[{"type":"thinking","thinking":"The last error: hp4's `ring_nf` — the goal `22525804431/62500000000 + Real.log(...)·(13432203/1250000) = 360091/1000000 + Real.log(...)·(13432203/1250000)` — ring_nf normalized but the constants differ: my hp4 claim `12.1284 * 0.886 * (LL + 0.03354) = 10.7457624 * LL + 0.360091` — check: 12.1284·0.886 = 10.7457624 ✓ (matches 13432203/1250000 = 10.7457624 ✓). And the constant: 12.1284·0.886·0.03354 = 10.7457624·0.03354 = 0.360392·... vs my 0.360091!! Let me recompute: 10.7457624·0.03354 = 0.360394·... — earlier I computed 12.1284·0.886·0.03351 = 0.360091 and 12.1284·0.886·0.03354 = 0.360394. I wrote 0.360091 in hp4 but with slack 0.03354 — MISMATCH. The ring_nf shows 22525804431/62500000000 = 0.36041·... = 0.3604128·... hmm: 22525804431/62500000000 = 0.36041287. And my claimed 0.360091 ≠ 0.3604129. So hp4's constant should be 0.3604129?? — the exact product 10.7457624·0.03354 = 0.36041287·... let me verify: 10.7457624·0.03354: 10.7457624·0.03 = 0.322372872; ·0.003 = 0.0322372872; ·0.0005 = 0.0053728812; ·0.00004 = 0.000429830496: total 0.322372872+0.0322372872+0.0053728812+0.000429830496 = 0.3604128709. So the constant is 0.3604129 (22525804431/62500000000 = 0.3604128709 ✓). My hp4 said 0.360091 — WRONG (that was the 0.03351 value). Fix hp4's constant to 0.3604129. Then the total: 0.4376331+0.3604129+35.17236+7.2425 = 43.212906 — vs my earlier estimate 43.2125836 (with 0.03351) — slightly higher. Total: 1+2.7+43.212906+0.09+0.6 = 47.602906. Need ⟺ f ≤ 0.397094. Pinch gives ≤ 0.3975 ✓ (margin 1.36 — the pinch bound 0.3975 covers 0.397094 ✓). Wait — the final linarith uses hpinch : f ≤ 0.3975 and needs f ≤ 48−47.602906 = 0.397094 ≤ 0.3975 ✓ — linarith: total ≤ 5.8516L+11.6321LL+47.602906 ≤ 10.5L+48 ⟺ 11.6321LL−4.6484L ≤ 0.397094 — linarith with hpinch (≤ 0.3975): 0.397094 ≤ 0.3975 ✓ — linarith derives it ✓ (a ≤ b ≤ c ⟹ a ≤ c).\n\nHmm wait — but does linarith verify 0.397094 ≤ 0.3975? The pinch bound is 0.3975 and the needed is 0.397094 — linarith with hpinch : f ≤ 0.3975 proves f ≤ 0.3975 ⊇ f ≤ 0.397094?? — NO — f ≤ 0.3975 does NOT imply f ≤ 0.397094 (0.3975 > 0.397094)!! The pinch says f ≤ 0.3975; I need f ≤ 0.397094 — WEAKER hypothesis doesn't give the stronger conclusion!! I need the pinch bound ≤ 0.397094, not 0.3975!!\n\nHmm wait — let me recompute: total ≤ 5.8516L + 11.6321LL + C where C = 1+2.7+43.212906+0.09+0.6 = 47.602906. Need total ≤ 10.5L+48 ⟺ 11.6321LL − 4.6484L ≤ 48−47.602906 = 0.397094. So the pinch must give 11.6321·LL − 4.6484·L ≤ 0.397094. My pinch lemma gives ≤ 0.3975 — INSUFFICIENT (0.3975 > 0.397094). ARGH — off by 4e-4!!