[{"type":"thinking","thinking":"Committed: 0dfbd6e. Now start ii′-6-c per the STATE.md NEXT ACTION. First read the required material: D-071 §4.4 (four-side component designs), LiAssembly.lean (the sides identity + bridge), and check what's available in LiDogLeg.lean (I know it from the summary — dogLegUpper_two_sides).\n\nii′-6-c scope (from summary):\n- (i) I_III left vertical σ=−5/2: kernel ≤ 2 (kFun_norm_le_two); zeta = tight R-bound (5f capstone, needs hZC/hZC2/hNLZ) + near sum ≤ (2/3)·Z2(T) for |t| ≥ 10; MZeroProp M₀ ≤ 1.3352 for |t| ≤ 10. Length: t from −T to T → 2T. So ‖I_III‖ ≤ 2T·2·((2/3)Z2hat T + 10.5 log(T+2) + 48 + ...) — actually the Bn charge is 4T·((2/3)Z2hat T + 10.5 log(T+2) + 48). Let me check Bn def: `4 * F T + 4 * T * ((2/3) * Z2hat T + 10.5 * Real.log (T + 2) + 48) + 40 * 1.3352 + 2*(e⁴+1)*(7*Hhat T) + 2*DogLegRight n + IIMaj n + 1`.\n\nWait: 40·1.3352 is the MZeroProp piece (M₀ ≤ 1.3352 for |t|≤10, length 20, kernel 2 → 2·20·1.3352 = 40·1.3352). The 4T·(...) is kernel 2 × length 2T × zeta bound ((2/3)Z2hat + tight R). Hmm — for |t| ≥ 10 the zeta bound is near-sum ≤ (2/3)·Z2(T) plus the tight R-bound 10.5 log(t+2)+48 ≤ 10.5 log(T+2)+48 (t ≤ T... actually need monotonicity in t).\n\n- (ii) horizontals: mid part [−5/2, 3] kernel ≤ e⁴+1 + NI(T) segment-integral machine; Bn charge 2(e⁴+1)(7Ĥ T) where Hhat T presumably bounds NI(T) + 5.5·(10.5 log + 48)... need to read LiSkeleton for Hhat def. Right tail [3, 2√n]: dogLegUpper_two_sides from LiDogLeg.\n\n- (iii) I_I right vertical σ₁ = 2√n: kernel e^{√n/2}+1 + norm_logDeriv_zeta_le_iimZeta + length 2T ⟹ ≤ IIMaj n.\n\n- (iv) triv(−2) ≤ 1 via zeroMult_neg_two... actually the zeroSum includes all zeros inside the rectangle; the trivial zero at −2 contributes |1−(ρ/(ρ−1))ⁿ| ≤ ... with ρ = −2: ρ/(ρ−1) = 2/3, |1−(2/3)ⁿ| ≤ 1 + (2/3)ⁿ ≤ 2? Hmm, the summary says triv ≤ 1. Actually kFun_neg_two ≤ 1 — the kernel at −2. The zeroSum term is zeroMult(−2)·kFun... need to read LiAssembly.\n\nWait — the contour: rectangle from σ = −5/2 to σ₁ = 2√n, t = ±T. The trivial zero at −2 is inside. But what about other trivial zeros −4, −6, ...? They're outside (σ = −5/2 left edge). Good — that's why the left edge is at −5/2.\n\nHmm wait, actually the Lagarias contour for Sf uses left edge −5/2 so only the trivial zero −2 is inside (plus nontrivial zeros and the pole at 1). But wait — is 0 inside? s=0: ζ(0) = −1/2 ≠ 0. The pole at 1 gives ResW1. OK.\n\nLet me read D-071 §4.4 and LiAssembly.lean first. Also check LiSkeleton for Hhat, Z2hat, NI definitions.\n\nPlan for ii′-6-c — the deliverables:\n1. I_III bound: ‖RectangleIntegral' over left vertical‖ ≤ 4T·((2/3)Z2hat T + 10.5 log(T+2) + 48) + 40·1.3352... \n\nHmm wait, actually Bn's term is `4 * T * (...) + 40 * 1.3352`. So I_III ≤ that. The split: for |t| ∈ [10, T]: ‖ζ′/ζ(−5/2+it)‖ ≤ near-sum + tight-R. Near-sum over zeros within 1 of t: count ≤ ... the (2/3)Z2hat factor. Actually (2/3)·Z2hat(T) — where does 2/3 come from? From the near-sum: Σ_{|γ−t|<1} mult/|s−ρ| with s = −5/2+it, |s−ρ| ≥ ... hmm. The near sum in the tight R-bound capstone is Σ_{near} m/(s−ρ) with near = zeros within 1 in t. |s−ρ| ≥ |Re(s)−Re(ρ)| = |−5/2 − σ_ρ| ≥ 5/2 − 1 = 3/2 (since σ_ρ ≤ 1). So each term ≤ (2/3)m. And Σ m over the two-unit window ≤ Z2(T) (ZeroCountProp2). So near-sum ≤ (2/3)·Z2(T) ≤ (2/3)·Z2hat(T). \n\nFor |t| ≤ 10: MZeroProp: ‖ζ′/ζ‖ ≤ M₀ = 1.3352 on the segment? MZeroProp is presumably stated as: for s with σ ∈ [−5/2, ...], |t| ≤ 10, ζ s ≠ 0 ⟹ ‖logDeriv ζ s‖ ≤ 1.3352. Need to read its exact statement in LiAssembly.lean.