[{"type":"thinking","thinking":"The final linarith (hbig's `linarith [hser2, hld2, hψ2, h1s, h1s1, hlogpi, h1, lowZeroConst_nonneg, dirichletC2_nonneg]` at ~3921) failed. The issue: hbig's statement — its GOAL is `‖logDeriv ζ s‖ + ‖seriesFun s‖ + ... ≤ 2·lowZeroConst + dirichletC2 + 400000·...` — and I have hser2 (‖seriesFun‖ ≤ ...), hld2 (‖logDeriv‖ ≤ ...), hψ2 (‖ψ/2‖ ≤ 31+12L), h1s, h1s1, hlogpi, h1 (the absorption 2900X + (200000L² + 300X) + ((31+12L)+2) ≤ 400000X). The linarith needs to combine: LHS = sum of the 6 norms ≤ [hser2's RHS] + [hld2's RHS] + 1 + 1 + 1 + [hψ2's RHS] = 2·lowZeroConst + 2900X + dirichletC2 + 200000L² + 300X + 3 + 31+12L = 2·lowZeroConst + dirichletC2 + (2900X + 200000L² + 300X + 34 + 12L). And h1 says 2900X + (200000L² + 300X) + ((31+12L)+2) ≤ 400000X: i.e. 2900X+200000L²+300X+33+12L ≤ 400000X. My sum has 34+12L (extra 1 from... let me recount: h1s : ‖1/s‖ ≤ 1; h1s1 : ‖1/(s−1)‖ ≤ 1; hlogpi : ‖logπ/2‖ ≤ 1: THREE ones: total +3. hψ2 : 31+12L. So non-X part: 31+12L+3 = 34+12L. h1's form: (31+12·L)+2 = 33+12L. MISMATCH by 1!! h1 as stated: `((31 + 12 * Real.log R) + 2)` — but the actual sum needs +3 (three ones). Fix h1's +2 → +3:\n`2900X + (200000L² + 300X) + ((31+12L) + 3) ≤ 400000X` — and the nlinarith for h1 must absorb 34+12L: 400000X − 3200X = 396800X ≥ 200000L² + 34 + 12L: X = A₀Ĉ₀L² ≥ 162: 200000L² ≤ 100000·2·... — the h5 inside h1 handles 200000·L² ≤ 100000·X (X ≥ 2L² ✓): remaining: 396800X − 100000X = 296800X ≥ 34+12L ⟸ 34+12L ≤ 34·X/162·... 34+12L ≤ 46·L·... ≤ 46·X·... X ≥ 2·81 = 162 ≥ 46 hmm: 34+12L ≤ 34+12·(L²/81)·162... simpler: 34 ≤ 34·X/162·... nlinarith with h4 : 1 ≤ X and hlogRnn: 34+12L ≤ 34·X + 12L·X·... hmm: 34 ≤ 34·X (X ≥ 1); 12L ≤ 12·X·L·... X·L ≥ L ✓ X ≥ 1: 12L ≤ 12·X·L — hmm X·L vs L: X·L ≥ L ⟺ X ≥ 1 ✓. So 34+12L ≤ 34X+12·X·L ≤ 46·X·L·... 34X + 12XL = X(34+12L) — circular. Just: 34+12L ≤ 46·X·L·... OK the nlinarith inside h1 already proved the +2 version (33+12L); the +3 version (34+12L) — the slack is enormous (h4 : 1 ≤ X and h5's 100000X headroom and h2 : 2 ≤ A₀Ĉ₀(R+2) and h3 : 81 ≤ L²): 34+12L vs the budget 400000X−3200X−100000X = 296800X ≥ 296800 ≥ 34+12L ⟸ 12L ≤ 296766 and L ≤ ... we don't have a numeric cap on L! L = log R — unbounded! Need 12L ≤ 296800·X·... : 12L ≤ 296800·A₀Ĉ₀L² ⟸ A₀Ĉ₀L ≥ 12/296800 ⟸ A₀Ĉ₀L ≥ 2·9 = 18 ✓: 12L ≤ (296800·18)·... hmm: 296800·A₀Ĉ₀L² ≥ 296800·18·L ≥ 12L ✓. nlinarith [h5, h4, h2, h3, hlogRnn, hA0, hĈR2pos] — h5 : 200000L² ≤ 100000X; h4 : 1 ≤ X; h2 : 2 ≤ A₀Ĉ₀(R+2); h3 : 81 ≤ L²: products nlinarith can find: X·L ≥ L·(2·81·... ) hmm — X ≥ 2·L²·... X = A₀Ĉ₀(R+2)·L² ≥ 2·L²: and X·L ≥ 2·L³·... 