\n\nLet me recheck the constant C: piece3's constant part: 12.1284·(0.433/12 + 0.886·0.03354 + 2.9) + 7.2425 = 12.1284·0.0360833·... wait — 12.1284·(0.433·(1/12)) = 12.1284·0.433/12 = 5.2515972/12 = 0.4376331 ✓. 12.1284·(0.886·0.03354) = 0.3604129 ✓. 12.1284·2.9 = 35.17236 ✓. Sum +7.2425: 0.4376331+0.3604129+35.17236+7.2425 = 43.212906 ✓. C = 1+2.7+43.212906+0.045+0.045+0.6 = 47.602906 ✓. Need pinch ≤ 0.397094.\n\nSo the pinch lemma's bound must be 0.397094 or smaller. My pinch proof gave f(L) ≤ −0.96203+0.001 = −0.96103 ≤ 0.3975 — the max of f is −0.967 (true) — I have margin: the certified evaluation gives f(2.5024) ≤ −0.96203 and the correction ≤ 0.001 — so f ≤ −0.96103 ≤ 0.397094 ✓✓ — plenty of margin (1.358). So I just need to state the pinch with bound 0.397094 instead of 0.3975 — the PROOF is identical (the bound is way above the max). Let me change pinch_tight_R's statement to ≤ 0.397094. Actually — to be safe with the final linarith, state ≤ 0.397094... hmm — but actually let me double check the total constant ONE more time to make sure 0.397094 is right:\n\npieces:\n1. dirichletC2 ≤ 1: constant 1.\n2. near: 0.6L+0.886LL+2.7.\n3. far: 12.1284·C0hat(t+1)+7.2425 with C0hat ≤ 0.433(L+1/12)+0.886(LL+0.03354)+2.9:\n   - L-coeff: 12.1284·0.433 = 5.2515972.\n   - LL-coeff: 12.1284·0.886 = 10.7457624.\n   - const: 12.1284·(0.433/12) + 12.1284·(0.886·0.03354) + 12.1284·2.9 + 7.2425\n     = 0.4376331 + 0.3604129 + 35.17236 + 7.2425 = 43.212906.\n4/5. 0.045+0.045 = 0.09.\n6. 6/t ≤ 0.6.\nTotal L-coeff: 0.6+5.2515972 = 5.8515972.\nTotal LL-coeff: 0.886+10.7457624 = 11.6317624.\nTotal const: 1+2.7+43.212906+0.09+0.6 = 47.602906.\nNeed: 5.8515972·L + 11.6317624·LL + 47.602906 ≤ 10.5·L + 48\n⟺ 11.6317624·LL − 4.6484028·L ≤ 0.397094.\nPinch form: my pinch is 11.6321·LL − 4.6484·L ≤ B. The ACTUAL coefficients (5.8515972, 11.6317624) — note: 10.5 − 5.8515972 = 4.6484028 and LL-coeff 11.6317624 — my pinch uses 11.6321/4.6484 — CLOSE but not exact!! 11.6321 vs 11.6317624 and 4.6484 vs 4.6484028 — the pinch's f(L) = 11.6321·LL − 4.6484·L ≥ the actual (11.6317624·LL − 4.6484028·L)?? — compare: f(L) − actual = (11.6321−11.6317624)·LL − (4.6484−4.6484028)·L = 0.0003376·LL + 0.0000028·L — the L-term is POSITIVE (0.0000028·L ≥ 0) — so f ≥ actual ⟹ pinch on f does NOT bound actual!!! The coefficients must match EXACTLY. Let me recompute the exact totals as rationals:\n\nL-coeff: 0.6 + 12.1284·0.433 = 0.6 + 5.2515972 = 5.8515972. 10.5 − 5.8515972 = 4.6484028.\nLL-coeff: 0.886 + 12.1284·0.886 = 0.886·(1+12.1284) = 0.886·13.1284 = 11.6317624·... 0.886·13.1284: 0.886·13 = 11.518; 0.886·0.1284 = 0.1137624: total 11.6317624 ✓.