.7322...: I_I ≤ IIMaj' where IIMaj' = IIMaj with 0.7322. \n\n(D) **Redefine IIMaj with the honest constant 0.7322** (0.57·e^{1/4} — or certified-rational-friendly: e^{1/4} ≤ 1.2841 ⟹ 0.732): \n  - I_I ≤ IIMaj_new ✓ (kernel honest + T ≤ √n+1).\n  - V_right (DL-6-ii): TRUE bound unchanged (DL-4 kernel valid t² ≥ n) — the old bound ≤ old-IIMaj/4 ≤ new-IIMaj/4 ✓ — BUT the Lean proof `vIntegral_Jlag_right_le_IIMaj_quarter` ends with the DEFINITIONAL accounting e4 : IIMaj/4 = ((√n+1)/2)(e+1)·0.57·2^{−(2√n−2)}. If IIMaj's constant changes, e4 breaks and must be repaired: V_right ≤ ((√n+1)/2)(e+1)·0.57·2^{−...} ≤ ((√n+1)/2)(e+1)·0.732·2^{−...} = new-IIMaj/4 — an extra ≤ step (0.57 ≤ 0.732) instead of the rfl-equality. MECHANICAL fix in LiDogLeg (one lemma).\n  - iimaj_bracket_bound < 4/3: RATIO — but is the proof literally constant-free? The certified product 4.56×1.2237×0.2124×1.1099: if 0.57 enters factor 0.2124 (e.g. 0.57·e^{−something}): 0.2124/0.57 = 0.3726... hmm e^{−1} = 0.3679, close but not exact. If the factor contains 0.57 it must ALSO contain its reciprocal (since the bracket is C-free) — need to read the proof to know. If the proof computes deriv2/IIMaj·(4n) symbolically with C cancelling, the literal 0.57 never appears and the proof is untouched. Let me check iimaj_bracket_bound's proof.\n  - IIMaj(n) ≤ 1 (used in lt_three): fatter IIMaj still ≤ 1 for n ≥ domain ✓ trivially (10^{-106}).\n  - Base points: IIMaj/4 ≤ 2590.08/4 — wait, that bound seems way too big for IIMaj(8e4)/4 ~ 10^{-107}. OH — I bet the base-point proof doesn't compute IIMaj tightly at all; it probably uses a crude universal bound like IIMaj(n) ≤ 10360 for n ≤ ... no. Let me look: \"IIMaj/4 ≤ 2590.08/4\" — hmm, maybe they bound IIMaj(8e4) ≤ 2590.08·... whatever: any certified bound with slack ≫ e^{1/4} survives the constant change — but the PROOF TEXT of the bound cites IIMaj's definition (unfold IIMaj; the 0.57 appears in the arithmetic). The arithmetic steps with 0.57 → 0.732: if the certified bound has slack > 1.284×, the same proof script goes through with literal replacements (the linarith/norm_num steps auto-adjust... no — the script has hardcoded intermediate rationals; they'd need re-derivation if tight, auto-pass if the final linarith has slack and intermediates are inequalities). Need to look.\n  - marginExpanded's mIIM block: (−1.5)·IIMaj — the bookkeeping (marginTilde_eq_expanded) is definitional ✓ unchanged; convexity coefficient stays 1.5 ✓✓; the mIIM row in the ψ-table: −1.5·IIMaj″(n) ≥ −1.5/(3n) = −1/(2n) ✓ UNCHANGED (the lt_three bound is on the NEW IIMaj — ratio-invariant + ≤ 1 check).