[{"type":"thinking","thinking":"Key findings:\n1. `dirichletC2 := ∑' n, ‖LSeries.term Λ 2 n‖` — the exact sum Σ Λ(n)/n² = −ζ′/ζ(2) ≈ 0.56996. D-056's piece 1 uses Z₂ ≤ 0.57. For the Lean fold, I need dirichletC2 ≤ 0.57 (a certified numeric bound — the E-028 V5 certification: partial sum to 10⁶ + tail 1.5e-5 = 0.569975 ≤ 0.5700). In Lean, bounding a tsum of LSeries norms ≤ 0.57 requires... hmm. That's a certified numeric computation — partial sums to 10⁶ in Lean?? That's heavy. Alternative: is there a closed form or a coarser bound? Σ Λ(n)/n² = −ζ′(2)/ζ(2). Could bound via ζ(2) = π²/6 and ζ′(2)... Mathlib has ζ(2). ζ′(2) — probably not. A cheaper bound: Σ Λ(n)/n² ≤ Σ (log n)/n² = −ζ′(2) ≈ 0.9375 — too big (the fold needs ≤ 0.57 with the current constants; 0.9375 would add 0.37 to the constant side — the fold had 47.13 + 0.37 = 47.5 ≤ 48 ✓ hmm actually the constant budget has room: total constant 47.13 vs 48 — 0.87 slack; adding 0.37 keeps ≤ 48 ✓!). Wait let me recompute: piece 1 ≤ −ζ′(2) = Σ (log n)/n²·... no: Λ(n) ≤ log n, so Σ Λ(n)/n² ≤ Σ log n/n² = −ζ′(2) = 0.9375. And does Lean/Mathlib have −ζ′(2)?? Unlikely. Σ log n / n² as a tsum bound — also not standard. Hmm.\n\nActually — the cleanest: piece 1's value only needs to fit the constant budget. The fold: Z₂ + [piece2 const 2.7] + [piece3 const 7.06] + [pieces 4–6: 0.38] + [fold slack]. Let me redo the full honest fold with the two-unit perimeter form:\n- piece 1: Z₂ (need SOME certified bound ≤ ~1)\n- piece 2: 0.6L + 0.886LL + 2.7 (two-unit form)\n- piece 3: 12.7279·C0hat(t+1) + 7.064 where C0hat(t+1) = 0.433L' + 0.886LL' + 2.9, L' = log(t+3)\n- pieces 4–6: ≤ 0.045 + 0.045 + 0.29 = 0.38\n\nL-side: piece2: 0.6L + 0.886LL; piece3: 12.7279·(0.433L' + 0.886LL') = 5.511L' + 11.277LL'. With L' = log(t+3) vs L = log(t+2): L' ≤ L + log((t+3)/(t+2)) ≤ L + 1/(t+2)·... ≤ L + 1/12 (t ≥ 10). Similarly LL' ≤ LL + small. Total L-side: 0.6L + 5.511(L + 1/12) + [0.886LL + 11.277(LL + 1/(12·...))] ≈ 6.111L + 12.163LL + (5.511/12 + 11.277/12·1/L·...) ≈ 6.111L + 12.163LL + 0.5.\nConstants: 0.57 + 2.7 + 7.064 + 0.38 = 10.714.\nTotal: 6.111L + 12.163LL + 0.5 + 10.714 = 6.111L + 12.163LL + 11.214.\nTarget: 10.5L + 48 ⟺ need 12.163·LL ≤ 4.389·L + 36.786. At the LL/L pinch (L = e, t ≈ 13.15): 12.163·1 ≤ 4.389·2.718 + 36.786 = 11.93 + 36.79 = 48.72 ✓ HUGE margin (36.6!). Wait — why so much room now? Because D-056's fold was ≤ 48 with the constant 47.8 — I mis-added: the constants: Z₂ 0.57 + piece2-const 2.7 + piece3-const 7.064 + pieces4–6 0.38 = 10.714 — and D-056 had 0.57 + [piece2 = ONE C₀: const 2.9] + 7.06 + 0.38 = 10.81?! And D-056's total constant was 47.8?! Let me recheck D-056's assembly: \"5.95 log(t+2) + 12.16 log log(t+2) + 39.8 + 7.97\" — the 39.8 = 13.73·2.9 = 39.82 (the C₀ constant × 13.73) and 7.97 = the other constants (0.57+7.06+0.09+0.29 = 8.01 ≈ 7.97). So D-056's constant = 13.73·2.9 + 8.01 = 47.84. My honest version: piece2's count is the two-unit form (const 2.7) instead of one C₀ (const 2.9) — SAVES 0.2 on the constant but ADDS 0.167 to the L-coefficient (0.6 vs 0.433). Piece 3 unchanged: 12.7279·C₀. So my L-coefficient: 0.6 + 12.7279·0.433 = 0.6 + 5.511 = 6.111 vs D-056's 5.95 (0.433·13.73). LL: 0.886 + 12.7279·0.886 = 12.163 ≈ 12.16 ✓ same. Constants: 2.7 + 12.7279·2.9 + 7.064 + 0.57 + 0.38 = 2.7 + 36.91 + 7.064 + 0.57 + 0.38 = 47.62 vs D-056's 47.84 — actually LOWER (the two-unit form's constant 2.7 < 2.9). So total: 6.111L + 12.163LL + 47.62 ≤ 10.5L + 48 ⟺ 12.163·LL ≤ 4.389·L + 0.38. At the pinch L = e: 12.163 ≤ 11.93 + 0.38 = 12.31 ✓ FITS by 0.15!! Let me double-check the pinch point: max of 12.163·log L − 4.389·L over L ≥ log 12 = 2.485: d/dL = 12.163/L − 4.389 = 0 ⟹ L* = 2.7716; value: 12.163·log(2.7716) − 4.389·2.7716 = 12.163·1.01947 − 12.1656 = 12.400 − 12.166 = 0.234 ≤ 0.38 ✓✓ FITS by 0.146 at the worst point. At L = 2.485 (t=10): 12.163·0.9103 − 4.389·2.485 = 11.072 − 10.907 = 0.165 ≤ 0.38 ✓ by 0.215. So the honest assembly fits 10.5/48 with margin ~0.15 at the pinch — TIGHT but provable (4–6× tighter than the D-056 fold's apparent margin, but the pinch is a smooth one-variable max — certifiable by calculus + interval evaluation).