[{"type":"thinking","thinking":"6-c-ii committed (6e5a98f). Update todos and proceed to 6-c-iii (I_I, F-E form).\n\n**6-c-iii RECIPE (from STATE.md):**\nNEW kernel lemma `kFun_right_vertical_norm_le_exp_sqrt_add_third` {n} (hn : 4 ≤ n) (t):\n‖kFun n (↑(2√n)+↑t·I)‖ ≤ exp(√n/2+1/3)+1 — |w/(w−1)| ≤ 1+1/(2√n−1)\n(norm_one_add_one_div_le + normSq monotone in t², worst t=0:\n(2·re(w−1)+1)/normSq(w−1) = (4√n−1)/((2√n−1)²+t²) ≤ (4√n−1)/(2√n−1)²\n⟹ |1+1/(w−1)| ≤ 1+1/(2√n−1) — sqrt of both sides, sq_le_sq₀), then\n(1+1/(2√n−1))ⁿ ≤ exp(n/(2√n−1)) (add_one_le_exp^ⁿ) and\nn/(2√n−1) ≤ √n/2+1/3 ⟺ n ≥ 4 (cross-multiply, pure nlinarith) ⟹\nexp mono. Then I_I ≤ 2T·(e^{√n/2+1/3}+1)·0.57·2^{−(2√n−2)} ≤\ne^{1/3}·IIMaj n (norm_logDeriv_zeta_le_iimZeta (n≥5); e^{√n/2+1/3}+1\n≤ e^{1/3}(e^{√n/2}+1); T ≤ √n+1; unfold IIMaj + ring).\n\nTarget statement: something like\n`vIntegral_Jlag_right_norm_le (hNLZ?) ... {n} (hn : 100000 ≤ n) {T} (hTn : √n ≤ T) (hTn1 : T ≤ √n+1) : ‖VIntegral (Jlag n) (2*√n) (-T) T‖ ≤ Real.exp (1/3) * IIMaj n`\n\nWait — I need to check what IIMaj is and the existing DL-4/DL-6 lemmas:\n- `kFun_right_vertical_norm_le_exp_sqrt_half_add_one` (DL-4): the w = 2√n kernel variant, e^{√n/2}+1 — but with the off-by-one: needs t² ≥ n or re ≥ 2√n+1. The F-E finding says this lemma's charge is off by one at t=0.\n- `norm_logDeriv_zeta_le_iimZeta` (DL-6): ‖ζ′/ζ(w)‖ ≤ 0.57·2^{−(2√n−2)} at re w = 2√n (n ≥ 5)? Need exact signature.\n- IIMaj def in LiSkeleton.\n- `norm_one_add_one_div_le` — the normSq core from LiContour (used in kFun_norm_le_two).\n\nLet me look at these signatures first. Also vIntegral_Jlag_right_norm_le_quarter (DL-6): V_right ≤ IIMaj/4 EXACTLY — that lemma presumably bounds the VIntegral at 2√n by IIMaj/4 using the OLD kernel charge... wait no — DL-6's vIntegral_Jlag_right_le_IIMaj_quarter — hmm, what does that prove exactly? If DL-6 already proved V_right ≤ IIMaj/4, why does 6-c-iii need a new I_I bound? \n\nOH — I see: DL-4/DL-6's V_right is the dog-leg's RIGHT vertical segment from T to U = (3/2)T at σ = 2√n (a SHORT segment, length U−T = T/2, used for the dog-leg). The I_I of the main rectangle is the FULL right vertical from −T to T at σ = 2√n (length 2T). Different segment! The F-E finding: the kernel bound kFun_right_vertical_norm_le_exp_sqrt_half_add_one (used for both) is off by one — DL-4's usage was saved because... wait, was DL-4's usage also affected by F-E? Let me re-read F-E:\n\n\"FINDING F-E (D-074 §9): the I_I kernel charge e^{√n/2}+1 is off by one at re w = 2√n near t = 0 (true value e^{√n/2+1/8}; latent in the original D-036 charge); REPAIR ADOPTED: I_I charged honestly as e^{1/3}·IIMaj (uniform kernel ≤ e^{√n/2+1/3}+1, n ≥ 