\n  - E-034's fold model: the DogLegRight/IIMaj values change by ×1.284 of a 10^{-107} quantity — NOTHING measurable. The fold fits trivially.\n  \nSo option (D) — fatten IIMaj's constant to 0.732 (certified: 0.57·e^{1/4} ≤ 0.57·1.2841 = 0.731937 ≤ 0.732) — is the cleanest: ONE definition change + DL-6-ii's e4 accounting gets one extra ≤ step + re-check bracket/lt_three/base-points proofs for literal 0.57 appearances.\n\nWait wait. Hold on. Before going down this path — let me double-check that the I_I kernel bound is REALLY broken, i.e. that kFun_shift_norm_le_exp_sqrt_half_add_one really can't be applied. The lemma: (hn : 0 < n) (hσ : 2√n ≤ s.re) : ‖kFun n (s+1)‖ ≤ e^{√n/2}+1. For w := 2√n + it, we'd need s + 1 = w, s.re = 2√n − 1 — fails hσ by exactly 1. ✓ broken. And DL-4 needs t² ≥ n ✓ broken for |t| < √n. And the sharp analysis confirms the charged bound is genuinely exceeded at t = 0 by the factor e^{1/8}-ish > 1. So yes — genuine (if microscopic) defect. This is Finding F-E.\n\nActually — hmm, one more check: maybe the ORIGINAL D-036 contour had the right edge at 2√n+1, not 2√n? LiAssembly's rectW = 2√n. D-071: σ₁ = 2√n. So the off-by-one is real.\n\nHmm, but actually — WAIT. Let me reconsider. Maybe instead of fattening the constant, note that Bn charges I_I as IIMaj n — and the FOLD only needs the TOTAL to fit Bn. The I_I piece's overrun is ~0.13·IIMaj ~ 10^{-108} at n=10⁵ — while the horizontal piece (2(e⁴+1)·7Ĥ) has the pinch slack E-030 measured at ratio ≤ 0.40 — i.e., the horizontals use only 40% of their charge! The overrun 0.13·IIMaj vs horizontal slack 0.6·2(e⁴+1)·7Ĥ ~ astronomically larger. But the fold in Lean is per-piece (each piece ≤ its charge), not cross-subsidized... The pinch inequality NI + 5.5(...) ≤ 7Ĥ has a factor-2.5 slack; adding 0.13·IIMaj·... no — cross-subsidy requires re-deriving the pinch with the extra term. The pinch proof: NI + 5.5·(10.5 log+48) ≤ 7·Ĥ. If we instead prove NI + 5.5(...) + (e^{1/4}−1+0.5)·IIMaj·... no wait. Hmm: the fold: sides ≤ [I_III charge] + [horizontal: 2(e⁴+1)(NI+5.5...) + 2·DogLegRight] + [I_I: e^{1/4}IIMaj] + [bridge 4F+1]. Bn = [I_III charge] + [2(e⁴+1)·7Ĥ] + [2·DogLegRight] + [IIMaj] + [4F+1]. The only mismatches: horizontal NI+5.5(...) vs 7Ĥ (slack ≥ 1.5×) and I_I: e^{1/4} vs 1 (overrun 0.284·IIMaj). Cross-subsidy: need 2(e⁴+1)·(7Ĥ − NI − 5.5(...)) ≥ 0.284·IIMaj. The LHS ~ 2·55.6·(0.6·7Ĥ) ~ 66·Ĥ ~ 66·(3Ĉ₀²) ~ 200·L² vs RHS ~ 10^{-107} at n=10⁵ and DECAYING. So the subsidy fits with grotesque slack — but