\n\nBUT — the piece-2 two-unit form (0.6L + 0.886LL + 2.7) is NOT derivable from the per-unit perimeter (2·C0hat = 0.866L + 1.77LL + 5.8 — the LL and constant terms blow the fold: LL-side would be 1.772 + 11.277 = 13.05, constants 5.8+36.91+7.06+0.95 = 50.72 — breaks 48 by 2.7 and the pinch: 13.05·LL ≤ (10.5−6.377)·L + (48−50.72) = 4.123L − 2.72: at L=2.485: LHS 13.05·0.91 = 11.88 vs RHS 10.25−2.72 = 7.53 — FAILS). So the two-unit perimeter form is REQUIRED. Confirmed the need for the new declared specialization.\n\nNow — the alternative interpretation: rather than a new perimeter item, note that the two-unit count bound 0.6L + 0.886LL + 2.7 could be declared as a SECOND clause of ZeroCountProp (the same declaration, two window shapes — both Trudgian specializations). Or a separate `ZeroCountProp2`. Either way it's a perimeter growth requiring user sign-off.\n\nHmm, wait — actually, hold on. Is there any chance the two-unit form is derivable from the one-unit form + a SHIFT? #{(t−1,t]} + #{(t,t+1]} ≤ C0hat(t) + C0hat(t+1). C0hat(t) + C0hat(t+1) vs the two-unit form 0.6L+0.886LL+2.7: C0hat(t)+C0hat(t+1) = 0.433·(log(t+2)+log(t+3)) + 0.886·(LL(t+2)+LL(t+3)) + 5.8 ≈ 0.866L + 1.772LL + 5.8. Too big (0.866 > 0.6). No.\n\nSo: 5f requires a new declared two-unit count. This is a perimeter growth (same Trudgian source). ALSO: I still owe the user the H(T)/3C0hat² clearance-coverage finding for ii′-6 (the near-sum count/clearance accounting on the contour) — that's a SEPARATE issue from piece 2, affecting marginTilde's baked 3·C0hat² and (2/3)·C0hat coefficients. Hmm — wait, let me re-examine whether that's real too, since it affects whether the baked margin survives at all.\n\nii′-6's H(T) (horizontal contour bound): |ζ′/ζ(σ+iT)| ≤ Σ_{|γ−T|<1} 1/|s−ρ| + |R(s)|. The near sum: count × clearance. L3 (exists_zero_avoiding_T) gives T ∈ [a, a+1] with clearance 1/(3·C0hat(a+1)) vs zeros with γ ∈ [a, a+1] ONLY. The near window |γ−T|<1 ⊂ (a−1, a+2). Zeros in (a+1, a+2) or (a−1, a): NO clearance from L3. So |ζ′/ζ| on the horizontal is NOT bounded by C₀/d_min as D-036 claimed — the honest bound needs the extended pigeonhole (clearance vs all zeros in (a−1, a+2), count ≤ 3·C0hat) ⟹ near sum ≤ 3C₀·(extended clearance 1/(9C₀+3))... \n\nHmm wait — actually, hold on: is that right? Let me redo: exists_avoiding_midpoint with the window [a, a+1] and k+1 midpoints where k = count bound: returns T ∈ [a, a+1] avoiding the k window zeros by 1/(2(k+1)). For the extended version: apply exists_avoiding_midpoint to the window [a−1/2, a+1/2]?? Hmm — but we need T at a specific height range for the contour (T ∈ [√n, √n+1]-ish for the height-window slack argument). Apply the midpoint lemma on [a, a+1] with the zero set = zeros in (a−1, a+2) (3-unit count ≤ 3C₀): midpoints in [a, a+1] — k₀+1 of them with k₀ = 3C₀ — each zero blocks ≤ 1 midpoint ⟹ T with clearance ≥ 1/(2(3C₀+1)) ≥ 1/(7C₀) from ALL zeros within (a−1, a+2) ⊇ |γ−T|<1 ✓. Then near sum ≤ count(|γ−T|<1) · 7C₀ ≤ 3C₀·7C₀ = 21·C₀². vs baked 3·C0hat². 7× worse on the C0hat² coefficient. As computed before, that kills the margin at 8×10⁴ (the crossover jumps past the certified base points and possibly past 5×10⁵!?).\n\nHmm, no wait — my earlier smarter split: near sum ≤ inner + outer: inner |γ−T| < 1/4: count ≤ C₀ (one unit), terms ≤ 7C₀ ⟹ 7C₀²; outer 1/4 ≤ |γ−T| < 1: terms ≤ 4, count ≤ 2C₀ ⟹ 8C₀. Total ≤ 7C₀² + 8C₀. vs baked 3C₀². The increment: 4C₀² + 8C₀. In the margin at n = 8×10⁴: the C0hat²-piece of B̃ is 2(e⁴+1)·7·(coefficient)·C0hat² — coefficient 3 → 7: increment 2(e⁴+1)·7·4·47.3 ≈ 111.2·28·47.3 ≈ 147 300?!? margin(8×10⁴) was ~15 764 — NEGATIVE. At n = 5×10⁵: C0hat² ≈ 54.9: increment ≈ 111.2·28·54.9 ≈ 170 900; margin(5×10⁵) = 2.0×10⁶ → 1.83M ✓ fine. The crossover: from 80 000 to...? The increment grows like C0hat² ~ (log T)² ~ (log n)² while S grows like n log n — the crossover moves modestly (the increment at crossover n ≈ 33 000: T ≈ 183: C0hat ≈ 6.7, C0hat² ≈ 45: increment ≈ 140 000 — vs margin at 33 000 ≈ 0 (crossover!). Hmm the crossover moves to where S − B − 140 000·(C0hat²/45)-ish = 0 — maybe n ≈ 60 000–120 000. N₀ was adopted 80 000 with margin 8001 — the new margin at 80 000 is NEGATIVE (increment 147 300 > 15 764). So N₀ must move to where margin > 0 — need recomputation, but plausibly ~1.5–2×10⁵ — still ≤ 5×10⁵ (the finite verification cap) ✓ workable. The convexity/secant argument: pos_of_convex_of_two_points with base points at N₀ and 10⁵... re-certify.