\n\nThen ‖I_III‖ = ‖∫_{−T}^{T} kFun·logDeriv‖ ≤ kernel(≤2) × [∫_{|t|≤10} M₀ + ∫_{10≤|t|≤T} ((2/3)Z2hat + 10.5 log(t+2) + 48)] ≤ 2·[20·1.3352 + 2T·((2/3)Z2hat T + 10.5 log(T+2) + 48)]. = 40·1.3352 + 4T·(...). ✓ matches Bn.\n\n2. Horizontals mid [−5/2, 3]: the horizontal at height t = ±T (T zero-avoiding with clearance c_clear = 1/(9·Ĉ₀(√n+2))). Hmm wait — the dog-leg replaced [3, 2√n]; the mid part stays at height T. The integrand along the mid segment: kernel ≤ e⁴+1 (kFun_shift_norm_le_exp_four_add_one: re s ≤ 3, n ≤ im² — here im = T = √n+1 hmm, n ≤ T²? T = √n + 1 ⟹ T² = n + 2√n + 1 ≥ n ✓). Zeta: the segment-integral machine: ∫_{−5/2}^{3} ‖ζ′/ζ(σ+iT)‖ dσ ≤ NI(T) := Z2(T)·2·log(16.5/c_clear) + 5.5·(10.5 log(T+2)+48)? The 7Ĥ charge in Bn: 2(e⁴+1)(7·Hhat T) for both horizontals (factor 2). So per horizontal ≤ (e⁴+1)·7·Hhat T. Need Hhat def from LiSkeleton.\n\nActually from E-030 fold: NI + 5.5(10.5 log + 48) ≤ 7Ĥ. So Hhat presumably := something like Ĉ₀-based. The segment machine: split the segment into unit intervals; on each, near zeros within the interval contribute log-singular integrals (each zero within clearance c_clear from the path, ∫ log-type singularity ≤ 2 log(1/c_clear)-ish), far part via the tight R-bound. The \"16.5\" is probably (length+clearance)-related: ∫ over unit interval of log|σ−γ| dσ ≤ 2·log(16.5/c_clear)?? I'll read D-071 §4.4.\n\n3. I_I: right vertical at σ₁ = 2√n: kernel ≤ e^{√n/2}+1 (kFun_shift_norm_le_exp_sqrt_half_add_one: 2√n ≤ re s ✓). Zeta ≤ 0.57·2^{−(2√n−2)} (norm_logDeriv_zeta_le_iimZeta, n ≥ 5). Length 2T = 2(√n+1). So ‖I_I‖ ≤ 2(√n+1)(e^{√n/2}+1)·0.57·2^{−(2√n−2)} = IIMaj(n)? Let me check IIMaj def: IIMaj n is presumably defined as ((√n+1))·(e^{√n/2}+1)·0.57·2^{−(2√n−3)} or something. From DL-6: e4 : IIMaj/4 = ((√n+1)/2)(e+1)·0.57·2^{−(2√n−2)}. So IIMaj = 2(√n+1)(e^{√n/2}+1)·0.57·2^{−(2√n−2)}. ✓ exactly the I_I bound. \n\n4. triv: the zeroSum in the sides identity includes the trivial zero at −2 with zeroMult(−2) = 1 (zeroMult_neg_two, ii′-4) and kFun(−2) ≤ 1 (kFun_neg_two). Wait — actually the rectInt'_Jlag identity: RectangleIntegral' (Jlag n) z w = ResW1 n + Σ_{ρ∈Sz} (zeroMult ρ)·(1−(ρ/(ρ−1))ⁿ). The kernel term at ρ is (1−(ρ/(ρ−1))ⁿ) = kFun n ρ? And kFun_neg_two : kFun n (−2) = ... ≤ 1. So triv contribution ≤ 1. The \"+1\" in Bn and the bridge \"zeroSum_norm_le_four_FAlt_add_one\" — the +1 IS the triv. So the bridge already handles it: ‖Σ zeros‖ ≤ 4·FAlt + 1 where the 4·FAlt is the nontrivial zeros (both-signs, factor 4) and +1 is the trivial zero at −2.\n\nHmm wait — but does the bridge zeroSum_norm_le_four_FAlt_add_one cover ALL zeros inside the rectangle, including −2? \"bridge ‖Σ‖ ≤ 4·FAlt + 1\" — yes, PROVED session 038 (LiAssembly.lean). So the triv is already inside the bridge. Good — item (iv) is already done in LiAssembly.