formalizing it means the pinch lemma becomes 2(e⁴+1)·NI-total + e^{1/4}IIMaj ≤ 2(e⁴+1)·7Ĥ + IIMaj, i.e. NI + 5.5(...) ≤ 7Ĥ − (e^{1/4}−1)·IIMaj/(2(e⁴+1)) — the pinch proof's margin dwarfs the subtracted term; the proof would need a tiny extra arithmetic step. This keeps IIMaj/Bn/convexity 100% untouched! But it couples the I_I overrun into the horizontal pinch — slightly ugly conceptually.\n\nvs option (D) — change Bn's I_I term to e^{1/4}·IIMaj: then Bn changes ⟹ marginTilde/marginExpanded lockstep (definitional ✓), convexity coefficient 1.5 → e^{1/4}+0.5 ✗ BREAKS the exact bridge. Dead end unless the bridge is re-worked (vIIM literal: 1.5·(1/(3n)) = 1/(2n); with 1.784: 1.784/(3n) ≤ 1/(2n) FALSE).\n\nvs option (D′) — fatten IIMaj's constant 0.57 → 0.732 (inside the definition): Bn's expression UNCHANGED (IIMaj n with the new def); convexity UNCHANGED (coefficient 1.5; lt_three via the ratio-invariant bracket + IIMaj ≤ 1 which survives); DL-6-ii's e4 gets one extra ≤ step; base-point proofs need a 0.57→0.732 literal check; marginExpanded mIIM block (−1.5·IIMaj) definitional ✓. This is the winner — IF the bracket proof is ratio-constant-free (must verify) and the base-point scripts survive (must verify).\n\nLet me also double check the NEW I_I bound proof with the fattened IIMaj: I_I ≤ 2T·(e^{√n/2+1/4}+1)·0.57·2^{−(2√n−2)} ≤ 2(√n+1)·(e^{√n/2}+1)·(e^{1/4}·0.57)·2^{−(2√n−2)} ≤ 2(√n+1)(e^{√n/2}+1)·0.732·2^{−(2√n−2)} = IIMaj_new n ✓ (using e^{√n/2+1/4}+1 ≤ e^{1/4}(e^{√n/2}+1) and 0.57·e^{1/4} ≤ 0.732; the kernel lemma: |w/(w−1)|ⁿ ≤ (2√n/(2√n−1))ⁿ ≤ e^{√n/2+1/4} — via normSq monotonicity in t² (worst at t=0) + the log-form n·log(1/(1−u)) ≤ n(u+u²) = √n/2 + 1/4 for u = 1/(2√n) ≤ 1/2. Hmm — formalizing log(1/(1−u)) ≤ u+u²: Real.log_le_sub_one_of_pos gives log x ≤ x−1; for log(1/(1−u)) = −log(1−u): need −log(1−u) ≤ u+u² ⟺ log(1−u) ≥ −u−u². Real.log_le_sub_one_of_pos at (1−u): log(1−u) ≤ −u — wrong direction. The other direction: log(1−u) ≥ −u/(1−u)? There are lemmas Real.neg_log_le... hmm. Or avoid logs: (1+1/(2√n−1))ⁿ ≤ e^{√n/2+1/4}: take the n-th root... or bound directly: (1 + 1/(2√n−1))ⁿ = exp(n·log(1+1/(2√n−1))) ≤ exp(n/(2√n−1)) via log(1+x) ≤ x ✓ (Real.log_le_sub_one_of_pos at 1+x: log(1+x) ≤ x ✓ EASY). But n/(2√n−1) ≤ √n/2 + 1/4?? computed above: n/(2√n−1) vs √n/2+1/4: multiply: 4n vs (2√n+1)(2√n−1) = 4n − 1: so n/(2√n−1) = 4n/(4(2√n−1)) and (√n/2+1/4) = (2√n+1)/4: n/(2√n−1) ≤ (2√n+1)/4 ⟺ 4n ≤ (2√n+1)(2√n−1) = 4n−1 ⟺ 0 ≤ −1 FALSE. Damn — off by an hair: 4n ≤ 4n−1 false. So n/(2√n−1) > √n/2+1/4 always. The 1+x ≤ e^x route overshoots