\n\nSo the ii′-6 H(T) issue is REAL and moves the margin significantly (but survivably). HOWEVER — scope check: 5f is ONLY the R-bound. The H(T) issue belongs to ii′-6 (L0 assembly). For 5f, the required decision is just the two-unit perimeter form. The H(T) issue I should DOCUMENT (journal + a repair-map entry) but not solve now.\n\nHmm, wait — but if the H(T) issue will force a margin re-fold at ii′-6 anyway, does the 10.5/48 R-bound still fit the (future) margin? The R-bound enters the margin via Hhat's linear term (10.5·log + 48) AND I_III's (2/3·C0hat + 10.5 log + 48). If ii′-6's near-sum accounting changes C0hat²'s coefficient 3 → 7, the margin's B-side grows by ~4·C0hat²·778 — and the crossover moves — but the R-bound's 10.5/48 stays VALID (it's a standalone analytic bound). The margin re-verification at ii′-6 will use the same R-bound. So 5f at 10.5/48 is safe to do now.\n\nHmm — although — wait. Should I double-check the 3C0hat² in Hhat once more against ii′-6's actual future need? D-036 §2: \"|ζ′/ζ(s)| ≤ C₀(T)/d_min + A log(T+2) + B₀ ≤ 3C₀(T)² + A log(T+2) + B₀ =: H(T)\" — the near sum C₀(T)/d_min with d_min ≥ 1/(3C₀(√n+1)): C₀(T)·3C₀(√n+1) ≈ 3C₀². The count C₀(T) for |γ−T|<1 (two-unit window — the piece-2 issue AGAIN: honest count is the two-unit form 0.6L+0.886LL+2.7 ≈ C₀ + 0.167L − 0.2, fine) AND the clearance coverage issue (out-of-window zeros — the 3C₀² → 7C₀²+8C₀ issue). So Hhat has BOTH issues. ii′-6 will need the full repair. NOT 5f's problem — 5f is the standalone R-bound.\n\nDECISION for 5f: proceed with the honest assembly at 10.5/48, which requires the two-unit count form. Present to the user:\n1. The piece-2 finding (D-056's one-C₀ usage for the two-unit window — a real but small assembly gap; the bound survives via the honest two-unit Trudgian form + the unfolded loglog pinch analysis with ~0.15 margin at the pinch).\n2. The required perimeter growth: a two-unit mult-weighted count bound (same Trudgian source, new specialization: 0.6·log(t+2) + 0.886·loglog(t+2) + 2.7 for |γ−t| ≤ 1, t ≥ 10).\n3. The documented ii′-6 finding (H(T) clearance coverage — 3C0hat² → ~7C0hat²+8C0hat honest; margin impact to be faced at ii′-6; N₀/crossover will move).\n\nActually hold on — before presenting, let me double-check the two-unit closed form once more, since it must be EXACT (it's a declared bound — well, it's a hypothesis, so it just needs to be TRUE and reasonably clean):\n\nTwo-unit mult-weighted count: #{ρ : |γ−t| ≤ 1, mult} = N(t+1) − N(t−1) ≤ F(t+1) − F(t−1) + S_b(t+1) + S_b(t−1) (Trudgian 2014: N(T) = F(T) + S(T), |S(T)| ≤ S_b(T) := 0.137 log T + 0.443 log log T + 1.588 for T ≥ e).\n\nF(t+1) − F(t−1): F(T) = (T/2π)(log(T/2π) − 1) + 7/8. F is convex-increasing (F'(T) = (1/2π)log(T/2π) > 0 for T > 2π). F(t+1)−F(t−1) = ∫_{t−1}^{t+1} F' ≤ 2·F'(t+1) = (1/π)·log((t+1)/2π) (F' increasing). ≤ (1/π)·log(t+2)... hmm need ≤ (1/π)·log(t+2) − 0.585-ish for the fold. Actually for the declaration I just need the CLOSED FORM to honestly majorize: claim N(t+1)−N(t−1) ≤ 0.6·log(t+2) + 0.886·loglog(t+2) + 2.7 for t ≥ 10:\n- F-part ≤ (1/π)·log((t+1)/2π) ≤ 0.31831·log(t+2) − 0.31831·log(2π)·(1 − ...) hmm: log((t+1)/2π) = log(t+2) − log(2π) − log((t+2)/(t+1)) ≤ L − 1.8379 (since log((t+2)/(t+1)) ≥ 0). So F-part ≤ 0.31831·L − 0.58492.\n- S-pair ≤ 0.274·L + 0.886·LL + 3.176 (as before: 0.137·2·L... wait: 0.137·(log(t+1)+log(t−1)) ≤ 0.137·2·log(t+1) ≤ 0.274·L? log(t+1) ≤ log(t+2) = L ✓. 0.443·(LL(t+1)+LL(t−1)) ≤ 0.443·2·LL = 0.886·LL ✓ (t−1 ≥ 9 > e so LL(t−1) ≥ 0... loglog(t−1) requires log(t−1) ≥ 0 ⟺ t ≥ 2 ✓; and LL monotone ✓). Constants: 2·1.588 = 3.176.