[{"type":"thinking","thinking":"The RvM constant is `max 1 (max (2 * A₁) K)` — symbolic (A₁ from half_count_large, K from a finite count) — no explicit numeric value. So A0count has no numeric bound in the package. Using it in 5f's numeric fold would require bounding it — a sub-project.\n\nSo for piece 3's low intervals, the clean route is a **numeric zero-free input**: no nontrivial zeros with |γ| ≤ 14 (say). Is that derivable cheaply? Hmm — actually, wait. Let me think about whether piece 3 even needs the low-window counts to be NUMERIC...\n\nThe far sum Σ_{|γ−t|≥1} m·(s₀−s)/((s−ρ)(s₀−ρ)): bound per term 4.5√2/(x(1+x)), x = |t−γ| ≥ 1. Split:\n- LEFT (γ < t−1, includes all negative γ and low positive): intervals (t−k−1, t−k], k = 1..K_max where the intervals bottom out at... γ ≥ −∞ — the intervals extend down indefinitely! (t−k−1, t−k] for k → ∞: γ → −∞. The per-interval count for low/negative heights: the A0count bridge gives count ≤ A₀·log(|height|+3) — SYMBOLIC A₀. The sum Σ_k A₀·log(k+5)/(k(k+1)) converges but has A₀ in it — the total would be (something)·A₀ — symbolic — BREAKS the numeric fold.\n\nHmm!! Wait — but D-056's piece 3 says \"left side ≤ C₀(t)\". How does that handle the deep-negative γ? For γ very negative, x = |t−γ| = t + |γ| — large — the intervals are (t−k−1, t−k] with k up to t+|γ| — the per-interval counts C₀(t−k) for t−k < 0... C₀ of a negative height? D-056's left-side bound ≤ C₀(t) treats the whole left side as intervals with count ≤ C₀(t) — the count of the window at height t−k for k > t is the count of zeros in NEGATIVE-height windows — by conjugation = count at |t−k|... The honest statement: the interval (t−k−1, t−k] for k > t contains zeros with negative ordinates; by conjugation symmetry their count equals the count in (k−t, k−t+1] (positive) ≤ C₀(k−t+1)... which GROWS with k. The per-term weight 4.5√2/(k(k+1)). So the left side Σ = Σ_{k=1}^{∞} count_k·4.5√2/(k(k+1)) where count_k ≤ C0hat-ish(min(k−t+1, t+1)-ish)... For k ≤ t−2: count ≤ C₀(t). For k > t: count ≤ C₀(k−t+1) ≤ 0.433·log(k−t+3)+... The tail Σ_{k>t} C₀(k−t+1)/(k(k+1)) ≤ Σ_j C₀(j+1)/((j+t)(j+t+1)) ≤ C₀-ish·Σ 1/(j+t)² ≤ C₀(j~small)... — C₀(j+1) for small j is the LOW count again (symbolic via A₀)!! Hmm no — C₀ as a closed form 0.433 log(j+3) + ... is a VALID upper bound only where the perimeter/Trudgian applies (j+1 ≥ 2). For j < 2: the zeros with |γ| ≤ 3-ish: count unknown (symbolic). BUT the per-term weight at those k is ≤ 4.5√2/((t)(t+1)) ≤ 4.5√2/110 ≈ 0.058 at t = 10. So the low-zero contribution to the far sum ≤ 0.058·(low mult count). If the low mult count is symbolic (lowZeroConst-ish or A₀-based), the fold breaks by a symbolic term × 0.058. Hmm — UNLESS we use the A0count bridge and accept A₀ symbolically in the bound... but then the final fold ≤ 10.5L + 48 needs A₀ gone.\n\nOK so no matter what, 5f needs SOME numeric handle on the low zeros. Cleanest: the classical fact \"no nontrivial zeros with |im| ≤ 14\" (or ≤ 3, enough for the intervals at play... wait — the negative side: intervals (t−k−1, t−k] for k ∈ (t−3, t+3] cover heights (−4, 2]: zeros there have |γ| ≤ 4. So the needed input: no zeros with |γ| ≤ 4 (or 14). This is a FINITE numeric fact (the first zero is at 14.13). How to prove in Lean cheaply? \n\nOption (z1): from RHUpTo + ... no (RHUpTo doesn't exclude low zeros).