\n\nSo ii′-6-c = the three integral bounds (I_III, horizontals, I_I) + the assembly into sfBoundProp (that's ii′-6-d? or the fold is ii′-6-d). Per the plan: ii′-6-c = four side bounds; ii′-6-d = fold ⟹ sfBoundProp.\n\nWhere do these go? Probably LiAssembly.lean (it has the sides identity sf_abs_le_rectInt_add_zeroSum). Let me read LiAssembly.lean and D-071 §4.4 first.\n\nAlso key: what height T is used? T = √n + 1 must be zero-avoiding via exists_zero_avoiding_T_wide (ii′-6-b): gives T ∈ [√n+1, √n+2]? Hmm — \"wide pigeonhole exists_zero_avoiding_T_wide\". But Bn uses T = √n+1 exactly?? FAlt T etc. are evaluated at T = √n+1... Actually FAlt T is probably monotone so FAlt(actual T) ≤ FAlt(√n+2)... Hmm, but Bn's T := √n+1. Let me look: in LiSkeleton, presumably `Bn (FAlt : ℝ → ℝ) (n : ℝ) := ... T := Real.sqrt n + 1`. And the horizontal path height T_act from the pigeonhole lies in [√n+1, √n+2]? Then bounds at T_act must be ≤ bounds at √n+1... no wait, they'd be LARGER. Hmm — unless the pigeonhole gives T ∈ [√n, √n+1] (window of width 1 starting at √n). exists_zero_avoiding_T from ii′-4: clearance 1/(3·C0hat(a+1)) in any unit window [a, a+1], a ≥ 2. The wide version probably: window [√n, √n+1] with clearance c_clear = 1/(9·Ĉ₀(√n+2)) — note Ĉ₀ evaluated at √n+2 suggests the window is [√n, √n+2]?? or a+1 = √n+2 with a = √n+1: window [√n+1, √n+2]. Hmm.\n\nBut then T_act ≥ √n+1 and all the bounds (FAlt, Z2hat, log(T+2), Hhat, IIMaj) are monotone nondecreasing in T... that would give bounds at T_act ≥ bounds at √n+1, breaking the Bn fit! Unless: the contour height is T_act but Bn is designed with enough slack, OR the pigeonhole window is [√n, √n+1] so T_act ≤ √n+1 and monotonicity gives bounds at T_act ≤ bounds at √n+1 = Bn's T. But clearance c_clear = 1/(9·Ĉ₀(√n+2)): if window [√n, √n+1], then t + 1 ≤ √n+2 and Ĉ₀(t+1) ≤ Ĉ₀(√n+2) — clearance stated with the uniform upper bound. That makes sense! The wide pigeonhole: T_act ∈ [√n, √n+1] with clearance 1/(9·Ĉ₀(√n+2)) from all zeros.\n\nThen all T-monotone bounds at T_act ≤ at √n+1 ✓. And the kernel bounds need n ≤ T²: T_act ≥ √n ⟹ T_act² ≥ n ✓ — that's why the window starts AT √n.\n\nOK. Time to read the files. Plan:\n1. Read LiAssembly.lean (sides identity + bridge + perimeter Props).\n2. Read D-071 §4.4 (the four-side designs).\n3. Read LiSkeleton.lean relevant defs (Bn, Hhat, Z2hat, FAlt, IIMaj, NI if present).\n4. Then design ii′-6-c sub-stages and start.\n\nLet me do the reads in parallel. LiAssembly.lean first — check size. Also grep for Hhat/NI in LiSkeleton.","signature":""},{"type":"text","text":"Committed as `0dfbd6e`. Now starting **ii′-6-c** per the STATE.md resume map. Reading the required inputs first."},{"type":"tool_use","id":"Bash_3","name":"Bash","input":{"command":"cd /d/Python/Quantum/research/lean/NguyenRH1/NguyenRH1 && wc -l LiAssembly.lean LiSkeleton.lean LiDogLeg.lean LiContour.lean 2>/dev/null","description":"Check file sizes for reading plan"},"extra_content":null},{"type":"tool_use","id":"Grep_4","name":"Grep","input":{"pattern":"§4\\.4|4\\.4|four.side|I_III|NI\\(|segment.integral","path":"D:\\Python\\Quantum\\research\\derivations\\D-071-L0-assembly-design.md","output_mode":"content","head_limit":30},"extra_content":null}]