\n- Total ≤ (0.31831 + 0.274)·L + 0.886·LL + (3.176 − 0.58492) = 0.59231·L + 0.886·LL + 2.5911.\n- Closed form: 0.6·L + 0.886·LL + 2.7: difference = (0.6−0.59231)·L + (2.7−2.5911) = 0.00769·L + 0.1089 ≥ 0 ✓ for all L ≥ 0. HONEST ✓.\n\nSo the declared two-unit form: #{mult-weighted, |γ − t| ≤ 1} ≤ 0.6·log(t+2) + 0.886·loglog(t+2) + 2.7 for t ≥ 10. (Hmm — should it be |t| for negative t too? The R-bound is stated for |t| ≥ 10 with σ ∈ [−5/2, 3] — the s = σ+it with t possibly negative. D-056 states |t| ≥ 10. The near window |γ−t|<1 for t negative uses the conjugate symmetry. For the Lean statement, handle t ≥ 10 and extend to |t| ≥ 10 by conjugation? The LiDisc capstones handled all |t| ≥ 10 — via nearZeros t with t possibly negative (they had the conj machinery: isNontrivialZero_conj). For 5f, following D-056: state for t ≥ 10 first? D-056 says |t| ≥ 10. The ii′-6 consumer needs |t| = T ≥ 10 positive heights (T = √n+1 > 0) — and the conjugate for the bottom horizontal. Let me design the Lean theorem for t ≥ 10 and note the t ≤ −10 extension via conj if needed. Actually the LiDisc capstone `logDeriv_riemannZeta_sub_nearZeros_le` was stated for 10 ≤ |t| (both signs) with the near sum over nearZeros t. For 5f the tight version should match: 10 ≤ |t|. The proof: t ≥ 10 direct; t ≤ −10 via conjugation (logDeriv ζ (conj s) = conj (logDeriv ζ s)? — riemannZeta_conj: ζ(conj s) = conj ζ(s); logDeriv conj: logDeriv ζ (conj s) = conj (logDeriv ζ s) — needs deriv under conj — Zeta23 might have it; the 5c/5d capstones handled both signs via logDeriv_riemannZeta_sub_nearZeros_le_left/right separately — for the LEFT half they reflected. For t < 0, nearZeros t = conj (nearZeros (−t)) presumably via isNontrivialZero_conj. This is mechanical but adds work. Design choice: prove for t ≥ 10; add the t ≤ −10 corollary via conjugation.)\n\nOK. Now — WAIT. One more simplification check: does 5f really need the two-unit form, or can the near window be SHRUNK? The near sum in R(s) is Σ_{|γ−t|<1} m/(s−ρ) — the window |γ−t|<1 is intrinsic to the R-bound's definition (it's what ii′-6 subtracts). Could 5f instead prove the R-bound for a DIFFERENT near window, e.g. |γ−t| < 1/2 (one unit)? Then the far sum includes 1/2 ≤ |γ−t| < 1 zeros with per-term ≤ 4.5√2/(x(1+x)) at x ∈ [1/2, 1] — the far-sum interval sum starts at k=... the |t−γ| ∈ [1/2,1] interval: per-term ≤ 4.5/(0.5·√1.25) = 8.05; count ≤ C₀; so the far sum gains ≤ 8.05·C₀... and piece 2 becomes ≤ C₀(t) (one unit!). Net: piece2 saves 0.167L−0.2, far sum gains 8.05·C₀ vs the k=1 interval's previous 4.5√2/(1·2)·C₀ = 3.18·C₀ — increment 4.87·C₀ ≈ 2.1·L. WORSE. No.\n\nSo the two-unit perimeter form is needed. But actually — hmm, ONE more alternative: bound the near sum via the SERIES difference directly without the count?? Σ_near m/|s₀−ρ| ≤ Σ_near m — no alternative.\n\nAlright — also quickly reconsider (β) once more (honest A=11.2, B₀=51 from the per-unit perimeter) vs (α) (two-unit declaration): (β) requires re-baking marginTilde/Hhat constants + re-verifying convexity (session-011 proof, ~2 tight numeric blocks) + re-certifying base points + N₀ crossover recomputation + paper text at ii′-7. (α) requires: one new declared hypothesis (same Trudgian source) + the pinch-point calculus lemma. (α) is much cheaper and keeps all certified artifacts valid. RECOMMEND (α).\n\nNote the perimeter growth is honest but must be declared in the paper's appendix at ii′-7 (the freeze allows this at ii′-7).\n\nHmm — one more thought on (α): is the two-unit form really a \"new\" perimeter item, or can it be folded into ZeroCountProp as a conjunction? ZeroCountProp currently: ∀ t ≥ 2, (window [t,t+1] mult-count) ≤ C0hat(t+1). The two-unit form: ∀ t ≥ 10, (#{|γ−t| ≤ 1} mult-count) ≤ 0.6 log(t+2) + 0.886 LL(t+2) + 2.7. These are two different shapes — cleanest as a separate def `ZeroCountProp2` (or a renamed combined prop). A separate def keeps the existing consumers untouched. The capstone signatures then take (hZC : ZeroCountProp) (hZC2 : ZeroCountProp2). The perimeter count goes from 4 to 5 declared items (same source family). At ii′-7, the paper's appendix declaration updates.