\nOption (z2): a new tiny perimeter declaration: \"no nontrivial zeros with |im| ≤ 14\" — this is part of the Platt–Trudgian certificate semantically (they located the zeros) but NOT implied by RHUpTo as encoded. It's the same CLASS as the other perimeter items (computational, published). Hmm — perimeter growth again.\nOption (z3): prove it from the argument principle numerically — the rectangle integral of ζ′/ζ over the box [−1, 2]×[−14, 14]... in Lean, with interval arithmetic — a real sub-build (~1 session+, needs numeric ζ evaluation in Lean — we don't have a numeric ζ oracle in Lean!).\nOption (z4): derive from a ζ-value fact in Mathlib: does Mathlib have ζ(0) = −1/2 and nonvanishing on [some interval]? No zero-free strip in the critical strip.\nOption (z5): avoid the low intervals entirely by NOT splitting there: bound the left-side far sum differently — pair each far zero's term with 1/|ρ|... hmm: Σ_{|γ−t|≥1} m/((|t−γ|)(1+|t−γ|)) — for the zeros with |γ| ≤ 3: ≤ Σ_{|γ|≤3} m/((t−3)(t−2)) ≤ (low count)·2/t² — the low count via... symbolic. Circular.\n\nHmm — wait, actually, hold on: maybe use the SERIES total instead. The far difference sum is bounded by Σ_all m/((x)(1+x))... Or bound the whole far sum via the CONVERGENT majorant Σ m/(1+γ²) (Zeta23's zero_sum_inv_sq — symbolic value?) — zero_sum_inv_sq gives SUMMABILITY, the sum's value is symbolic. Hmm.\n\nActually — hmm, wait. Let me reconsider. For the low intervals, do we even need their count to be ZERO or numeric? The far-sum total must be ≤ 12.73·C₀ + 7.06 — NUMERIC. Every piece must be numeric. The low intervals contribute (low count)·(their weight). If the low count is symbolic, no numeric fold. So YES — need a numeric low-zero input.\n\nThe honest cheapest: a small perimeter declaration `NoLowZeros : ∀ ρ, IsNontrivialZero ρ → 14 ≤ |ρ.im|` — justified as part of the Platt–Trudgian-class computational certificate (it's the classical first-zero location, machine-verified many times; arguably implied by the Platt–Trudgian computation which located all zeros to 3×10¹²). It's a new declared item.\n\nHmm — wait, actually. Is it implied by RHUpTo + the N(T) machinery? N(T) for T = 14: F(14) ≈ (14/6.28)·log(14/6.28) − 14/6.28 + 0.875 ≈ 2.229·0.802 − 2.229 + 0.875 ≈ 1.788 − 1.354 = 0.434. |S(14)| ≤ S_b(14) = 0.137·2.64 + 0.443·0.97 + 1.588 ≈ 0.362 + 0.430 + 1.588 = 2.38. N(14) ≤ 0.434 + 2.38 = 2.81 — doesn't give 0. So no, the Trudgian count can't exclude low zeros (the S_b slack is too big). The first-zero exclusion needs the actual zero-location computation. So it's a genuine new perimeter item (Platt-class: Platt–Trudgian's certificate DOES include the zero locations to 3×10¹² — declaring \"no nontrivial zeros with 0 < |im| ≤ 14\" is a weak consequence of their certificate).\n\nHmm hold on, wait — actually does it? Platt–Trudgian verified RH up to 3×10¹²: all zeros with 0 < γ ≤ 3×10¹² have re = 1/2. This doesn't exclude zeros at LOW heights either — it's an on-line statement, not a count. The zero LOCATIONS (first at 14.13) come from the zero-counting part of such computations (locating sign changes of Z(t)). Platt's 2017/Platt–Trudgian 2021 computations did isolate the zeros (that's HOW you verify RH — count zeros via Turing's method and locate them). So \"no zeros with |γ| < 14\" is a certified consequence of the Platt-class computation — same trust class. But as a LEAN hypothesis it's new.\n\nHmm, wait — alternatively, is \"no zeros with |γ| ≤ 14\" provable from RHUpTo 3×10¹² + the RvM count + ...? No — RHUpTo is purely on-line.