\n\nSo the pinch should be: 11.6317624·LL − 4.6484028·L ≤ 0.397094·... — the max of THAT f: f'(L) = 11.6317624/L − 4.6484028 = 0 at L* = 11.6317624/4.6484028 = 2.5023392·... — slightly different L* — the max ≈ −0.967-ish as before. To keep the SAME pinch machinery, let me round the coefficients: use LL-coeff 11.6318 and L-coeff 5.8516:\n- 11.6317624 ≤ 11.6318 (round up ✓).\n- 5.8515972 ≤ 5.8516 (round up ✓).\nThen total ≤ 5.8516·L + 11.6318·LL + 47.602906 ⟺ 11.6318·LL − (10.5−5.8516)·L = 11.6318·LL − 4.6484·L ≤ 0.397094.\nAnd the pinch: 11.6318·LL − 4.6484·L ≤ 0.397094 — with L* = 11.6318/4.6484 = 2.5023124 — and the max ≈ −0.967 ✓ margin 1.36.\n\nHmm — but the pinch's tangent point 2.5024 has f′(2.5024) = 11.6318/2.5024 − 4.6484 ≤ 0 ⟺ 11.6318 ≤ 4.6484·2.5024 = 11.63215616 ✓ (margin 3.6e-4 — better than before). And f(2.5024) = 11.6318·log(2.5024) − 4.6484·2.5024 ≤ 11.6318·0.9173 − 11.63215616 = 10.669848 − 11.632156 = −0.962308·... = −0.962308 ≤ −0.96203 ✓ (better). So the SAME pinch proof works with coefficients (11.6318, 4.6484) — and the tangent point 2.5024 with f′ ≤ 0 certified ✓.\n\nActually — even cleaner: keep the pinch as-is with (11.6321, 4.6484) but note 11.6317624 ≤ 11.6321 and 4.6484028 ≥ 4.6484 — the actual f_actual(L) = 11.6317624·LL − 4.6484028·L vs f(L) = 11.6321·LL − 4.6484·L: f_actual = f − 0.0003376·LL − 0.0000028·L — sign of the difference is MIXED (LL ≥ 0, L ≥ 0: f_actual = f − (positive + positive) ⟹ f_actual ≤ f ✓!!). WAIT: f − f_actual = (11.6321−11.6317624)·LL − (4.6484028−4.6484)·L = 0.0003376·LL − 0.0000028·L — sign?? 0.0003376·LL ≥ 0 and −0.0000028·L ≤ 0: f − f_actual could be positive or negative depending on L!! For L large, the −0.0000028·L dominates ⟹ f < f_actual ⟹ pinch on f does NOT bound f_actual for large L!!! ARGH. For example L = 100: 0.0003376·LL − 0.0000028·L = 0.0003376·4.6 − 0.00028 = 0.00155−0.00028 = 0.00127 > 0: f > f_actual ✓ good. L = 10000: 0.0003376·9.2 − 0.028 = 0.0031−0.028 < 0: f < f_actual ✗ bad. So for large L the actual is LARGER than my pinch function — the pinch on (11.6321, 4.6484) does NOT bound the actual for large L!!\n\nSo I MUST use the exact coefficients (11.6317624, 4.6484028) or a dominating form. Let me round UP both to be safe? Rounding LL-coeff UP helps the LL side; the L-side: 10.5 − (L-coeff): rounding the total's L-coeff UP (5.8515972 → 5.8516) makes 10.5 − that SMALLER (4.6484 — but the actual total's L-coefficient is 5.8515972 ≤ 5.8516 so the total ≤ 5.8516·L + ... ✓ safe). And LL-coeff: 11.6317624 ≤ 11.6318 (round up ✓ safe). Then the needed pinch: 11.6318·LL − 4.6484·L ≤ 48 − 47.602906 = 0.397094. With f(L) = 11.6318·LL − 4.6484·L: EXACTLY the (11.6318, 4.6484) form ✓ and its max ≈ −0.962 ≤ 0.397094 ✓ margin 1.36.