\n\nHmm, or... actually — reconsider ONE more time whether the two-unit bound can be DERIVED from ZeroCountProp by a smarter split: #{|γ−t|≤1} = #{(t−1, t−1/2]} + #{(t−1/2, t]} + #{(t, t+1/2]} + #{(t+1/2, t+1]} — half-unit windows, each ≤ the unit bound C0hat — 4·C0hat?? worse. No: sub-window counts ≤ full-window count (monotonicity of the count in the window inclusion!) — each half-window ⊆ a unit window ⟹ ≤ C0hat. 4 half-windows ≤ 4·C0hat. Worse than 2·C0hat. No gain. The perimeter only speaks about full unit windows. CONFIRMED: need the declaration.\n\nLet me also double check that the honest two-unit form is indeed TRUE (it's a declaration, but it must be mathematically valid — the Trudgian S_b bound applies for T ≥ e ✓; F' increasing for T > 2π ✓; the mult-weighted N(t+1)−N(t−1) = Σ_{t−1 < γ ≤ t+1} m ✓ — N(T) counts with multiplicity by definition ✓; |γ−t| ≤ 1 vs (t−1, t+1]: the open/closed endpoints don't matter for the bound ✓).\n\nAnd the pinch lemma: for L ≥ log 12 (t ≥ 10): 12.163·log L ≤ 4.389·L + 0.38 — hmm let me restate with cleaner constants. The full assembly: |R| ≤ 6.111·L + 12.163·LL + 47.62 where L = log(t+2), LL = log L. Wait — but I should double-check the piece-3 right-side count: intervals (t+k, t+k+1] for k = 1..⌈t⌉-ish have counts ≤ C0hat(t+k+1) ≤ C0hat(2t+1)?? The window start t+k ≥ t+1 ≥ 11 ≥ 2 ✓ perimeter applies. C0hat(2t+1) = 0.433·log(2t+3) + 0.886·LL(2t+3) + 2.9 ≤ 0.433·(L + log 2·(1+...)) hmm: log(2t+3) vs L = log(t+2): (2t+3)/(t+2) = 2 − 1/(t+2) ≤ 2 ⟹ log(2t+3) ≤ L + log 2 = L + 0.693. LL(2t+3) = log(log(2t+3)) ≤ log(L + 0.693) ≤ LL + 0.693/L ≤ LL + 0.28 (L ≥ 2.485). So C0hat(2t+1) ≤ C0hat(t+1) + 0.433·0.693 + 0.886·0.28 = C0hat + 0.30 + 0.248 = C0hat + 0.548 ≤ C0hat + 0.55. (D-056 said +0.58 — consistent.) Piece 3 ≤ 4.5√2·[C0hat(t+1) + (C0hat(t+1) + 0.55) + 0.53]·... wait the tail: \"k > t tail ≤ C₀(2t)/(t+1) + 0.433/t ≤ 0.53\" — let me trust D-056's 0.53 but redo: tail = Σ_{k > t} 4.5√2·count_k/(k(k+1)) + the negative-γ side. Hmm — actually the per-interval sum Σ_{k=1}^{K} 1/(k(k+1)) = 1 − 1/(K+1) ≤ 1 — the LEFT side intervals (k = 1..~t−1): ≤ C0hat(t+1)·(1). The RIGHT side (k = 1..~t): ≤ C0hat(2t+1)·1. The TAIL (k > t): Σ_{k>t} C0hat(t+k+1)/(k(k+1)) — with C0hat(t+k+1) ≤ C0hat(2k+1) ≤ 0.433·log(2k+3) + ... ≤ 0.433·(log k + 0.7) + ... — D-056's bound: ≤ C₀(2t)/(t+1) + 0.433/t ≈ 0.53 at t=10 decreasing. For the Lean proof this tail needs a clean summable bound: Σ_{k>t} C0hat(t+k+1)/(k(k+1)) ≤ Σ_{k>t} [0.433·log(t+k+1) + 0.886·LL(t+k+1) + 2.9]/(k(k+1)) ≤ [0.433·log(2t+... hmm — for k > t: t+k+1 ≤ 2k+... ≤ 3k: log(t+k+1) ≤ log(3k) = log 3 + log k. And Σ_{k>t} log k/(k(k+1)) ≤ Σ_{k>t} log k/k² ≤ ∫_{t−1}... ≤ (log t + 1)/t·... At t ≥ 10: ≤ (log 10+1)/10 ≈ 0.33. Plus 2.9·Σ_{k>t} 1/(k(k+1)) = 2.9/(t+1) ≤ 0.27. Plus the LL piece ≤ 0.886·LL(3t)/... small. Tail ≤ ~0.7 at t = 10 decreasing. D-056 used 0.53. For safety use 0.75. Then piece 3 ≤ 4.5√2·[C0hat(t+1) + C0hat(t+1) + 0.55 + 0.75] = 6.36396·[2·C0hat(t+1) + 1.30] = 12.7279·C0hat(t+1) + 8.273. Hmm — slightly bigger constant than D-056's 7.06 (my tail is cruder). Total constants: 0.57 + 2.7 + 8.273 + 0.38 = 11.92; the C0hat constant piece: 12.7279·2.9 = 36.91. Total constant: 36.91 + 11.92 = 48.83 — vs 48!! OVER by 0.83. Hmm. Tighten the tail: at t ≥ 10: Σ_{k>t} log(3k)/(k(k+1)): bound Σ_{k>t} 1/(k(k+1)) = 1/(⌊t⌋+1) ≤ 1/10.5 hmm t ≥ 10 ⟹ k ≥ 11 (k > t ≥ 10, k integer ≥ 11): Σ_{k≥11} 1/(k(k+1)) = 1/11 ≈ 0.0909. Σ_{k≥11} log k/(k(k+1)) ≤ Σ_{k≥11} log k/k² ≤ ∫_{10}^∞ (log x)/x² dx = [−(log x + 1)/x]_{10}^∞ = (log 10 + 1)/10 = 0.3303. So tail ≤ 0.433·(0.3303 + log 3·0.0909) + 0.886·LL-ish·0.0909 + 2.9·0.0909 ≤ 0.433·(0.3303+0.0999) + 0.886·(log log(3·11))·0.0909 + 0.2636 ≈ 0.433·0.4302 + 0.886·1.206·0.0909 + 0.2636 ≈ 0.1863 + 0.0971 + 0.2636 ≈ 0.547. Hmm ≈ 0.55. So tail ≤ 0.55 (not 0.75). Piece 3 constant: 6.36396·(0.55 + 0.55) = 6.36396·1.10 = 7.0004 — wait: piece 3 = 4.5√2·[C₀ + (C₀+0.55) + 0.55] = 4.5√2·(2C₀ + 1.10) = 12.7279·C₀ + 7.0004. Constants: 0.57 + 2.7 + 7.00 + 0.38 = 10.65; C0hat-constant: 36.91. Total: 47.56 ≤ 48 ✓ (0.44 room). L-side: 6.111L + 12.163LL. Pinch check: need 12.163·LL − 4.389·L ≤ 0.44 for L ≥ 2.485: max at L* = 12.163/4.389 = 2.7716: value 12.163·1.01947 − 4.389·2.7716 = 12.400 − 12.166 = 0.234 ≤ 0.44 ✓ (1.9× margin). At L = 2.485: 0.165 ≤ 0.44 ✓. OK — FITS with ~0.2–0.44 margin. Tight but certifiable. (These are all my hand estimates — the Lean proof will need the constants done exactly; there's enough room.)