4), the overrun absorbed by the fold's free dropped 2π (e^{1/3} ≤ 1.4 ≤ 2π) — NO change to Bn/IIMaj/convexity/pinch/base points\"\n\nAnd from the NGUYEN.md entry: \"the ii′-4 lemma needs re w ≥ 2√n+1; DL-4's variant needs t² ≥ n; at t = 0 the true kernel is (2√n/(2√n−1))ⁿ = e^{√n/2+1/8}\". So DL-4's variant `kFun_right_vertical_norm_le_exp_sqrt_half_add_one` — what are its hypotheses? If it needs t² ≥ n, then DL-4's own usage (V_right over t ∈ [T, U] with T ≥ √n ⟹ t² ≥ n ✓) is FINE — the dog-leg V_right segment has |t| ≥ T ≥ √n so t² ≥ n ✓. The F-E issue only bites for the FULL I_I segment which includes t near 0. OK good — DL-4/DL-6 stand.\n\nSo 6-c-iii needs:\n1. NEW kernel lemma: `kFun_right_vertical_norm_le_exp_sqrt_add_third` {n : ℕ} (hn : 4 ≤ n) (t : ℝ) : ‖kFun n (↑(2 * Real.sqrt n) + ↑t * Complex.I)‖ ≤ Real.exp (Real.sqrt n / 2 + 1/3) + 1.\n2. The I_I bound: `vIntegral_Jlag_right_norm_le_... {n} (hn : 100000 ≤ n) {T} (hTn : √n ≤ T) (hTn1 : T ≤ √n+1) : ‖VIntegral (Jlag n) (2 * Real.sqrt n) (-T) T‖ ≤ Real.exp (1/3) * IIMaj n`.\n\nWait — but what's the exact charged form in Bn? The F-E repair says \"I_I charged honestly as e^{1/3}·IIMaj\" and the fold absorbs e^{1/3} ≤ 2π. Actually in the fold (6-d): |Sf| ≤ (1/2π)(I_III + Horiz + e^{1/3}IIMaj) + (4F+1) ≤ ... ≤ (1/2π)(I_III-charge + Horiz-charge) + IIMaj + 4F + 1. So the 6-c-iii deliverable is ‖VIntegral (Jlag n) (2√n) (−T) T‖ ≤ e^{1/3} · IIMaj n.\n\nHmm wait — actually let me check: is it ≤ e^{1/3}·IIMaj or the raw 2T(e^{√n/2+1/3}+1)·0.57·2^{−(2√n−2)}? The named deliverable should be the e^{1/3}·IIMaj form (that's what the fold consumes). But per the recipe: \"Then I_I ≤ 2T·(e^{√n/2+1/3}+1)·0.57·2^{−(2√n−2)} ≤ e^{1/3}·IIMaj n (norm_logDeriv_zeta_le_iimZeta (n≥5); e^{√n/2+1/3}+1 ≤ e^{1/3}(e^{√n/2}+1); T ≤ √n+1; unfold IIMaj + ring)\".\n\nSo the lemma: \n```lean\ntheorem vIntegral_Jlag_right_norm_le_exp_third_IIMaj {n : ℕ} (hn : 100000 ≤ n)\n    {T : ℝ} (hTn : Real.sqrt n ≤ T) (hTn1 : T ≤ Real.sqrt n + 1) :\n    ‖VIntegral (Jlag n) (2 * Real.sqrt n) (-T) T‖ ≤ Real.exp (1/3) * IIMaj n\n```\n\nHmm — what does IIMaj look like? Need its def. IIMaj n = (√n+1)·(e^{√n/2}+1)·2^{−(2√n−2)}? Let me check LiSkeleton. The recipe says \"T ≤ √n+1; unfold IIMaj + ring\". So IIMaj n probably = (√n + 1) * (Real.exp (√n/2) + 1) * 2^{-(2√n-2)}? But then 2T·(e^{√n/2+1/3}+1)·0.57·2^{−(2√n−2)} ≤ e^{1/3}·(√n+1)(e^{√n/2}+1)2^{−(2√n−2)} ⟺ 2T(e^{√n/2+1/3}+1)·0.57 ≤ e^{1/3}(√n+1)(e^{√n/2}+1) — with T ≤ √n+1: suffices 2·0.57·(e^{√n/2+1/3}+1) ≤ e^{1/3}(e^{√n/2}+1) ⟺ 1.14·(e^{√n/2+1/3}+1) ≤ e^{1/3}(e^{√n/2}+1). Now e^{√n/2+1/3}+1 ≤ e^{1/3}(e^{√n/2}+1) ⟺ e^{√n/2+1/3}+1 ≤ e^{√n/2+1/3}+e^{1/3} ⟺ 1 ≤ e^{1/3} ✓ TRUE. So e^{√n/2+1/3}+1 ≤ e^{1/3}(e^{√n/2}+1) ✓ and 1.14 ≤ ... wait we need 1.14·X ≤ Y where X ≤ e^{1/3}(e^{√n/2}+1) =: Y. 1.14·Y?? No: suffices 1.14·X ≤ Y with X ≤ Y/1? Hmm: X := e^{√n/2+1/3}+1 ≤ e^{1/3}(e^{√n/2}+1) =: Y. Need 1.14·X ≤ e^{1/3}(e^{√n/2}+1) = Y. Since 1.14 ≤ e^{1/3}·?? — 1.14·X ≤ 1.14·Y ≤ ??? need 1.14·Y ≤ Y ⟺ 1.14 ≤ 1 FALSE. So the 0.57·2 = 1.14 factor must be absorbed elsewhere: norm_logDeriv_zeta_le_iimZeta's 0.57 — hmm.\n\nLet me recompute: I_I ≤ 2T·(e^{√n/2+1/3}+1)·0.57·2^{−(2√n−2)}. Target: e^{1/3}·IIMaj n with IIMaj n = (√n+1)·(e^{√n/2}+1)·2^{−(2√n−2)}? — need to check IIMaj's actual def. Maybe IIMaj n = 2·(√n+1)·(e^{√n/2}+1)·2^{−(2√n−2)} (with a factor 2 built in for the two sides?) or with 0.57... The DL-6 charge was \"IIMaj/4\" for V_right with ‖ζ′/ζ‖ ≤ 0.57·2^{−(2√n−2)} and kernel e^{√n/2}+1 and length (U−T) = T/2 ≤ (√n+1)/2: V_right ≤ (U−T)·(e^{√n/2}+1)·0.57·2^{−(2√n−2)} ≤ (√n+1)/2·(e^{√n/2}+1)·0.57·2^{−(2√n−2)}. For this to be IIMaj/4: IIMaj = 2·(√n+1)·(e^{√n/2}+1)·0.57·2^{−(2√n−2)}? Then IIMaj/4 = (√n+1)/2·(...)·0.57·2^{−...} ✓. So IIMaj n ≈ 1.14·(√n+1)·(e^{√n/2}+1)·2^{−(2√n−2)}.\n\nThen I_I ≤ 2T·(e^{√n/2+1/3}+1)·0.57·2^{−(2√n−2)} with T ≤ √n+1: ≤ 2(√n+1)(e^{√n/2+1/3}+1)·0.57·2^{−(2√n−2)} = IIMaj·(e^{√n/2+1/3}+1)/(e^{√n/2}+1) ≤ IIMaj·e^{1/3} ✓✓ — using (e^{√n/2+1/3}+1) ≤ e^{1/3}(e^{√n/2}+1). \n\nSo the final lemma: ‖VIntegral (Jlag n) (2√n) (−T) T‖ ≤ Real.exp (1/3) * IIMaj n. The proof:\n- VIntegral_norm + norm_integral_le_integral_norm: ≤ ∫_{-T}^{T} ‖Jlag n (↑(2√n)+↑t·I)‖ dt.\n- Integrand ≤ K·L where K = e^{√n/2+1/3}+1 (kernel, NEW lemma, ALL t) and L = 0.57·2^{−(2√n−2)} (norm_logDeriv_zeta_le_iimZeta, n ≥ 5 — check signature; does it need hypotheses on t? DL-6 proved it at re = 2√n for... let me check its exact statement).\n- So integrand ≤ K·L constant; integral ≤ 2T·K·L via integral_mono_on + integral_const. Wait — need norm of product: ‖Jlag‖ = ‖kFun‖·‖logDeriv‖ ≤ K·L via mul_le_mul ✓ (L ≥ 0).\n- Then 2T·K·L ≤ e^{1/3}·IIMaj via T ≤ √n+1, the X ≤ e^{1/3}Y lemma, unfold IIMaj.\n\nFor the integral bound: ∫_{-T}^{T} c dt = 2T·c. integral_mono_on needs -T ≤ T ✓ and pointwise bound on Icc(-T,T).\n\nThe integrability: same path-continuity argument but at re = 2√n: Jlag continuous on the path (kFun differentiable — s ≠ 1 since re = 2√n ≥ 632 ≠ 1... wait need s ≠ 1: re = 2√n ≥ 2·√100000 ≈ 632 — if s = 1 then re = 1 contradiction ✓; logDeriv continuous: ζ(s) ≠ 0 at re = 2√n — ζ has no zeros with re > 1 — riemannZeta_ne_zero_of_one_le_re (banked: takes ONLY re). re s = 2√n ≥ 632 ≥ 1 ✓.\n\nSo I need a `zeta_ne_zero_right_vertical` helper: riemannZeta (↑(2√n) + ↑t·I) ≠ 0 via riemannZeta_ne_zero_of_one_le_re.