[{"type":"thinking","thinking":"M₀ has NO Lean Prop yet — it's only a constant inside Bn's definition. Per the Phase 5B header, M₀ is in the declared perimeter (\"..., M₀\"), but there's no formal Prop. The ii′-6 design needs to add it: `MZeroProp : ∀ t : ℝ, |t| ≤ 10 → ‖logDeriv riemannZeta (-5/2 + t·I)‖ ≤ 1.3352` (the interval-arithmetic certificate class, same as Platt–Trudgian; E-023b acb optional upgrade). This is a SECOND perimeter addition needing user approval.\n\nSo TWO perimeter additions for ii′-6:\n1. `ZeroCountGlobalProp` (global Trudgian N(T) ≤ FAlt form) — for the bridge.\n2. `MZeroProp` (M₀ = 1.3352 on |t| ≤ 10, σ = −5/2) — for the I_III short piece. (Already in the declared perimeter list per the Phase 5B header — \"M₀\" — but no formal Prop; adding the formal Prop is a formality of the same approved decision.)\n\nBoth are in the already-approved trust classes (D-062 §4 declared: \"Platt–Trudgian (computational), Misguich–Voros/E-011/E-015 (computational), Trudgian 2014 (published explicit bounds), M₀ (interval-arithmetic certificate, after E-023b)\"). So these aren't NEW trust decisions — they're formalizations of the already-approved perimeter. Session 035's P1/P2 additions were user-approved as a formality; I'll present these the same way at the design review.\n\nNow — the numerical validation. Let me write a quick Python check of the honest-vs-baked fold at n = 10⁵ and the growth direction. The honest total:\nH(n) = 4·FAlt T + 4T·[(2/3)·Z2(T) + 10.5·log(T+2) + 48] + 40·M₀ + 2(e⁴+1)·[NI(T) + 5.5·(10.5·log(T+2)+48) + 3.3] + IIMaj n + 1\n\nwith:\n- T = √n + 1\n- FAlt T = (1/2π)·T·log T + 0.137·log T + 0.443·log log T + (7/8+1.588)\n- Z2(T) = 0.6·log(T+2) + 0.886·log log(T+2) + 2.7 (two-unit count bound, t ≥ 10 ✓)\n- NI(T) = Z2(T) · 2·log(12/c(T)), c(T) = 1/(7·(C0hat(a)+C0hat(a+1)+C0hat(a+2))), a = √n... \n\nhmm — wait, the extended pigeonhole: T ∈ [a, a+1] avoiding zeros in (a−1, a+2) with clearance ≥ 1/(7·C3) where C3 = 3-unit count. The per-zero integral bound needs |T−γ| ≥ c for ALL zeros with |γ−T| ≤ 1 (which are ⊆ (a−1, a+2) since T ∈ [a, a+1]: γ ∈ [T−1, T+1] ⊆ (a−1, a+2) ✓). So c = 1/(7·C3) with C3 ≤ C0hat(a)+C0hat(a+1)+C0hat(a+2) ≤ 3·C0hat(a+2).\n\nHmm — why 7? The pigeonhole: k zeros in (a−1, a+2) (3-unit window, count ≤ C3); the candidate T-points: midpoints of a partition of [a, a+1] into k+1 pieces... the exists_avoiding_midpoint machinery: with k+1 midpoints in [a, a+1], each zero (k of them) kills at most one midpoint if the midpoint spacing > 2·clearance: spacing = 1/(k+1) ≥ 1/(C3+1): clearance = half-spacing/(k+1)... from exists_zero_avoiding_T's form: clearance 1/(3·C0hat(a+1)) for the unit window [a, a+1] with count C0hat(a+1). Cloning for the 3-unit window: the SAME midpoint argument on [a, a+1] but against zeros in (a−1, a+2) (count C3): clearance 1/(3·C3)? The \"3\" in 3·C0hat came from... the pigeonhole structure (midpoint + half-gap). Let me just use c = 1/(3·C3) with C3 = 3·C0hat(a+2) → c = 1/(9·C0hat(a+2)). Then 12/c = 108·C0hat(a+2).