\n\nAlso pieces 4–6 need their honest forms: piece 4/5: |1/s − 1/s₀| ≤ |s₀−s|/(|s||s₀|) ≤ 4.5/t² ≤ 0.045 ✓; piece 6: the ψ-difference — Lemma D (digamma remainder) + the log-difference bound: ½|log(s/2) − log(s₀/2)| + ½|1/s − 1/s₀| + ½(|r(s/2)|+|r(s₀/2)|) — D-056 got ≤ 0.29. Lemma D in Lean: `norm_digamma_sub_log_le_of_two_le_abs_im` (5d) gives ‖ψ(w) − log w‖ ≤ 3 for |im w| ≥ 2 — much weaker than Lemma D's 1.2/|w|² but here we only need the DIFFERENCE ψ(s/2) − ψ(s₀/2) = log(s/2) − log(s₀/2) + O(1/|s|²): |log(s/s₀)| = |log(s/s₀)| with s/s₀ near 1: |log(s/s₀)| ≤ 2·|s/s₀ − 1| ≤ 2·|s₀−s|/|s₀| ≤ 2·4.5/|t| ≤ 9/t. And the remainder difference: via 5d's ‖ψ−log‖ ≤ 3: |ψ(s/2)−ψ(s₀/2)| ≤ |log(s/2)−log(s₀/2)| + 6 — the +6 is WAY too big (piece 6 budget 0.29!). So 5f needs the SHARP digamma remainder (Lemma D: |r(w)| ≤ 1.2/|w|²) — which is D-039's content... which D-056 §1 PROVED on paper (the Binet + recurrence argument) but is it in Lean? 5d's bound is the crude ‖ψ(w) − log w‖ ≤ 3. Lemma D (1.2/|w|²) is NOT in Lean (D-056 §1 was a paper proof; the Lean side has only the 5d crude bound). So 5f also needs Lemma D in Lean — the Binet second formula + the recurrence error E(w) — that's a real analytic sub-build (~0.5 session per D-056 §1's structure: Binet integral bound + the max-principle argument for E(w)... the max-principle on g(u)/u⁴ — Lean has the maximum modulus machinery ✓ but this is work).\n\nHmm — OR avoid Lemma D: bound the ψ-DIFFERENCE directly via the digamma series: ψ(s/2) − ψ(s₀/2) = Σ_j [1/(s₀/2+j) − 1/(s/2+j)] = (s/2 − s₀/2)·Σ_j 1/((s/2+j)(s₀/2+j))... Σ_j 1/((s/2+j)(s₀/2+j)) ≤ Σ_j 1/|s₀/2+j|²-ish ≤ Σ 1/(t/2)²-ish — the series Σ_{j≥0} 1/((t/2)²+j²)-ish ≈ (π/t)-ish... Hmm: ψ(w₁) − ψ(w₂) = Σ_{n≥0} [1/(w₂+n) − 1/(w₁+n)] (the Weierstrass series difference — termwise) = (w₁−w₂)·Σ 1/((w₁+n)(w₂+n)). With |w₁−w₂| = |s−s₀|/2 ≤ 2.25 and |1/((w₁+n)(w₂+n))| ≤ 1/(|im|−...)²... For w = s/2, im w = t/2 ≥ 5: |w+n| ≥ t/2; |w₀+n| ≥ t/2: term ≤ 4/t². Σ over n ≥ 0 of 4/t² DIVERGES. Need the smarter form: |1/((w₁+n)(w₂+n))| ≤ 1/|w₀+n|²·(1+...)... Σ_n 1/|w+n|² for w with im = t/2: Σ_n 1/((t/2)² + (n+re w)²) ≈ ∫ dx/((t/2)²+x²) = π/(t/2) = 2π/t. So the difference ≤ 2.25·(2π/t)·(1+o(1)) ≈ 14/t·... hmm at t = 10: ≈ 1.4 — still bigger than 0.29·2 = 0.58 budget (piece 6 is ½|ψ-diff| ≤ 0.29 ⟹ |ψ-diff| ≤ 0.58). 2.25·2π/10 = 1.41 > 0.58. Hmm. But the sharp version: ψ(w) = log w − 1/(2w) + O(1/|w|²): the difference ψ(w)−ψ(w₀) = log(w/w₀) − (1/2)(1/w − 1/w₀) + O(1/|w|²): |log(w/w₀)| ≤ 1.1·|w/w₀ − 1| = 1.1·|w−w₀|/|w₀| ≤ 1.1·2.25/5 = 0.495 at t=10; the 1/(2w) difference: (1/2)·|w−w₀|/(|w||w₀|) ≤ 0.5·2.25/25 = 0.045; the r-difference: 2·1.2/|w|²·... ≤ 2.4/25 = 0.096. Total |ψ-diff| ≤ 0.495+0.045+0.096 = 0.636; half: 0.318 — hmm slightly over D-056's 0.29 but the log part: |log(w/w₀)| with w/w₀ = (s/2)/(s₀/2) = s/s₀: |s/s₀| = |s|/|s₀| ∈ [0.55/1, ...]... σ ∈ [−5/2,3], s₀ = 2+it: |s−s₀| = |σ−2| ≤ 4.5; |s₀| ≥ |t|: |log(s/s₀)| ≤ |s/s₀ − 1|/(1−|s/s₀−1|)-ish ≤ 1.1·(4.5/10) = 0.495 at t=10. So ½|ψ-diff| ≤ 0.318 vs D-056's 0.29. Over by 0.03 — because D-056's 0.23 for ½|log(s/s₀)| used |arg(s/s₀)| ≤ 4.5/|t| hmm: |log(s/s₀)|² = (log|s/s₀|)² + arg²; D-056: \"½|log(s/s₀)| ≤ 0.23 at |t| = 10\" — via |log|s/s₀|| ≤ 0.02 and |arg| ≤ 4.5/|t| = 0.45: |log(s/s₀)| ≤ √(0.02² + 0.45²) ≈ 0.45; ½·0.45 = 0.225 ≤ 0.23 ✓. My 1.1·|Δ| bound gave 0.495 (lossy by the 1.1 + only-modulus). The arg form: |arg(s/s₀)| ≤ |s−s₀|/|s₀|·(1+...) = 4.5/10 = 0.45 (arg ≤ tan-ish). OK so with the arg/modulus split, ½|log| ≤ 0.226, ½|1/(2w)-diff| ≤ 0.023, ½|r-diff| ≤ 0.048: total 0.297 ≈ 0.29 ✓ D-056's. So piece 6 needs the ARG bound on log(s/s₀) + the sharp digamma remainder r bound 1.2/|w|². The r bound (Lemma D) is the Binet argument — a real sub-build.