[{"type":"thinking","thinking":"One failure: e^5.65 ≤ 285 fails with the crude exp_hi (my upper bound of e^5.65 is 288.18 > 285). Use log 285 ≥ 5.6 instead: e^5.6 = e^5·e^0.6: e^0.6 upper: exp_hi(0.6, 16) = (1/(1−0.0375))^16 = (1/0.9625)^16: ln = 16·0.038216 = 0.61146, e^0.61146 = 1.843. (1/0.9625)^16 ≈ 1.8431. e^5.6 ≤ 148.414·1.8431 = 273.56 ≤ 285 ✓ 4% margin. So use log 285 ≥ 5.6 (for the hbase2 lower bound on Ĉ₀/Ĥ at n=80000). Adjust: Ĉ₀ ≥ 0.433·5.6 + 2.9 = 5.3248 (log log ≥ 0). 4T[...] ≥ 4·283·[(2/3)(5.3248) + 10.5·5.6 + 48] = 1132·[3.5499 + 58.8 + 48] = 1132·110.3499 = 124916. Ĥ ≥ 3·5.3248² + 58.8 + 48 = 85.07 + 106.8 = 191.87; 7Ĥ+3.3 ≥ 1346.4; 2(e⁴+1) ≥ 2·55.5978 = 111.1956; product ≥ 149714. Bn ≥ 124916 + 149714 + 53 = 274683. margin(80000) ≤ 338670 − 274683 + 1 = 63988 < 86210 ✓ still fine.\n\nLet me also double check e^11.5 ≤ 100000 with margin 0.49% — fine (exact rational).\n\ne^11.263 ≤ 80000: 2.43% ✓. e^11.3 ≥ 80000: 1.03% ✓. All good except 5.65 → replaced by 5.6 (4%+).\n\nAlso record: e^4 ∈ [54.5978, 54.6015] ✓ (margins tiny but exact-rational — the certified e1lo/e1hi are themselves exact; the pow computation is exact rational arithmetic; norm_num will handle).\n\nWait — one concern about e^4 ≥ 54.5978: margin 0.00064% — exact rational comparison 2.7182818283^4 = 54.59815003... ≥ 54.5978 ✓ TRUE with exact arithmetic. Fine.\n\nAlso need for hbase2's margin(100000) lower: log 100000 ≥ 11.5 ✓ (0.49%), slower(100000) ≥ 50000·11.5 − 1.41894·100000 − 4 where (1+log2+logπ)/2 ≤ (1 + 0.6931472 + 1.1448)/2 = 1.4189476 — hmm: upper on the coefficient: log 2 ≤ 0.6931472, log π ≤ 1.1448: (1+0.6931472+1.1448)/2 = 1.41897360. slower(100000) ≥ 50000·11.5 − 1.41897360·100000 − 4 = 575000 − 141897.36 − 4 = 433098.64. Bn(100000) ≤ 346893 (computed). margin ≥ 433098.64 − 346893 + 1 = 86206.6 ✓. And margin(80000) ≤ 63988 < 86206 ✓ (slack 22000).\n\nFor hbase (margin(80000) > 0): slower(80000) ≥ 40000·11.263 − 1.4189736·80000 − 4 = 450520 − 113517.9 − 4 = 336998.1; Bn ≤ 327037; margin ≥ 9962 ✓.\n\nAll consistent. Now — the remaining question for the hbase lower bound on slower(80000): coefficient upper 1.4189736 uses log π ≤ 1.1448 (margin 0.007% — the tightest link). Fine as computed.\n\nLet me now also verify my Bn(80000) upper bound components numerically (the 327037 figure) with exact certified constants in Python — simulate the certified chain:\n- T ≤ 284 (via √80000 ≤ 283; 283² = 80089 ≥ 80000 ✓)\n- log T ≤ 5.7 (T ≤ 284 ≤ e^5.7: e^5.7 ≥ 298.87 ✓ 4.5%)\n- log log T ≤ 1.75 (log T ≤ 5.7 ≤ e^1.75 = 5.7546 ✓)\n- log(T+2) ≤ 5.7 (T+2 ≤ 286 ≤ 298.87 ✓)\n- 1/(2π) ≤ 1/(2·3.1415926535) = 0.15915500...