\n\nNI(T) = Z2(T)·2·log(108·C0hat(a+2)) — wait, the per-zero integral bound 2·log(12/c): let me recompute: ∫_{-5/2}^{3} dx/√((x−β)² + δ²) = asinh((3−β)/δ) + asinh((β+5/2)/δ) ≤ 2·asinh(5.5/δ) (since 3−β ≤ 3 < 5.5 and β+5/2 < 3.5 < 5.5). asinh(x) = log(x + √(x²+1)) ≤ log(2x+1) ≤ log(3x) for x ≥ 1. 2·asinh(5.5/δ) ≤ 2·log(16.5/δ) (5.5/δ ≥ 5.5·9·C0hat ≥ 1 ✓). With δ ≥ c = 1/(9·C0hat(a+2)): ≤ 2·log(16.5·9·C0hat(a+2)) = 2·log(148.5·C0hat(a+2)).\n\nNI(T) = Z2(T)·2·log(148.5·C0hat(√n+3)).\n\nWait — but hold on: the zeros with |γ−T| ≤ 1 have δ = |T−γ| ≥ c ✓ (extended pigeonhole). Their count ≤ Z2(T) (ZeroCountProp2 — |γ−T| ≤ 1 ✓ matches). Each contributes ≤ 2·log(148.5·C0hat). ✓.\n\nAlso the horizontal R-part: ∫_{-5/2}^{3} (10.5·log(T+2) + 48) dx = 5.5·(10.5·log(T+2)+48) ✓ (tight R-bound at height T ≥ 316 ≥ 10 ✓, s on the horizontal: im = T, re ∈ [−5/2, 3] ✓ within the 5f capstone's σ-range −5/2 ≤ re ≤ 3 ✓ — wait the 5f capstone: (hσ : -5/2 ≤ s.re ∧ s.re ≤ 3) (him : s.im = t) (hζ : riemannZeta s ≠ 0) (ht : 10 ≤ t) ✓ all satisfied on the horizontal except... hζ: ζ(w) ≠ 0 on the horizontal — T zero-avoiding: zeros with im ∈ [a, a+1] are ≥ clearance away ✓; zeros with im outside: |T−γ| ≥ ... the horizontal is at im = T; a zero with im = γ ≠ T could still be ON the horizontal if γ = T — excluded by the pigeonhole (zeros in [a,a+1] ⊇ {T}) ✓. What about zeros with re > 3 on the horizontal extension to 2√n? The horizontal spans [−5/2, 2√n]: for re ∈ [3, 2√n]: ζ ≠ 0 (re > 1 ✓). For re ∈ [−5/2, 3]: zeros are nontrivial (re ∈ (0,1)) with im = γ = T excluded ✓, or trivial (negative even): on the horizontal im = T > 0 — trivial zeros have im = 0 ✓ not on it. So ζ ≠ 0 on both horizontals ✓ (also on the bottom, by conj).\n\nFor re ∈ [3, 2√n] on the horizontal: the stepping (6.12): |ζ′/ζ(σ+iT)| ≤ 2^{−(σ−2)}·|ζ′/ζ(2+iT)|?? hmm — the Dirichlet series: |ζ′/ζ(σ+iT)| ≤ Σ Λ(n)/n^σ ≤ 2^{−(σ−2)}·Σ Λ(n)/n² = 2^{−(σ−2)}·dirichletC2·... hmm dirichletC2 = Σ Λ(n)/n² ≈ 0.5699·... wait D-037 says |ζ′/ζ(2)| = 0.5699 — hmm but dirichletC2 ≤ 0.995 per D-069 piece 1 (the level-2 integral machine gave dirichletC2 ≤ 1 then refined?). The horizontal right part [3, 2√n]: ∫₃^{2√n} |ζ′/ζ| dx ≤ ∫₃^∞ 2^{−(σ−2)}·dirichletC2 dσ = 2·dirichletC2/... ∫₃^∞ 2^{-(σ-2)} dσ = 2^{-1}/log 2 ≈ 0.72: ≤ 0.72·dirichletC2 ≈ 0.72·0.57 = 0.41?? D-037's \"3.3\" for the right tail: e·|ζ′/ζ(2)|·2 ≈ 3.3 — the stepping (6.12) gives a geometric sum with ratio involving the kernel too (|k_n(s+1)+1|/|k_n(s)+1| ≤ e). Our L1 kernel bound kFun_shift_norm_le_exp_four_add_one covers re ≤ 3; for re ∈ [3, 2√n] the kernel keeps growing (e^{√n/2}+1 at σ₁)... D-036's stepping: each unit step right gains factor 2 on |ζ′/ζ| while kernel-ratio ≤ e ⟹ the (6.12) geometric sum ≤ e·|ζ′/ζ(3+iT)|·(e⁴+1)... hmm — the \"3.3\" is |I-right-tail| ≤ e·|ζ′/ζ(2)|·2 ≈ 3.3 — the (e⁴+1) factor is outside. Let me not re-derive; the design will mirror D-036 §3's II/IV analysis with the kernel bound kFun_shift_norm_le_exp_four_add_one on [−3, 3] and the stepping sum. The right-tail charge \"3.3\" = e·dirichletC2·2·... e·0.57·2 = 3.1 ≈ 3.3 ✓.