\n\nOK so 5f needs TWO perimeter additions:\n(P1) ZeroCountProp2 (two-unit mult-weighted window count, closed form 0.6L + 0.886LL + 2.7, t ≥ 10) — Trudgian N(T)/S(T) specialization.\n(P2) NoLowZeros (no nontrivial zeros with |im| ≤ 14 — or some explicit height) — Platt-class zero-location certificate.\n\nBoth are honest, same-class-as-existing-perimeter declarations. But the perimeter is the trusted surface and growing it needs the user's sign-off — especially (P2).\n\nHmm, wait — actually, let me reconsider (P2). Is there a way to bound the low-zero count numerically from EXISTING proved material? The mult-count of zeros with |γ| ≤ 3: via Jensen on a disc (LiDisc machinery!): scaledZeta_jensen_count gives a NUMERIC bound? The Jensen count: ≤ log((9/4)|t|²)/log(33/32) — for the disc at 2+it with |t| ≤ 3: log((9/4)·9)/log(33/32) = log(20.25)/0.03077 ≈ 3.008/0.0308 ≈ 97.8. So the low count ≤ 98 NUMERICALLY (the Jensen bound is fully proved in LiDisc — no perimeter!). The low-interval contribution: zeros with |γ| ≤ 3, in the far sum their x = |t−γ| ≥ t−3 ≥ 7: per-term ≤ 4.5√2/(7·8) = 0.1137; count ≤ 98: contribution ≤ 11.15?!? WAY too big (the far-sum budget is ~12.73·C₀ ≈ 60–100, but the constant side...). Hmm wait — 11.15 on the constant side: total constant 47.56 + 11.15 = 58.7 > 48. BREAKS. The Jensen count 98 is too lossy (true count is 0). \n\nRefine: the low zeros' far-sum contribution with the x-weight: Σ_{|γ|≤3} m·4.5√2/(x(1+x)) with x ≥ t−3: ≤ (count)·4.5√2/((t−3)(t−2)). At t = 10: 98·6.364/42 = 14.8. Too big. At larger t it decays (1/t²) but the fold's t=10 pinch is where it matters. So the Jensen route breaks the constants. NEED the true zero-free input (P2).\n\nHmm, unless... bound the low count via the A0count bridge: count of zeros in [−3, 3] ≤ 3·A₀·8·log(14) ≈ 63·A₀ — A₀ symbolic. No.\n\nWhat about a TIGHTER Jensen: the disc of radius 5/2 at 2+it covers re ≥ −1/2... for the low zeros a disc centered at 2 (t=0) radius ~4: Jensen with the disc bound ‖ζ‖ ≤ (3/4)|t|²... at t=0 the disc bound is ‖ζ‖ ≤ (3/4)·|t|² — at t = 0 that's 0?! The LiDisc bound needed |t| ≥ 10. For a disc at height 0: ‖ζ‖ on B(2, 4): the max of |ζ| there — a numeric fact... could be computed from the EM representation (LiRBound's level-2 form!) at finitely many... — the EM bound: ‖ζ(s)‖ ≤ 1/2 + 1/|s−1| + |s(s+1)|·∫₁^∞ B̃₂(x)x^{−σ−2} — at re ≥ −1/2: the integral ≤ ∫₁^∞ (1/12)x^{−3/2} = 1/6-ish; so ‖ζ(s)‖ ≤ 1/2 + 1/(dist to 1) + |s(s+1)|/6 — on the circle |s−2| = 4: |s(s+1)| ≤ 6·7 = 42; 1/|s−1| ≤ 1/3 (nearest point s = −2: |−2−1| = 3); so ‖ζ‖ ≤ 0.5 + 0.34 + 7 = 7.84 on the circle. Jensen count of zeros in B(2, 3.5): ≤ log(7.84/|ζ(2)|)/log(4/3.5)·... |ζ(2)| = π²/6 ≈ 1.645: log(7.84/1.645) = log 4.77 = 1.562; radius ratio 4/3.5: log(8/7) = 0.1335: count ≤ 1.562/0.1335 = 11.7. So ≤ 11 zeros (with mult) in B(2, 3.5) ⊇ {|γ| ≤ 3}-ish. Contribution: 11·0.1137 = 1.25 at t=10 — still too big (budget-wise: constant side 47.56+1.25 = 48.8 > 48 ✗). Tighter disc: B(2, 3.2) count ≤ log(4.77)/log(4/3.2) = 1.562/0.223 = 7.0 ⟹ 7·(weight at x ≥ t−3.2+...) ≈ 7·6.364/((6.8)(7.8)) ≈ 0.84 at t=10. Still 0.84 over a budget of 0.44. Hmm. Tighter still: B(2, 2.5): log(4.77)/log(4/2.5) = 1.562/0.470 = 3.32 ⟹ count ≤ 3.32 — the zeros with |γ| ≤ √(2.5²−1) ≈ 2.3: contribution 3.32·6.364/((t−2.3)(t−2.3+1)) at t=10: 3.32·6.364/(7.7·8.7) = 0.315 ≤ 0.44 ✓ FITS! So a Jensen count ≤ 3.32 for zeros in B(2, 2.5) + the far-weight gives ≤ 0.32 at t=10 — but WAIT: zeros with 2.3 < |γ| ≤ 3 fall OUTSIDE this disc — they're in the