\n\nHmm — but is the LL/L rounding really safe? The total is ≤ (rounded-coefficient combination): 5.8515972 ≤ 5.8516 ⟹ 5.8515972·L ≤ 5.8516·L (L ≥ 0 ✓); 11.6317624 ≤ 11.6318 ⟹ 11.6317624·LL ≤ 11.6318·LL (LL ≥ 0 ✓ — LL = log log(t+2) ≥ log log 12 ≥ 0 ✓ since log 12 ≥ 2.4848 ≥ 1 ✓). So total ≤ 5.8516L + 11.6318LL + 47.602906 ✓ safe. And the pinch with (11.6318, 4.6484): f_actual-total ≤ that form ⟺ 11.6318·LL − 4.6484·L ≤ 0.397094 — and its max at L* = 11.6318/4.6484 = 2.5023124: f(L*) = 11.6318·(log L* − 1) ≤ 11.6318·(0.9173−1)·... hmm — the tangent point for the (11.6318, 4.6484) pinch: f′(2.5024) = 11.6318/2.5024 − 4.6484 ≤ 0 ⟺ 11.6318 ≤ 4.6484·2.5024 = 11.63215616 ✓ (margin 3.6e-4 ✓). f(2.5024) = 11.6318·log 2.5024 − 4.6484·2.5024 ≤ 11.6318·0.9173 − 11.63215616 = 10.6698477 − 11.63215616 = −0.9623085 ≤ −0.96203 ✓ (better than −0.96203 even).\n\nSo the pinch: `pinch_tight_R {L} (hL : log 12 ≤ L) : 11.6318 * Real.log L - 4.6484 * L ≤ 0.397094` — with pinchF/pinchF' updated to 11.6318. And the tangent evaluation: f(2.5024) ≤ −0.96203: 11.6318·0.9173 = 10.66984774·... compute: 11.6318·0.9173 = 10.6698477·... 11.6318·0.9173: 11.6318·0.9 = 10.46862; ·0.0173 = 0.20123: total 10.66985. And 4.6484·2.5024 = 11.63215616: f = 10.66985−11.63215616 = −0.96231 ≤ −0.96203 ✓.\n\nLet me now update pinchF/pinchF' coefficients (11.6321 → 11.6318), pinchF_at_critical's norm_num (11.6318·0.9173 = 10.66984774 — norm_num will compute), pinchF'_at_critical (11.6318 ≤ 4.6484·2.5024 = 11.63215616 ✓ norm_num), the h3's 4.6474·2.5024 ≤ 11.6318·... hmm — hcorr's h6 : 4.6474 ≤ 11.6318/2.5024 ⟺ 4.6474·2.5024 ≤ 11.6318: 4.6474·2.5024 = 11.62965376 ≤ 11.6318 ✓ (margin 2.1e-3 ✓). And the final assembly's coefficient forms: hp3 : 12.1284·0.433·(L+1/12) = 5.2515972·L + 0.4376331 ✓ (keep); hp4 : 12.1284·0.886·(LL+0.03354) = 10.7457624·LL + 0.3604129 ✓; and the final fold needs the coefficients rounded: total ≤ 5.8516·L + 11.6318·LL + 47.602906 ≤ 10.5L+48 ⟺ 11.6318LL − 4.6484L ≤ 0.397094 — with 5.8515972 ≤ 5.8516 and 11.6317624 ≤ 11.6318 — the linarith needs those rounding facts as hypotheses:\n```\nhave hr1 : (5.8515972 : ℝ) ≤ 5.8516 := by norm_num\nhave hr2 : (10.7457624 : ℝ) ≤ 11.6318 := by norm_num\nhave hr3 : (11.6317624 : ℝ) ≤ 11.6318 := by norm_num\n```\n— linarith with the coefficient comparisons (linear in atoms L, LL ✓).\n\nHmm wait — actually let me reconsider the LL coefficient: the total LL-coefficient from the pieces: piece2's 0.886 + piece3's 12.1284·0.886·... — piece3's LL-coefficient is 12.1284·0.886 = 10.7457624 (from hp4). So total LL-coefficient = 0.886 + 10.7457624 = 11.6317624 ✓. And in hp4, the LL-coefficient displayed: 12.1284·0.886·(LL+0.03354) = 10.7457624·LL + 0.3604129 ✓ (hp4's claim with the constant 0.3604129 — needs fixing from my wrong 0.360091).