12L ≤ 296800X ⟸ 12 ≤ 296800·A₀Ĉ₀L ⟸ A₀Ĉ₀L ≥ 12/296800 ⟸ A₀Ĉ₀L ≥ 18 (from h2·L ≥ 2·9): so 296800·A₀Ĉ₀L·L ≥ 296800·18·L·... nlinarith multiplies pairs: h2 × (L ≥ 9): A₀Ĉ₀·L ≥ 18·... then × h5-form... let me just strengthen the hints: give h2L : (18 : ℝ) ≤ A0count * C0hat (R + 2) * Real.log R := by nlinarith [h2, hlogR, hA0, hĈR2pos] — h2L·(X-scale): 12L ≤ 296800·X·... X·18·... hmm: 296800·X·... OK let me compute the needed final directly:\ngoal: 2900X + (200000L² + 300X) + (34 + 12L) ≤ 400000X.\n⟺ 200000L² + 34 + 12L ≤ 396800X.\n200000L² ≤ 100000X (h5). Remaining: 34 + 12L ≤ 296800X.\n12L ≤ 12·X·L·... : L ≤ X·L·... hmm: L ≤ L² (hL2-like: L ≥ 9 ⟹ L ≤ L²·... give hLL : Real.log R ≤ (Real.log R)^2 := by nlinarith [hlogR]); then 12L ≤ 12L² = (12/2)·2L² ≤ 6·A₀Ĉ₀L² = 6X·... 2L² ≤ X (from h5's h6 : 2 ≤ A₀Ĉ₀(R+2): X = A₀Ĉ₀(R+2)·L² ≥ 2L² ✓): 12L ≤ 12L² ≤ 6·2L² ≤ 6X ✓. And 34 ≤ 34·X/162·... 34 ≤ X ⟸ X ≥ 2·81 = 162 ≥ 34 ✓: 34 ≤ 34/162·X ≤ X. So 34+12L ≤ X + 6X = 7X ≤ 296800X ✓✓.\nSo the nlinarith needs: h5 (200000L² ≤ 100000X), hLL (L ≤ L²), h3 (81 ≤ L²), h2 (2 ≤ A₀Ĉ₀(R+2)), and nonnegs. Give those:\n```\nhave hLL : Real.log R ≤ (Real.log R) ^ 2 := by nlinarith [hlogR]\nhave h1 : 2900 * A0count * C0hat (R + 2) * (Real.log R) ^ 2 +\n    (200000 * (Real.log R) ^ 2 + 300 * A0count * C0hat (R + 2) * (Real.log R) ^ 2) +\n    ((31 + 12 * Real.log R) + 3)\n    ≤ 400000 * A0count * C0hat (R + 2) * (Real.log R) ^ 2 := by\n  have h2 : (2 : ℝ) ≤ A0count * C0hat (R + 2) := (as fixed)\n  have h3 : (81 : ℝ) ≤ (Real.log R) ^ 2 := by nlinarith [hlogR, hlogRnn]\n  have h4 : (1 : ℝ) ≤ A0count * C0hat (R + 2) * (Real.log R) ^ 2 := by\n    nlinarith [h2, h3, hA0, hĈR2pos]\n  have h5 : 200000 * (Real.log R) ^ 2\n      ≤ 100000 * A0count * C0hat (R + 2) * (Real.log R) ^ 2 := by\n    nlinarith [h2, pow_two_nonneg (Real.log R)]\n  nlinarith [h5, h4, h2, h3, hLL, hlogRnn, hA0, hĈR2pos]\n```\nCheck nlinarith can finish: goal: 3200X + 200000L² + 34+12L ≤ 400000X ⟺ 200000L²+34+12L ≤ 396800X. Facts: h5 : 200000L² ≤ 100000X; hLL : L ≤ L²; h2 : 2 ≤ A₀Ĉ₀; h3 : 81 ≤ L²; h4 : 1 ≤ X; nonnegs. Products: (A₀Ĉ₀−2)·(L²) ≥ 0 ⟹ X ≥ 2L² ✓; (L²−L)·(...) : 12L ≤ 12L²; 12L² ≤ 6·2L² ≤ 6X ✓; 34 ≤ 34·X hmm: X ≥ 1 (h4): 34·1 ≤ 34·X·... 34 ≤ 34X ✓ (34·(X−1) ≥ 0 ✓ from h4). Sum: 200000L² + 34 + 12L ≤ 100000X + 34X + 6X + ... hmm: 12L ≤ 12L² ≤ 6·(2L²) ≤ 6X: and 34 ≤ 34X: total ≤ 100000X + 6X + 34X = 100040X ≤ 396800X ✓. nlinarith should find with the hints. Actually the X ≥ 2L² fact: from h2 (2 ≤ A₀Ĉ₀) and L² ≥ 0: (A₀Ĉ₀−2)·L² ≥ 0 ⟹ A₀Ĉ₀L² ≥ 2L² ✓ — nlinarith product of (h2) with (L² ≥ 0) ✓.\n\nLet me apply: fix h1's +2 → +3 with the hLL addition.","signature":""},{"type":"text","text":"The final absorption is off by one (three `1`s, not two). Fixing h1 with the L ≤ L² hint."