far sum with x = |t−γ| ∈ [t−3, t−2.3]: their count via the LARGER disc B(2, 3.5) minus... the annulus count: ≤ 11.7 (B(2,3.5) count) with weight 4.5√2/((t−3)(t−2)) = 0.1137 at t=10: ≤ 1.33. Hmm — the annulus zeros (2.3 < |γ| ≤ 3) — do any exist? In reality NO (first zero 14.13). The Jensen count of the annulus = count(B(2,3.5)) − count(B(2,2.5))-ish — Jensen gives upper bounds only: 11.7 for the big disc INCLUDING the small one. Annulus ≤ 11.7. Contribution ≤ 11.7·0.1137 = 1.33. Over budget again. To kill it: need the annulus empty — zero-free. ARGH.\n\nActually — the cleanest: the Jensen count on a disc containing ALL of {|γ| ≤ 3}: if that count is < 4 (say ≤ 3.32 via B(2, 2.5))... but zeros at |γ| ∈ (2.3, 3] aren't covered. Cover {|γ| ≤ 3} with a disc where Jensen gives < 1: count < 1 ⟹ count = 0 ⟹ ZERO-FREE! Jensen count < 1 needs log(M/|ζ(2)|)/log(R/r) < 1 ⟺ M/|ζ(2)| < R/r. With M = 7.84 (on |s−2| = 4), |ζ(2)| = 1.645: M/|ζ(2)| = 4.77; R/r = 4/r: need r > 4/4.77 = 0.84. So for the disc B(2, r) with r ∈ (0.84, ...): Jensen count ≤ log(4.77)/log(4/r) < 1 ⟺ r > 4/4.77 = 0.839. So: the count in B(2, 0.85) ≤ log(4.77)/log(4/0.85) = 1.562/1.549 = 1.008 — hmm > 1 barely. B(2, 0.9): 1.562/log(4/0.9) = 1.562/1.493 = 1.046 — worse (bigger disc). B(2, 0.84): log(4/0.84) = 1.560: 1.562/1.560 = 1.001. Hmm the ratio M/|ζ(2)| = 4.77 ≈ R/r needs r ≈ 0.84 — count ≤ 1.001 — just OVER 1. To get count < 1: need a bigger outer disc (better ratio): outer R = 5: M on |s−2| = 5: ‖ζ‖ ≤ 0.5 + 1/4 + |s(s+1)|/6 ≤ 0.75 + 8·9/6 = 12.75; M/|ζ(2)| = 7.75; log 7.75 = 2.048; inner r: need log(5/r) > 2.048 ⟺ r < 5/7.75 = 0.645. So B(2, 0.64) has count ≤ 2.048/log(5/0.64) = 2.048/2.056 = 0.996 < 1 ⟹ ZERO-FREE in B(2, 0.64) ⟹ no zeros with |γ| ≤ √(0.64² − ...) hmm — B(2, 0.64) covers |γ| ≤ √(0.64² − 1²)?? re ρ ∈ (0,1): ρ ∈ B(2, 0.64) requires |γ| ≤ 0.64... so zero-free for |γ| ≤ 0.64 only. The strip zeros with |γ| ∈ (0.64, 3]: not covered. Bigger discs have worse M. The disc approach can't reach |γ| ≤ 3 (M grows like |s(s+1)|/6 ~ R²/6, log M ~ 2 log R, and log(R/r) ~ log R — ratio → 2 — count → ~2·(log|ζ(2)|⁻¹+...)/1 — can't get below ~1.5–2 for r ~ 3). So Jensen CANNOT prove the low-zero emptiness to |γ| ≤ 3. It can only give count ≤ 2-ish for |γ| ≤ 3 — contribution to the far sum ~2·0.11 = 0.22 ≤ 0.44 budget ✓?? Let me redo: the best Jensen disc covering {|γ| ≤ 3}: need r ≥ √(9+1) ≈ 3.16 (re ρ ≤ 1: |ρ−2| ≤ √(1+γ²)): r = 3.2, outer R = 6: M on |s−2| = 6: ≤ 0.5+1/5+10·11/6 = 0.7+18.3 = 19; M/|ζ(2)| = 11.55; log = 2.447; log(6/3.2) = 0.628; count ≤ 2.447/0.628 = 3.90. ⟹ ≤ 3.9 for |γ| ≤ 3. Far-weight at t=10: x ≥ 10−3 = 7: 4.5√2/(7·8) = 0.1137 ⟹ contribution ≤ 3.9·0.1137 = 0.443 — JUST at the 0.44 budget!! KNIFE-EDGE. And my M estimate is crude (|s(s+1)| ≤ 10·11 — the max of |s(s+1)| on |s−2|=6 at s = 2+6i... |s| = |2+6i| = 6.32, |s+1| = 7.16: product 45.3/6 = 7.55; hmm |s(s+1)| max on the circle: s = −2−6i? |s| = 6.32, |s+1| = |−1−6i| = 6.08: 38.4. At s = 2+6i: 6.32·7.16 = 45.3. At s = 8: |s(s+1)| = 72. At s = −4: 4·3 = 12. So max ≈ 72 (s = 8): M ≤ 0.5 + 1/7 + 72/6 = 12.64. M/|ζ(2)| = 7.68, log = 2.039. count ≤ 2.039/0.628 = 3.25. Contribution 3.25·0.1137 = 0.369 ≤ 0.44 ✓.) So the Jensen route MIGHT fit the budget with a careful M — but every constant here needs exact Lean verification (the EM bound on a circle, |ζ(2)| = π²/6, the log evaluations...) and it only gives count ≤ 3.25 for |γ| ≤ 3 — whereas the TRUTH is 0. The contribution 0.37 vs budget 0.44 — knife-edge again, and it eats the pinch margin.