\n\nAlternatively, piece 6 crude-but-sufficient: need ½|ψ(s/2) − ψ(s₀/2)| ≤ C₆ with the total fitting. Budget for piece 6: 48 − (other pieces). Other pieces at the pinch: Z₂ 0.57 + piece2 2.7 + piece3-const 7.00 + pieces4–5 0.09 = 10.36; C0hat-const 36.91; total-without-6: 47.27. Room for piece 6: 48 − 47.27 = 0.73. So piece 6 ≤ 0.73 suffices!! My crude 0.318–0.636... the ½|ψ-diff| ≤ 0.318 (arg-split version) fits; the cruder 0.636/2... wait ½·0.636 = 0.318 ✓ also fits (0.318 ≤ 0.73). And even cruder: via 5d's ‖ψ(w) − log w‖ ≤ 3: ½|ψ-diff| ≤ ½|log(w/w₀)| + ½·6 = 0.226 + 3 = 3.23 — too big. So the middle route: the digamma SERIES difference |ψ(w)−ψ(w₀)| ≤ |w−w₀|·Σ_n 1/|w+n||w₀+n| with Σ ≤ 2π/t·(1+o(1)) — the o(1) needs care — gives ½·2.25·(2π/t + small) ≈ ½·2.25·0.66 = 0.74 at t=10 — JUST over 0.73! Knife-edge again. The ARG-split + series route: ½|log(w/w₀)| ≤ 0.226 + ½·|series-remainder-difference|... \n\nHmm, actually the cleanest: |ψ(w) − ψ(w₀)| ≤ |log w − log w₀| + |(ψ−log)(w)| + |(ψ−log)(w₀)|. The (ψ−log) remainder via the SERIES: ψ(w) − log w = −1/(2w) − Σ_{n≥1} 1/((w+n)·n)·... the Weierstrass form: ψ(w) = −γ − 1/w + Σ_{n≥1}(1/n − 1/(w+n))... and log w = ... the identity ψ(w) = log w − 1/(2w) − Σ... not clean.\n\nOK — enough. The point: piece 6 has budget 0.73 and multiple routes land ≤ 0.32–0.74 — the sharp Lemma D route (0.29) is the paper's; a Lean-friendly route via the digamma series difference exists but needs the Σ 1/|w+n|² ≤ 2π/t·(1+o(1)) estimate (the 5d/5e-ii machinery has the separated reciprocal sums — `digamma_recip_sum_le` (C) gives Σ‖(w+j)⁻¹‖ ≤ 2/c + 2 + 2·log m for dist ≥ c — that's for the near-axis; the strip version Σ 1/((t/2)²+(n+u)²) ≤ π/(t/2)+... via arctan-FTC (5d's pattern) ✓ machinery exists).\n\nSo 5f's full piece list in Lean:\n1. dirichletC2 ≤ 0.57 — NEEDS a certified tsum bound! Σ Λ(n)/n² ≤ 0.57. In Lean: the tsum is symbolic. Options: (a) partial sum to N + tail bound — a finite computation: Σ_{n≤N} Λ(n)/n² + tail ≤ 0.57 with the tail (log N+1)/N... E-028 used N = 10⁶. In Lean, Σ_{n≤10⁶} Λ(n)/n² is a HUGE finite computation — decide-able but slow (10⁶ terms with real division... as rationals? Λ(n) is log p or 0 — the sum over prime powers: Σ_p log p ·(1/p² + 1/p⁴ + ...) = Σ_p log p/(p²−1)... the partial sum to 10⁶ as a rational number — evaluating in Lean by norm_num/decide is infeasible (10⁶ terms). Native_decide with floats?? Not trustworthy for the audit (well — native_decide IS trusted-ish... no, native_decide adds an axiom-free trust in the compiler — the project avoids it; the LiBasePoints used certified interval arithmetic via proved rational bounds). Alternative: (b) bound dirichletC2 by a clean analytic expression: Σ Λ(n)/n² = −ζ′(2)/ζ(2). Does Mathlib have ζ′(2)? No. (c) Coarser: Σ Λ(n)/n² ≤ Σ log n/n² =: −ζ′(2) ≈ 0.9375 — still needs a tsum bound. Σ_{n≥2} log n/n² ≤ ∫_1^∞ log x/x² + (max log x/x² at x = e^{1/2})... Σ_{n≥2} log n/n² ≤ (log 2)/4 + ∫_2^∞ log x/x² dx = 0.173 + (log 2 + 1)/2 = 0.173 + 0.847 = 1.02 — too big (need ≤ 0.57 + the 0.73 budget slack ≈ ≤ 1.3 — wait piece 1's budget: total constants 48 − (36.91 + 2.7 + 7.00 + 0.38) = 48 − 46.99 = 1.01! So piece 1 ≤ 1.01 suffices!! Σ Λ(n)/n² ≤ Σ_{n≥2} log n/n² ≤ 1.02 — KNIFE-EDGE over. Hmm. Sharper: Σ_{n≥2} log n/n²: exact ≈ 0.9375. Integral bound with the first terms exact: (log 2)/4 + (log 3)/9 + ∫_3^∞ log x/x² = 0.1733 + 0.1221 + (log 3+1)/3 = 0.1733+0.1221+0.6995 = 0.9949 ≤ 1.01 ✓ FITS by 0.015. KNIFE-EDGE but the first-terms-exact version is a clean Lean-provable bound (two exact rationals-logs + the integral (log 3 + 1)/3 via integration by parts — Mathlib has ∫ log x/x²? Hmm — `Real.integral_...`? Might need to derive: d/dx[−(log x + 1)/x] = (log x)/x² ✓ so ∫_3^N = (log 3+1)/3 − (log N+1)/N → (log 3+1)/3. Provable via FTC (intervalIntegral with the explicit antiderivative — doable: `intervalIntegral.integral_deriv_eq_sub` or by the hasDeriv pattern). And the tail series ≤ integral needs the decreasing-ness of log x/x² for x ≥ 3 (deriv ≤ 0) — `AntitoneOn.integral_le_sum` reversed: for decreasing f ≥ 0, Σ_{n≥N+1} f(n) ≤ ∫_N^∞ f — Mathlib: `Summable`/`integral_le_sum`... there are lemmas like `Real.sum_le_integral_of_decreasing`? For tsum: `tsum_le_integral...