},{"type":"tool_use","id":"Bash_240","name":"Bash","input":{"command":"cd \"D:/Python/Quantum/research/lean/NguyenRH1\" && python - << 'PYEOF'\np = 'NguyenRH1/LiCircleBound.lean'\nsrc = open(p, encoding='utf-8').read()\nold = '''    have h1 : 2900 * A0count * C0hat (R + 2) * (Real.log R) ^ 2 +\n        (200000 * (Real.log R) ^ 2 + 300 * A0count * C0hat (R + 2) * (Real.log R) ^ 2) +\n        ((31 + 12 * Real.log R) + 2)\n        ≤ 400000 * A0count * C0hat (R + 2) * (Real.log R) ^ 2 := by\n      have h2 : (2 : ℝ) ≤ A0count * C0hat (R + 2) := by\n        have h1c : (1 : ℝ) ≤ C0hat (R + 2) := by linarith [hĈ2, hĈR2]\n        have h2c : C0hat (R + 2) ≤ A0count * C0hat (R + 2) :=\n          calc C0hat (R + 2) = 1 * C0hat (R + 2) := by ring\n            _ ≤ A0count * C0hat (R + 2) :=\n                mul_le_mul_of_nonneg_right A0count_ge_one hĈR2pos.le\n        linarith [hĈ2, hĈR2, h2c]\n      have h3 : (81 : ℝ) ≤ (Real.log R) ^ 2 := by nlinarith [hlogR, hlogRnn]\n      have h4 : (1 : ℝ) ≤ A0count * C0hat (R + 2) * (Real.log R) ^ 2 := by\n        nlinarith [h2, h3, hA0, hĈR2pos]\n      have h5 : 200000 * (Real.log R) ^ 2\n          ≤ 100000 * A0count * C0hat (R + 2) * (Real.log R) ^ 2 := by\n        have h6 : (2 : ℝ) ≤ A0count * C0hat (R + 2) := h2\n        nlinarith [h6, pow_two_nonneg (Real.log R)]\n      nlinarith [h5, h4, h2, h3, hlogRnn, hA0, hĈR2pos]'''\nnew = '''    have h1 : 2900 * A0count * C0hat (R + 2) * (Real.log R) ^ 2 +\n        (200000 * (Real.log R) ^ 2 + 300 * A0count * C0hat (R + 2) * (Real.log R) ^ 2) +\n        ((31 + 12 * Real.log R) + 3)\n        ≤ 400000 * A0count * C0hat (R + 2) * (Real.log R) ^ 2 := by\n      have h2 : (2 : ℝ) ≤ A0count * C0hat (R + 2) := by\n        have h1c : (1 : ℝ) ≤ C0hat (R + 2) := by linarith [hĈ2, hĈR2]\n        have h2c : C0hat (R + 2) ≤ A0count * C0hat (R + 2) :=\n          calc C0hat (R + 2) = 1 * C0hat (R + 2) := by ring\n            _ ≤ A0count * C0hat (R + 2) :=\n                mul_le_mul_of_nonneg_right A0count_ge_one hĈR2pos.le\n        linarith [hĈ2, hĈR2, h2c]\n      have h3 : (81 : ℝ) ≤ (Real.log R) ^ 2 := by nlinarith [hlogR, hlogRnn]\n      have h4 : (1 : ℝ) ≤ A0count * C0hat (R + 2) * (Real.log R) ^ 2 := by\n        nlinarith [h2, h3, hA0, hĈR2pos]\n      have h5 : 200000 * (Real.log R) ^ 2\n          ≤ 100000 * A0count * C0hat (R + 2) * (Real.log R) ^ 2 := by\n        have h6 : (2 : ℝ) ≤ A0count * C0hat (R + 2) := h2\n        nlinarith [h6, pow_two_nonneg (Real.log R)]\n      have hLL : Real.log R ≤ (Real.log R) ^ 2 := by nlinarith [hlogR]\n      nlinarith [h5, h4, h2, h3, hLL, hlogRnn, hA0, hĈR2pos]'''\nassert old in src, \"h1 not found\"\nsrc = src.replace(old, new)\nopen(p, 'w', encoding='utf-8').write(src)\nprint(\"done\")\nPYEOF\nexport PATH=\"$HOME/.elan/bin:$PATH\" && lake env lean NguyenRH1/LiCircleBound.lean 2>&1 | head -20","description":"Fix the h1 absorption (+3, hLL hint); recompile","timeout":300000},"extra_content":null}]