\n\nvs (P2): declare \"no nontrivial zeros with |im| ≤ 14\" (Platt-class certificate — it IS part of what Platt–Trudgian-style computations certify; Turing's method explicitly counts and locates zeros). Then the low intervals are EMPTY and the issue vanishes. MUCH cleaner.\n\nSo the honest 5f needs TWO perimeter additions: (P1) the two-unit count, (P2) the low-zero exclusion. Both same-class (published computational certificates). This is a perimeter growth from 4 to 6 items — significant, needs user sign-off.\n\nHmm, wait — actually, let me reconsider (P2) ONE more time. Does the far sum REALLY hit the low zeros? The far sum Σ_{|γ−t|≥1} m(s₀−s)/((s−ρ)(s₀−ρ)) — for t ≥ 10, the zeros with |γ| ≤ 3 have x = |t−γ| ≥ 7, contributing ≤ (count)·4.5√2/(7·8)·... — with (P2) count = 0: contribution 0 ✓. Without (P2): symbolic. YES, (P2) or a numeric low-count is required.\n\nHmm — OR: restate the far sum split to avoid low zeros: split at |γ−t| ≥ 1 into 1 ≤ |γ−t| ≤ 3 and |γ−t| > 3?? No... The low zeros are unavoidable in the far sum (they're part of seriesFun). Their contribution must be bounded numerically — either via (P2) (count 0) or a numeric count bound (Jensen — knife-edge 3.25 with heavy verification).\n\nActually — WAIT. Alternative for the low-zero contribution without ANY new input: bound Σ_{|γ|≤3} m/(x(1+x)) ≤ Σ_{|γ|≤3} m/(x²) ≤ Σ_{|γ|≤3} m·2/(1+γ²)·(1+γ²)/x²... and Σ_{|γ|≤3} m/(1+γ²) is part of Zeta23's zero_sum_inv_sq total (symbolic sum value). No.\n\nOR use the LiSkeleton/E-011-style: the first zeros are at 14.13 — the E-008/E-011 zero tables computed 10⁴ zeros — but those are NUMERICS, not Lean facts. To get the fact into Lean: declaration (perimeter) or a verified-computation Lean proof (out of scope).\n\nSo (P2) it is — as a declared perimeter item. Actually — hold on, one more: could RHUpTo be REPLACED/STRENGTHENED to include the low-zero exclusion? RHUpTo 3×10¹² (Platt–Trudgian) — the semantic content of the Platt–Trudgian certificate: \"all zeros with 0 < γ ≤ 3×10¹² have re = 1/2 AND there are exactly N of them at the expected locations\" — the Turing-method certificate includes the zero COUNT matching N(T) — which implies no zeros below the first located one. So (P2) is a consequence of the FULL Platt–Trudgian certificate. Declaring it separately under the same source is honest. (Paper appendix at ii′-7 will list it.)\n\nAlright — the design is now clear enough to present to the user. The 5f plan:\n\n**New perimeter declarations (2, same source classes as existing):**\n- (P1) `ZeroCountProp2`: the two-unit mult-weighted window count: ∀ t ≥ 10, Σ_{|γ−t|≤1} m ≤ 0.6·log(t+2) + 0.886·loglog(t+2) + 2.7 (Trudgian N(t+1)−N(t−1) + S_b pair specialization).\n- (P2) `NoLowZeros`: ∀ ρ nontrivial, 14 ≤ |ρ.im| (Platt-class zero-location certificate; the classical first-zero exclusion).\n\n**5f build (LiTightR.lean):**\n1. The reference-point difference identity from partialFraction_identity at s and s₀ = 2+it (B cancels): logDeriv ζ s = logDeriv ζ s₀ + Σ m(1/(s−ρ)−1/(s₀−ρ)) − (1/s−1/s₀) − (1/(s−1)−1/(s₀−1)) − ½(ψ(s/2)−ψ(s₀/2)). Needs: the series difference termwise = Σ m(s₀−s)/((s−ρ)(s₀−ρ)) + summability of the difference series (the seriesSummable machinery + the difference identity — a tsum subtraction argument).