by exactly the gap 4n vs 4n−1: n/(2√n−1) = √n/2·(2√n/(2√n−1)) = √n/2·(1+1/(2√n−1)) = √n/2 + √n/(2(2√n−1)) = √n/2 + 1/(4−2/√n) ≈ √n/2 + 1/4 + 1/(8√n)+... So e^{√n/2+1/4} misses by e^{1/(8√n)} → need e^{√n/2+1/4+ε}. With the log-form bound n·log(1+1/(2√n−1)) ≤ n·(u+u²) where the actual u here: log(1+1/(2√n−1)) = log(2√n/(2√n−1)). Set v := 1/(2√n−1): log(1+v) ≤ v ✓ gives n·v = n/(2√n−1) (too big). The u+u² form: log(2√n/(2√n−1)) = −log(1−u), u = 1/(2√n) ≤ u + u² needs −log(1−u) ≤ u+u². Is there a Mathlib lemma? `Real.log_le_sub_one_of_pos : 0 < x → log x ≤ x − 1`. For the lower bound on log(1−u)... `Real.log_le_sub_one_of_pos` at x = 1/(1−u)... circular. There's `Real.add_one_le_exp` : x + 1 ≤ e^x. From e^x ≥ 1+x: e^{−u−u²} ≥ 1 − u − u²... want 1−u ≤ e^{−u}·e^{−u²}?? Actually want −log(1−u) ≤ u + u² ⟺ 1−u ≥ e^{−u−u²}. Since e^{−x} ≥ 1 − x (from e^x ≥ 1+x at −x): e^{−u−u²} ≥ 1 − u − u². But 1−u ≥ 1−u−u² ✓ TRUE! Wait that's the WRONG direction: we want 1−u ≥ e^{−u−u²}, and we know e^{−u−u²} ≥ 1−u−u² — that doesn't give it (1−u ≥ 1−u−u² but e^{−u−u²} could be anywhere ≥ 1−u−u²). Hmm. Use e^x ≥ 1+x+x²/2 (for x ≥ 0): want e^{u+u²} ≥ 1/(1−u) = 1+u+u²+u³+... e^{u+u²} ≥ 1 + (u+u²) + (u+u²)²/2 = 1+u+u²+(u²+2u³+u⁴)/2 ≥ 1+u+u²+u³ ✓ for the first 4 terms... but 1/(1−u) has all terms. For u ≤ 1/2: 1/(1−u) = 1+u+u²+u³(1+u+...) = 1+u+u²+u³/(1−u) ≤ 1+u+u²+2u³. And e^{u+u²} ≥ 1+(u+u²)+(u+u²)²/2+(u+u²)³/6 ≥ 1+u+u²+u²/2·... let me just: need 1+u+u²+2u³ ≤ e^{u+u²}. e^{u+u²} ≥ 1 + (u+u²) + (u+u²)²/2 + (u+u²)³/6 = 1 + u + u² + (u²+2u³+u⁴)/2 + (u³+...)/6 = 1+u+(3/2)u²+(7/6)u³+u⁴/2+... ≥ 1+u+u²+2u³ ⟺ (1/2)u² + (7/6)u³ + u⁴/2 ≥ 2u³ ⟺ (1/2)u² + u⁴/2 ≥ (5/6)u³ ⟺ u/2 + u³/2 ≥ 5u²/6 ⟺ (mult 6/u): 3 + 3u² ≥ 5u ⟺ true for u ≤ (5−√(25−36))/... discriminant 25−36 < 0 ⟹ ALWAYS true ✓. So for u ∈ [0, 1/2]: 1/(1−u) ≤ e^{u+u²}, i.e. −log(1−u) ≤ u+u² ⟺ log(1/(1−u)) ≤ u+u² ✓ PROVABLE (via Real.exp comparison: need 1/(1−u) ≤ exp(u+u²): show (1/(1−u)) ≤ the 3-term exp partial + positivity... hmm — actually cleaner in Lean: use `Real.add_one_le_exp` iteratively or `Real.sum_le_exp_of_nonneg` (the partial-sum machine from LiBasePoints)! exp x ≥ Σ₀ᴺ xⁱ/i! — LiBasePoints has the partial-sum lower-bound machine (used for e^{0.5} ≥ 1.6487). So: exp(u+u²) ≥ 1+(u+u²)+(u+u²)²/2+(u+u²)³/6 ≥ 1/(1−u) via the polynomial argument (needs (1−u)(poly) ≥ 1: multiply out). All doable but fiddly.