\n\nNow the kernel lemma. Statement:\n```lean\ntheorem kFun_right_vertical_norm_le_exp_sqrt_add_third {n : ℕ} (hn : 4 ≤ n) (t : ℝ) :\n    ‖kFun n (↑(2 * Real.sqrt n) + ↑t * Complex.I)‖ ≤ Real.exp (Real.sqrt n / 2 + 1/3) + 1\n```\n\nProof per recipe:\n|w/(w−1)| ≤ 1+1/(2√n−1) where w = ↑(2√n)+↑t·I.\n- |w/(w−1)| = |1 + 1/(w−1)| ≤ 1 + 1/|w−1| (norm_one_add_one_div_le? — that's the lemma used in kFun_norm_le_two: `norm_div_sub_one_le_one`? Let me check what kFun_norm_le_two used: `norm_div_sub_one_le_one hs : ‖s / (s - 1)‖ ≤ 1` for re ≤ 1/2. Different. For our case we need the normSq computation directly: ‖w/(w−1)‖² = normSq w / normSq (w−1) = ((2√n)²+t²)/((2√n−1)²+t²). The recipe: \"norm_one_add_one_div_le + normSq monotone in t², worst t=0: (2·re(w−1)+1)/normSq(w−1) = (4√n−1)/((2√n−1)²+t²) ≤ (4√n−1)/(2√n−1)² ⟹ |1+1/(w−1)| ≤ 1+1/(2√n−1) — sqrt of both sides, sq_le_sq₀\".\n\nLet me find `norm_one_add_one_div_le` in LiContour — it computes ‖1 + 1/z‖ ≤ something via normSq. Let me look at the actual lemma and the DL-4 kernel lemma (kFun_right_vertical_norm_le_exp_sqrt_half_add_one) — its proof is the closest template (it's the same shape with a different constant, and IT used the t² ≥ n or re ≥ 2√n+1 hypothesis).\n\nLet me read DL-4's kernel lemma proof first — that's the template. Also norm_logDeriv_zeta_le_iimZeta's statement, IIMaj's def, and the ii′-4 norm_one_add_one_div_le.","signature":""},{"type":"text","text":"6-c-ii committed (`6e5a98f`). Moving to 6-c-iii (I_I, F-E form). First, the templates: DL-4's kernel lemma, DL-6's iimZeta bound, and the IIMaj definition."},{"type":"tool_use","id":"TodoWrite_69","name":"TodoWrite","input":{"todos":[{"content":"6-c-ii: I_III left-vertical bound vIntegral_Jlag_left_norm_le","status":"completed","activeForm":"Proving 6-c-ii I_III left-vertical bound"},{"content":"6-c-iii: I_I right-vertical bound (F-E form)","status":"in_progress","activeForm":"Proving 6-c-iii I_I bound"},{"content":"6-c-iv: mid-horizontal segment machine","status":"pending","activeForm":"Proving 6-c-iv mid-horizontal machine"},{"content":"6-c-v: horizontal assembly (mid + DL-5 tails)","status":"pending","activeForm":"Assembling 6-c-v horizontal bound"},{"content":"6-d: fold ⟹ sfBoundProp (then commit, update archive)","status":"pending","activeForm":"Proving 6-d fold to sfBoundProp"}]},"extra_content":null},{"type":"tool_use","id":"Grep_70","name":"Grep","input":{"pattern":"theorem kFun_right_vertical_norm_le_exp_sqrt_half_add_one|theorem norm_logDeriv_zeta_le_iimZeta|theorem norm_one_add_one_div_le|def IIMaj|theorem vIntegral_Jlag_right_le_IIMaj_quarter","path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1","output_mode":"content","-A":8},"extra_content":null}]