\n2. Piece 1: ‖logDeriv ζ s₀‖ ≤ dirichletC2 ≤ 0.995 (first-terms + integral bound on Σ Λ(n)/n²... wait — dirichletC2 = Σ ‖LSeries.term Λ 2 n‖ = Σ Λ(n)/n². The bound ≤ 0.995 via (log 2)/4 + (log 3)/9 + (log 3+1)/3 — needs Λ(n) ≤ log n and the integral Σ_{n≥4} log n/n² ≤ ∫_3^∞ log x/x² = (log 3+1)/3 — the decreasing-tail integral comparison (Mathlib: `Summable`... the lemma form: for f decreasing nonneg, Σ_{n≥N} f n ≤ f N + ∫_N^∞ — `tsum_le_integral`? hmm — there's `MeasureTheory...` no — for real decreasing f: `Real.tsum_le_integral...`? There might be something like `summable... integral_test` — the integral test gives equivalence not inequality. The inequality Σ_{n≥N+1} f(n) ≤ ∫_N^∞ f for decreasing f: `AntitoneOn...`? I recall `Real...`? If missing, prove via per-term f(n+1) ≤ ∫_n^{n+1} f and tsum of integrals = the big integral. ~30 lines.)\n   Hmm — actually even simpler: dirichletC2 ≤ 0.995 might be avoidable: Σ Λ(n)/n² where Λ(n) ≤ log n·[n prime power]... the cheapest: Λ(n) ≤ log n for all n (ArithmeticFunction.vonMangoldt_le_log? — exists in Mathlib ✓ I believe) ⟹ dirichletC2 ≤ Σ_{n≥2} log n/n² — and that tsum ≤ 0.995 via the integral comparison. OK.\n3. Piece 2: near sum at s₀ ≤ the two-unit count (P1) — needs |s₀−ρ| ≥ 1 for near zeros (re ρ < 1 ⟹ |s₀−ρ| ≥ 2−re ρ > 1... ≥ 1 via |re part| ≥ 1: |s₀−ρ| ≥ |re(s₀−ρ)| = 2 − re ρ ≥ 1 ✓ (re ρ < 1 ⟹ 2 − re ρ > 1 ≥ 1 — actually ≥ 1 suffices: abs_re_le_norm).\n4. Piece 3: the far sum — the big one. Termwise: ‖(s₀−s)/((s−ρ)(s₀−ρ))‖ ≤ 4.5/(|t−γ|·√(1+(t−γ)²))·... hmm D-056 used |s−ρ| ≥ |t−γ| and |s₀−ρ| ≥ √(1+(t−γ)²) ≥ (1+|t−γ|)/√2. Then per-interval grouping + counts. In Lean: the interval machinery (like LiCircleBound's dyadic/band counts but with the TIGHT C0hat perimeter counts). The negative-γ side: via conjugation (isNontrivialZero_conj exists) or direct: intervals (t−k−1, t−k] for k > t cover negative heights — the count of zeros in (a, a+1] for a < −1... the perimeter ZeroCountProp requires t ≥ 2! Negative-height windows: via conjugation, count in (−b−1, −b] = count in [b, b+1)... (conj maps γ → −γ; the window (−b−1, −b] maps to [b, b+1)). So the negative-side interval counts via the positive-side perimeter + conj. LiCircleBound's shell_crit did exactly this (the conj for γ < 0 — banked). The count for windows starting below 2 (i.e., a ∈ (−2, 2)): EMPTY by (P2) (|γ| ≥ 14 > window range) — the intervals (a, a+1] with a+1 ≤ 14... hmm wait — the LEFT intervals (t−k−1, t−k] for k ≥ 1: when t−k ≥ 14... all fine; when t−k−1 < 2: those intervals cover heights < 3 < 14 ⟹ empty by (P2) IF the interval is within (−14, 14): (t−k−1, t−k] with t−k ≤ 14... for the negative side via conj, heights |γ| ≥ 14 ⟹ intervals (t−k−1, t−k] empty when t−k ≤ 14 AND t−k−1 ≥ −14. For t−k−1 < −14 (deep negative): count via conj = count in (|t−k−1|, |t−k|]-ish positive windows ≥ 14 ✓ perimeter applies. So the empty zone: windows inside (−14, 14): contribution 0 by (P2) ✓.\n   The interval sum: Σ_k count_k·4.5√2/(k(k+1)) ≤ 4.5√2·[C0hat(t+1)·Σ_{1≤k≤t−2} 1/(k(k+1)) + 0 (empty zone) + C0hat(2t+1)·Σ_{...} + tail]. The bookkeeping: left side k=1..⌈t⌉−2-ish ≤ C0hat(t−... hmm the count for window (t−k−1, t−k] is ≤ C0hat(t−k) ≤ C0hat(t) ≤ C0hat(t+1) (mono, needs t−k−1 ≥ 2 ⟺ k ≤ t−3; for k ∈ (t−3, t+13]: empty; for k > t+13: conj side). This is getting intricate but it's the same FLAVOR as LiCircleBound's shell machinery (which took ~2 sessions). Estimate: 1–1.5 sessions for piece 3.\n5. Pieces 4–5: |1/s − 1/s₀| ≤ 4.5/|t|² ≤ 0.045 (|s| ≥ |t|, |s₀| ≥ |t|... ‖1/s − 1/s₀‖ = |s₀−s|/(|s||s₀|) ≤ 4.5/t² ≤ 4.5/100 = 0.045 ✓ elementary).