\n\nSIMPLER: skip the log form entirely — bound the kernel directly against e^{√n/2+1/3}?? or just use a slightly bigger certified constant in IIMaj: kernel ≤ (1+1/(2√n−1))ⁿ ≤ e^{n/(2√n−1)} and n/(2√n−1) ≤ √n/2 + 1/3 for n ≥ ?: ⟺ 3n ≤ (√n/2+1/3)(3)(2√n−1)... let me redo: n/(2√n−1) ≤ √n/2 + c ⟺ n ≤ (√n/2+c)(2√n−1) = n − √n/2 + 2c√n − c ⟺ 0 ≤ (2c−1/2)√n − c ⟺ √n ≥ c/(2c−1/2). For c = 1/3: (2/3−1/2) = 1/6: √n ≥ (1/3)/(1/6) = 2 ⟺ n ≥ 4 ✓!! So n/(2√n−1) ≤ √n/2 + 1/3 for n ≥ 4, via the CLEAN 1+x ≤ e^x route (Real.add_one_le_exp — one line)! Then kernel ≤ e^{√n/2+1/3}, and the IIMaj constant: 0.57·e^{1/3} ≤ 0.57·1.3957 = 0.79555 ≤ 0.8 — a CLEAN constant 0.8!\n\nSo option (D′′): IIMaj constant 0.57 → 0.8. Check downstream slack:\n- lt_three's IIMaj(n) ≤ 1: fattened IIMaj(8e4) ~ 10^{-106}·(0.8/0.57) ~ still 10^{-106} ✓.\n- bracket < 4/3: ratio — need to verify 0.57 doesn't literally appear (or appears and cancels).\n- base points: IIMaj/4 ≤ 2590.08/4 — hmm wait, actually now I realize I should check what that 2590.08 was. Let me grep LiBasePoints for IIMaj.\n- DL-6-ii e4: extra ≤ step with (0.57 ≤ 0.8)·monotone ✓ mechanical.\n- V_right's actual bound (0.57-constant) ≤ new IIMaj/4 ✓ since 0.57 ≤ 0.8.\n- marginExpanded/mK: mIIM = (−1.5)·IIMaj — definitional ✓; mK constant — does IIMaj's 0.57 appear in mK? mK = −40·1.3352 − 2(e⁴+1)(7(3·2.9²+48)) − 16.76 − 1 + 1 — no IIMaj ✓ (it's folded as the mIIM block).\n- E-034: IIMaj values ×1.404 of 10^{-106} — immaterial; fold fits trivially ✓.\n\nAnd the I_I bound proof (6-c-iii): kernel ≤ e^{√n/2+1/3}+1 (new lemma, DL-4-style but uniform in t via normSq monotone + add_one_le_exp + the n ≥ 4 arithmetic) — actually simpler: (1+1/(2√n−1))ⁿ requires |1+1/(w−1)| ≤ 1+1/(2√n−1): |1+1/(w−1)|² = 1 + (2·re(w−1)+1)/|w−1|² = 1 + (4√n−1)/((2√n−1)²+t²) ≤ 1 + (4√n−1)/(2√n−1)². And (1+1/(2√n−1))² = 1 + (4√n−1)/(2√n−1)² ✓ EXACTLY — so |1+1/(w−1)| ≤ 1+1/(2√n−1) (sqrt monotone) ✓ — this is norm_one_add_one_div_le (the ii′-4 normSq core) + monotone in t. Then (1+1/(2√n−1))ⁿ ≤ e^{n/(2√n−1)} (add_one_le_exp ^ n) ≤ e^{√n/2+1/3} (the n ≥ 4 arithmetic + exp monotone) ✓ CLEAN.\n\nThen 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+1)(e^{√n/2}+1)·0.8·2^{−...} = IIMaj_new n ✓ (0.57·e^{1/3} ≤ 0.8: e^{1/3} ≤ 1.4 ⟺ e ≤ 1.4³ = 2.744 — certified: e ≤ 2.7182818286 ≤ 2.744 ✓ via Real.exp_one_lt_d9 and norm_num; so 0.57·1.4 = 0.798 ≤ 0.8 ✓ — actually simpler: e^{1/3} ≤ 1.4 via exp_le_inv_sub_sixteen? or Real.exp_one_lt_d9: e^{1/3} = e^{1/3}... hmm, Real.rpow or exp (1/3): exp (1/3) ≤ 1.4 ⟺ 1/3 ≤ log 1.4 ⟺ ... cleanest: exp(1/3)³ = e ≤ 2.7183 ≤ 2.744 = 1.4³ and x ↦ x³ monotone: Real.exp_le... : exp(1/3) ≤ 1.4 ⟸ exp(1/3)³ ≤ 1.4³; exp(1/3)³ = exp(1) via ← Real.exp_nat_mul: (exp (1/3))^3 = exp (3·(1/3)) = exp 1 ✓. Then (exp (1/3))³ ≤ 1.4³ ⟹ exp(1/3) ≤ 1.4 via pow_le_pow_iff... `pow_le_pow_left₀` reversed: a³ ≤ b³ ⟹ a ≤ b for a,b ≥ 0 — `pow_le_pow_iff_left₀` or le_of_pow_le_pow_left₀. Exists: le_of_pow_le_pow_left₀ (hn : n ≠ 0) (hy : 0 ≤ y) : x^n ≤ y^n → x ≤ y ✓.)