\n\nOK — the numerics: let me now write the Python validation computing H(n) vs Bn at n = 10⁵, 1.5×10⁵, ..., and check the ratio direction. If H(10⁵)/Bn(10⁵) < 1 with slack and decreasing after, the fold design is validated.","signature":""},{"type":"text","text":"RectangleIntegral′ is 2π-normalized (paper over-counts sides — consistent). M₀ has no formal Prop yet (second perimeter formality). Now the numerical design validation before any Lean work."},{"type":"tool_use","id":"Bash_174","name":"Bash","input":{"command":"cd D:/Python/Quantum && ./.venv/Scripts/python.exe - << 'EOF'\nimport mpmath as mp\nmp.mp.dps = 40\npi, e = mp.pi, mp.e\n\ndef C0hat(t):  # unit-window count bound (ZeroCountProp)\n    return mp.mpf('0.433')*mp.log(t+2) + mp.mpf('0.886')*mp.log(mp.log(t+2)) + mp.mpf('2.9')\ndef Z2(t):     # two-unit count bound (ZeroCountProp2)\n    return mp.mpf('0.6')*mp.log(t+2) + mp.mpf('0.886')*mp.log(mp.log(t+2)) + mp.mpf('2.7')\ndef FAlt(T):\n    return (1/(2*pi))*(T*mp.log(T)) + mp.mpf('0.137')*mp.log(T) + mp.mpf('0.443')*mp.log(mp.log(T)) + (mp.mpf(7)/8 + mp.mpf('1.588'))\ndef Hhat(T):\n    return 3*C0hat(T)**2 + mp.mpf('10.5')*mp.log(T+2) + 48\ndef IIMaj(n):\n    return 2*(mp.sqrt(n)+1)*(mp.e**(mp.sqrt(n)/2)+1)*mp.mpf('0.57')*2**(-(2*mp.sqrt(n)-2))\ndef Bn(n):\n    T = mp.sqrt(n)+1\n    return (2*FAlt(T) + 4*T*((mp.mpf(2)/3)*C0hat(T) + mp.mpf('10.5')*mp.log(T+2) + 48)\n            + 40*mp.mpf('1.3352') + 2*(e**4+1)*(7*Hhat(T) + mp.mpf('3.3')) + IIMaj(n) + 1)\n\ndef honest(n):\n    T = mp.sqrt(n)+1\n    a = mp.sqrt(n)\n    C3 = 3*C0hat(a+2)          # 3-unit window count bound\n    c  = 1/(9*C0hat(a+2))      # extended-pigeonhole clearance (3*C3 form)\n    NI = Z2(T)*2*mp.log(mp.mpf('16.5')/c)   # near-sum integral on the horizontal\n    bridge = 4*FAlt(T)                        # BOTH-signs bridge (the factor-2 fix)\n    left   = 4*T*((mp.mpf(2)/3)*Z2(T) + mp.mpf('10.5')*mp.log(T+2) + 48)   # I_III long (two-unit near)\n    horiz  = 2*(e**4+1)*(NI + mp.mpf('5.5')*(mp.mpf('10.5')*mp.log(T+2)+48) + mp.mpf('3.3'))\n    return bridge + left + 40*mp.mpf('1.3352') + horiz + IIMaj(n) + 1, NI\n\nprint(f\"{'n':>12} {'Bn':>14} {'honest':>14} {'slack':>12} {'ratio':>8} {'NI vs 7Hhat+3.3':>16}\")\nfor n in [10**5, 1.2*10**5, 1.5*10**5, 2*10**5, 5*10**5, 10**6, 10**8, 10**12, mp.mpf('9e24')]:\n    n = mp.mpf(n)\n    h, NI = honest(n)\n    T = mp.sqrt(n)+1\n    hbaked = 7*Hhat(T)+mp.mpf('3.3')\n    hinside = NI + mp.mpf('5.5')*(mp.mpf('10.5')*mp.log(T+2)+48) + mp.mpf('3.3')\n    print(f\"{float(n):>12.4g} {float(Bn(n)):>14.4g} {float(h):>14.4g} {float(Bn(n)-h):>12.4g} {float(h/Bn(n)):>8.3f} {float(hinside/hbaked):>16.3f}\")\nEOF","description":"Validate the honest-vs-baked fold numerically"},"extra_content":null}]