\n6. Piece 6: the ψ-difference — needs the sharp remainder (Lemma D, 1.2/|w|²) OR the series-difference route with budget 0.73. The series route: ψ(w) − ψ(w₀) = Σ_{j≥0} [1/(w₀+j) − 1/(w+j)] hmm — the Weierstrass series: ψ(w) = −γ − 1/w + Σ_{j≥1} (1/j − 1/(w+j))? — difference: ψ(w)−ψ(w₀) = −(1/w − 1/w₀) − Σ_{j≥1}(1/(w+j) − 1/(w₀+j)) = −(1/w−1/w₀) + Σ_{j≥1} (w−w₀)/((w+j)(w₀+j)). With |w−w₀| ≤ 2.25, w₀ = s₀/2 = 1 + it/2, w = s/2: |(w+j)(w₀+j)| ≥ |im|²-ish... Σ_{j≥1} 1/|w+j||w₀+j| ≤ Σ 1/((t/2)² + (j+re w)²)-ish ≤ ∫... ≤ π/(t/2)·... ≈ 2π/t + 4/t². At t=10: 0.628+0.04 = 0.668; ×2.25 = 1.50. Over the 0.73 budget!! Hmm. The sharper per-term: |1/(w+j) − 1/(w₀+j)| = |w₀−w|/|w+j||w₀+j| — summed ≈ 2.25·2π/t = 1.41 at t=10 — the TRUE ψ-difference is ≈ |log(w/w₀)| ≈ 0.45 — the series bound is 3× lossy. Because the log(s/s₀) part IS the answer — the series difference sums to the log. So the sharp route (Lemma D) is needed: ψ(w) = log w − 1/(2w) + r(w), |r| ≤ 1.2/|w|²: ψ-diff = log(w/w₀) − (1/2)(1/w−1/w₀) + r-diff: |log(w/w₀)| ≤ 0.45-ish (arg/modulus split), |(1/2)(1/w−1/w₀)| ≤ (1/2)·2.25/(5·5) = 0.045, |r-diff| ≤ 2.4/25 = 0.096: total ≤ 0.45+0.045+0.096 = 0.591; ½·0.591 = 0.296 ≤ 0.73 ✓. So Lemma D (the 1.2/|w|² remainder) IS needed in Lean. That's D-056 §1's Binet + max-principle argument — a genuine analytic sub-build (~0.5–1 session: Binet's second formula is NOT in Mathlib! Building it: Binet's formula for log Γ or ψ... Mathlib has the Bohr–Mollerup / Weierstrass forms but Binet's integral? Probably not. Hmm — alternatives to Binet: the sharp remainder |ψ(w) − log w + 1/(2w)| ≤ 1.2/|w|² could be proved via the Euler–Maclaurin series with the Bernoulli remainder (the 5d EM machinery!) — EM to order 2 with the explicit remainder: ψ(w) − log w + 1/(2w) = −∫... the B̃₂ integral: from the level-2 EM in LiRBound (b2Fun machinery EXISTS!) — ψ(w) − log w + 1/(2w) = ∫_0^∞ B̃₂(x)/(x+w)³·... hmm — the EM representation for ψ: ψ(w) = log w − 1/(2w) − ∫₀^∞ B̃₂(x)/(x+w)³ dx (roughly) with |B̃₂| ≤ 1/12: |∫| ≤ (1/12)·∫₀^∞ dx/|x+w|³ ≤ (1/12)·(π/(2|im w|²))-ish... at |im| ≥ 5: ≤ (1/12)·(π/50)·... ≈ 0.005 — and the budget for |r| is 1.2/|w|² ≈ 0.048 — fits with 10× room. So Lemma D via the level-2 EM (LiRBound machinery exists!) is feasible: ~0.5 session.)\n\nSo the FULL 5f build is more like 2–3 sessions with two new perimeter declarations. AND the D-056 piece-2 finding + the ii′-6 H(T) finding must be documented (D-069 + journal; the H(T) repair goes on the ii′-6 map).\n\nThis is a significant scope expansion vs the \"~1 session\" estimate in STATE.md. Per the pacing directive and the fork rule, I should present this to the user now: the findings + the perimeter growth + the revised estimate. Let me write it up concisely and ask.\n\nActually — the user already approved the mult-weighted strengthening (P0) with full context. P1 and P2 are the same class of decision. Let me present the complete 5f design with the two new declarations and the two findings, and ask for approval to proceed (or whether they'd prefer a different tradeoff).","signature":""},{"type":"text","text":"The 5f design is complete, and it surfaced two real findings that need your call before I build. Summary:\n\n## Finding 1 (5f-blocking): D-056's piece 2 uses one C₀(t) for a two-unit window\nThe near sum Σ_{|γ−t|<1} m/|s₀−ρ| runs over the **two-unit** window (t−1, t+1), but D-056's assembly charges it one C₀(t) (the one-unit Trudgian form). Honest accounting:\n\n- The per-unit perimeter (even after today's mult-weighted fix) gives 2·Ĉ₀(t+1) — that fold is ≈ 11.2·log + 51, which **breaks the baked 10.5/48** constants.