\n- FAlt(T) ≤ 0.1591551·284·5.7 + 0.137·5.7 + 0.443·1.75 + 2.463 \n  = 0.1591551·1618.8 = 257.639... + 0.7809 + 0.77525 + 2.463 = 261.658; 2F ≤ 523.32\n- Ĉ₀ ≤ 0.433·5.7 + 0.886·1.75 + 2.9 = 2.4681+1.5505+2.9 = 6.9186\n- 4T·[(2/3)·6.9186 + 10.5·5.7 + 48] = 1136·[4.6124+59.85+48] = 1136·112.4624 = 127757.3\n- Ĥ ≤ 3·6.9186² + 10.5·5.7 + 48 = 143.601+59.85+48 = 251.451\n- 2(e⁴+1)(7Ĥ+3.3) ≤ 2·55.6015·(7·251.451+3.3) = 111.203·1763.457 = 196103.6\n- IIMaj ≤ 4.56·285·2 = 2599.2 (s+1 ≤ 284+1 = 285; e terms ≤ 1)\n- Bn ≤ 523.32+127757.3+53.408+196103.6+2599.2+1 = 327037.8\nmargin(80000) ≥ 336998.1 − 327037.8 + 1 = 9961.3 > 0 ✓\n\nBn(100000): T ≤ 318 (317² = 100489 ≥ 100000 ✓), log T ≤ 5.8, log log ≤ 1.77:\n- FAlt ≤ 0.1591551·318·5.8 + 0.137·5.8 + 0.443·1.77 + 2.463 = 0.1591551·1844.4 = 293.51 + 0.7946+0.7841+2.463 = 297.55; 2F ≤ 595.1\n- Ĉ₀ ≤ 0.433·5.8+0.886·1.77+2.9 = 2.5114+1.56822+2.9 = 6.97962\n- 4T·[...] = 1272·[4.65308+60.9+48] = 1272·113.55308 = 144439.5\n- Ĥ ≤ 3·48.7151+60.9+48 = 146.145+108.9 = 255.045\n- 2(e⁴+1)(7Ĥ+3.3) ≤ 111.203·1788.615 = 198879.3\n- IIMaj ≤ 4.56·319·2 = 2909.28\n- Bn ≤ 595.1+144439.5+53.408+198879.3+2909.28+1 = 346877.6\nmargin(100000) ≥ 433098.64 − 346877.6 + 1 = 86222 ✓\n\nAnd margin(80000) upper for hbase2: slower(80000) ≤ 40000·11.3 − 1.4165736·80000 − 4 = 452000 − 113325.89 − 4 = 338670.1; Bn(80000) ≥ 274683 (recomputed with log ≥ 5.6); margin ≤ 338670.1 − 274683 + 1 = 63988.1 < 86222 ✓ slack 22000.\n\nGreat — the plan is fully consistent. Record all of this in D-062 §5 as the ready-made discharge plan.\n\nOne more check on the tighter points: e^11.5 ≤ 100000 uses exp_hi(0.5,16) = (1/(1−0.03125))^16 = (32/31)^16. In Lean: (32/31 : ℝ)^16 — norm_num can compute and compare to rationals. The chain: Real.exp (0.5) = (Real.exp (0.5/16))^16 ≤ (1/(1−0.03125))^16 needs: exp(y) ≤ 1/(1−y) for 0 ≤ y < 1: from Real.add_one_le_exp (−y): 1 − y ≤ exp(−y) = (exp y)⁻¹ ⟹ exp y ≤ 1/(1−y) (needs 1−y > 0; one_div_le_one_div_of_le inverted: (exp y)⁻¹ ≥ 1−y > 0 ⟹ exp y ≤ (1−y)⁻¹ by le_inv... `le_inv₀`? inv_le_inv... there are lemmas: le_of_one_div_le_one_div... I'll find at write time.) And (exp(0.5/16))^16 = exp(16·(0.5/16)) = exp(0.5) via Real.exp_nat_mul? Real.exp_nat_mul (x : ℝ) (n : ℕ) : Real.exp (n * x) = (Real.exp x) ^ n — I believe this exists (Real.exp_nat_mul). Then rewrite n * x = 0.5 by norm_num. ✓\n\nAlso e^11: exp(11) = exp(11·1) = (exp 1)^11 via Real.exp_nat_mul with 11·1. Then (exp 1)^11 ≤ 2.7182818286^11 via pow_le_pow_left (Real.exp_one_lt_d9) and norm_num computes the rational pow ✓.\n\nOK. Now write D-062. Also update D-061 (Lean tracking file) — let me check its current structure quickly. And check draft.tex formal verification appendix. Let me read both (D-061 partially, draft.tex appendix).","signature":""},{"type":"text","text":"One failure (`e^5.65 ≤ 285`) — replaced by `log 285 ≥ 5.6` (4% margin); recomputed chain stays consistent (margin(80 000) ≤ 63 988 < 86 210 ≤ margin(10⁵)). The full discharge plan is validated. Checking D-061 and the paper's verification appendix before writing D-062."},{"type":"tool_use","id":"Bash_93","name":"Bash","input":{"command":"grep -n \"Formal verification\\|section\" D:/Python/Quantum/research/paper/draft.tex | head -20; echo ---; grep -n \"^## \\|^# \" D:/Python/Quantum/research/derivations/D-061*.md | head","description":"Locate paper verification appendix and D-061 structure"},"extra_content":null}]