\n\nThis is Finding F-E with repair. The right protocol per the project: log it in D-074 (or a new finding note), and it ALSO needs to go in the ii′-7 paper correction queue (F-E: I_I kernel off-by-one — constant 0.57 → 0.8 in IIMaj; margin/theorem unaffected, slack astronomical).\n\nHmm — wait, actually, let me reconsider once more whether to fatten IIMaj vs charge Bn's I_I term as-is but prove I_I ≤ IIMaj_new... The DECISION hinges on how painful the IIMaj def change is. Files touching IIMaj: LiSkeleton (def, DogLegRight, Bn, marginTilde, marginExpanded mIIM, iimaj_bracket_bound, IIMaj_deriv2_lt_three, marginTilde_convex), LiBasePoints (base-point bounds), LiDogLeg (DL-6-ii e4 accounting + vIntegral_Jlag_right_le_IIMaj_quarter statement + hIntegral_Jlag_low_le_dogLegUpper statements use IIMaj/4 — statements UNCHANGED since they say IIMaj; only the PROOF e4 inside DL-6-ii), E-034 script (numeric — update the constant; the fold still fits).\n\nLet me now verify the two risky proof sites: (1) iimaj_bracket_bound — does 0.57 literally appear? (2) LiBasePoints' IIMaj bounds. Grep.","signature":""},{"type":"text","text":"**Found a defect before writing any code** — the I_I kernel bound is off by one. `kFun_shift_norm_le_exp_sqrt_half_add_one` needs `re s ≥ 2√n` at the *shifted* point (i.e. `re w ≥ 2√n+1`), and DL-4's variant needs `t² ≥ n` — but the main rectangle's right vertical runs at `re w = 2√n` through `t ∈ [−T, T]`, including `t = 0`. At `t = 0` the true kernel is `(2√n/(2√n−1))ⁿ ≈ e^{√n/2+1/8} > e^{√n/2}` — the charged `IIMaj` (constant 0.57) is genuinely exceeded. This is a latent defect in the original D-036 charge too (Finding F-E, sibling of F-A/F-B). The repair: 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 ≤ 0.8`). Checking how load-bearing the literal 0.57 is downstream."},{"type":"tool_use","id":"Grep_29","name":"Grep","input":{"pattern":"0\\.57|IIMaj","path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiBasePoints.lean","output_mode":"content","-n":true,"head_limit":40},"extra_content":null},{"type":"tool_use","id":"Grep_30","name":"Grep","input":{"pattern":"iimaj_bracket_bound|IIMaj_deriv2_lt_three","path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiSkeleton.lean","output_mode":"content","-n":true,"head_limit":20},"extra_content":null}]