`? Might need the partial-sum form + limits. Routine but real work.)\n   Actually — simpler: does the fold even need dirichletC2 ≤ 0.57? The R-bound theorem is |R| ≤ 10.5L + 48. Piece 1 enters the CONSTANT side. Budget 1.01. So I need dirichletC2 ≤ 1.01, provable via the first-two-terms + integral bound (≈ 0.995). OK — feasible. Note: this produces the bound ‖logDeriv ζ (2+it)‖ ≤ dirichletC2 ≤ 0.995 — slightly weaker than Z₂ = 0.57 but FITS the budget.\n\n2. The two-unit count (perimeter growth — USER DECISION).\n\n3. Piece 3 far sum: the per-interval machinery — significant Lean work (the interval decomposition of the far sum Σ_{|γ−t|≥1} m·(s₀−s)/((s−ρ)(s₀−ρ)) — needs: termwise bound 4.5√2/(x(1+x)) [x = |t−γ|]; the split into unit intervals on both sides + the negative-γ side via conjugation; per-interval mult-counts from the perimeter (left: ≤ C0hat(t+1) via mono for windows in [2, t]; the LOW windows (< 2): no zeros?? — need the zero-free input |γ| ≥ 14.13... WAIT — do we have that in Lean? The nontrivial zeros: |γ| ≥ 14.13 is the verified RH-to-height-3×10¹²'s weaker cousin — the zero-free region |t| ≤ 14-ish is... RHUpTo is a PERIMETER hypothesis (Platt–Trudgian) — available! hRH : RHUpTo 3×10¹² gives zeros on the line up to that height — but does it give the LOW bound |γ| ≥ 14? RHUpTo says zeros with |γ| ≤ 3×10¹² have re = 1/2 — it does NOT say there are no zeros with |γ| < 14!! The first-zero statement \"no nontrivial zeros with |γ| ≤ 14\" is a separate classical fact (backlund: the zeros start at 14.13). Is it in Zeta23 or our package? lowZeroConst — defined as Σ over lowZeros of mult·(1+1/‖ρ‖) — lowZeros := zeros with ‖ρ‖ ≤ some bound? Let me check the lowZeros def — from session-031: \"lowZeroConst\" built from the finite ball — if it's NONZERO, then the low windows have a count input via lowZeroConst (finite, unconditional). Let me check its definition/value.\n\nHmm — actually for piece 3, the LEFT intervals (t−k−1, t−k] for k = 1..⌈t−2⌉: windows starting at t−k−1 ≥ ... ≥ 2 requires k ≤ t−3; for k ∈ (t−3, t): windows starting in [0, 2) — the zeros there: none (14.13) — OR use lowZeroConst if it covers them. But actually — hmm, simpler: for the left side, intervals with t−k−1 < 2 contain only zeros with γ < 3 < 14.13 — if we have the zero-free fact, those intervals are EMPTY. If we don't, we need SOME count. Let me check what zero-free inputs exist: grep \"14.13\" or \"zeroFree\" in Zeta23/our package.\n\nAlso — wait, hmm: is it even true that the perimeter ZeroCountProp's range (t ≥ 2) suffices for all the left intervals? k from 1 to ~t: window start t−k−1 ranges from t−2 down to ~−1. Windows with start < 2: (t−k−1, t−k] with t−k−1 < 2 ⟺ k > t−3. For t ≥ 10: k ∈ (7, 10): windows like (1,2], (0,1], (−1,0]... zeros with γ ∈ (−1, 3]: NONE (|γ| ≥ 14.13). So need the zero-free fact. If unavailable: these intervals' zeros are part of the |γ| ≤ 3 ball — the lowZeros finset. lowZeroConst is a SUM (mult·(1+1/‖ρ‖)) — from it, the COUNT of lowZeros ≤ lowZeroConst/(1+1/‖ρ‖)-ish... messy. Check if Zeta23 has the 14.13 zero-free region.\n\nLet me grep for it.","signature":""},{"type":"text","text":"Now checking what zero-free/low-zero inputs exist (piece 3's low intervals need them) and the `lowZeros`/`lowZeroConst` definitions."},{"type":"tool_use","id":"Bash_116","name":"Bash","input":{"command":"cd D:/Python/Quantum/research/lean/NguyenRH1/NguyenRH1; grep -n \"def lowZeros\\|def lowZeroConst\\|lowZeros :=\" LiCircleBound.lean | head -4; grep -rn \"14.13\\|14.134\\|zeroFree\\|zero_free\\|no.*zero.*below\\|zeros.*empty\" LiSkeleton.lean LiCircleBound.lean LiDisc.lean | head -8; echo ===Z23===; grep -rn \"14.13\\|14.134\" D:/Python/Quantum/zeta-23-lean/Zeta23/ --include=\"*.lean\" -l | head -3","description":"Find lowZeros definition and zero-free inputs"},"extra_content":null}]