\n- But Trudgian's F/S-pair applied directly to the **two-unit** window N(t+1)−N(t−1) gives 0.6·L + 0.886·LL + 2.7 (the S-pair doesn't double), and then the full honest assembly is **6.111·L + 12.163·LL + 47.6 ≤ 10.5·L + 48 — it FITS**, worst point the loglog pinch at t ≈ 13 (margin ≈ 0.15–0.4, certifiable). So 10.5/48 survives, but only with the two-unit count form, which the current perimeter can't derive (2·Ĉ₀ is too lossy).\n\n## Finding 2 (documented, ii′-6's to face): H(T)'s near sum has the same gap plus a clearance-coverage gap\nD-036's H(T) = 3·Ĉ₀² + … uses (i) the one-unit count for |γ−T|<1 and (ii) L3's clearance, which only covers zeros *inside* the pigeonhole window — out-of-window zeros can sit arbitrarily close to T. Honest version is ≈ 7·Ĉ₀² + O(Ĉ₀). This moves marginTilde's baked 3·Ĉ₀² — the crossover/base points will need re-verification at ii′-6 (margin survives per my estimates, but N₀ may move within the 5×10⁵ cap). Not 5f's scope; documented for the repair map.\n\n## 5f needs two new perimeter declarations (same source classes as existing)\n- **(P1) `ZeroCountProp2`**: mult-weighted two-unit count Σ_{|γ−t|≤1} m ≤ 0.6·log(t+2) + 0.886·loglog(t+2) + 2.7 (t ≥ 10) — Trudgian N(T)/S(T) specialization (honesty-checked: honest form 0.592·L + 0.886·LL + 2.591, so the closed form has 0.008·L + 0.11 slack).\n- **(P2) `NoLowZeros`**: no nontrivial zeros with |im| ≤ 14 — Platt-class zero-location certificate (Turing-method computations locate the zeros; the far sum's low intervals are otherwise only symbolically bounded — Jensen gives ≈ 3.25 vs the true 0, which breaks the constants at the t = 10 pinch).\n\nWith those, 5f = the reference-point difference identity (B cancels, from `partialFraction_identity`) + 6 pieces: dirichletC2 ≤ 0.995 (first terms + integral bound), piece 2 via P1, the far-sum interval machine (biggest piece, ~LiCircleBound-shell flavor), the elementary 4/5, and the sharp digamma remainder via the existing level-2 EM machinery (Lemma D, |r| ≤ 1.2/|w|²). Revised estimate: ~2–3 sessions, not 1."},{"type":"tool_use","id":"AskUserQuestion_120","name":"AskUserQuestion","input":{"questions":[{"question":"5f requires two new perimeter declarations (P1 two-unit Trudgian count, P2 low-zero exclusion) to keep the baked 10.5/48 constants. Proceed?","header":"5f plan","options":[{"label":"Approve P1+P2, build 5f (Recommended)","description":"Declare both (same source classes as the existing perimeter: Trudgian N(T)/S(T) specialization + Platt-class zero-location certificate). Keeps all baked constants and certified artifacts valid; 5f ~2–3 sessions. Findings 1+2 documented (D-069 + repair map)."},{"label":"Only P1, avoid P2","description":"Try to discharge the low-zero intervals via a Jensen numeric count (~3.25 for |γ| ≤ 3, knife-edge vs the 0.44 budget, heavy EM-on-a-circle verification) instead of declaring the zero-location input."},{"label":"Re-fold constants instead","description":"No new declarations; accept honest A ≈ 11.2, B₀ ≈ 51 from the per-unit perimeter and re-bake marginTilde/Hhat + re-verify convexity + re-certify base points (and face the H(T) repair now too, since it also moves constants)."},{"label":"Pause 5f, discuss","description":"Hold here; review the findings (D-056 piece 2, H(T) clearance) before deciding."}]}]},"extra_content":null}]