\n\nAnd the total's L-coefficient: piece2's 0.6 + piece3's 12.1284·0.433 = 0.6+5.2515972 = 5.8515972 ✓ (hp3's claim).\n\nSo the final linarith in the capstone:\n```\n          have hr1 : (5.8515972 : ℝ) ≤ 5.8516 := by norm_num\n          have hr2 : (10.7457624 : ℝ) ≤ 11.6318 := by norm_num\n          linarith [hp, hp2, hp3, hp4, hp5, h6t, hpinch, hr1, hr2]\n```\nwith hpinch : 11.6318·LL − 4.6484·L ≤ 0.397094 — and the goal: 1 + (0.6L+0.886LL+2.7) + (12.1284·C0hat+7.2425) + 0.045+0.045+6/t ≤ 10.5·|t|+... — wait — the calc's second step has the pieces already assembled: `1 + (0.6·log(t+2)+0.886·loglog(t+2)+2.7) + (12.1284·C0hat(t+1)+7.2425) + 0.045+0.045+6/t` — and hp rewrites 12.1284·C0hat(t+1) ≤ 12.1284·(conversion) — hp2/3/4 expand — the total after substitution: 1 + (0.6L+0.886LL+2.7) + (5.2515972·L+0.4376331 + 10.7457624·LL+0.3604129 + 35.17236 + 7.2425) + 0.09+0.6 — linarith with hpinch (11.6318LL − 4.6484L ≤ 0.397094) + hr1/hr2 (the roundings) ✓ should close: the LHS ≤ 5.8515972·L + 11.6317624·LL + 47.602906 and linarith needs: that ≤ 10.5L+48 ⟺ 11.6317624LL − 4.6484028L ≤ 0.397094: from hpinch (11.6318LL − 4.6484L ≤ 0.397094): 11.6317624LL − 4.6484028L = (11.6318LL − 4.6484L) − 0.0000376·LL − 0.0000028·L ≤ 0.397094 − 0.0000376·LL − 0.0000028·L ≤ 0.397094 ⟸ −0.0000376·LL − 0.0000028·L ≤ 0 ⟺ 0.0000376·LL + 0.0000028·L ≥ 0 ✓ (LL, L ≥ 0 ✓). linarith with LL ≥ 0 (hL : log 12 ≤ L ⟹ L ≥ 2.4848 ≥ 0 and LL ≥ log 2.4848 ≥ 0·... hmm — LL = log log(t+2) ≥ log(2.4848) ≥ 0 ⟸ 2.4848 ≥ 1 ✓ — linarith needs LL ≥ 0 as a fact: Real.log_nonneg (log(t+2) ≥ 1 ⟸ log(t+2) ≥ 2.4848 ≥ 1): provide hLLnn : 0 ≤ Real.log (Real.log (t + 2)) := Real.log_nonneg (by linarith [hL] : (1:ℝ) ≤ Real.log (t+2)). And L ≥ 0 similarly.\n\nOK let me now write the fixes:\n1. pinchF/pinchF' coefficients → 11.6318.\n2. pinchF_at_critical's norm_num target: 11.6318·0.9173 = 10.66984774 — the h4 : (11.6321:ℝ)·0.9173 = 10.67012533 → change to (11.6318)·0.9173 = 10.66984774 — verify: 11.6318·0.9173 = 10.6698477·... compute: 11.6318×0.9173 = 11.6318×0.9 = 10.46862; ×0.017 = 0.1977406; ×0.0003 = 0.00348954: total 10.46862+0.1977406+0.00348954 = 10.66985014?? — let me recompute: 11.6318×0.9173 = 11.6318×0.9 = 10.46862; 11.6318×0.0173 = 0.20123014; total 10.66985014. Hmm — two different computations (0.6698477 vs 0.6698501) — let me be precise: 11.6318 × 0.9173: 11.6318 × 9173 / 10000 = (11.6318 × 9173)/10000. 11.6318×9000 = 104686.2; ×173 = 2012.30·... 11.6318×100 = 1163.18; ×73 = 849.12: ×173 = 1163.18+849.12 = 2012.30. Total 104686.2+2012.30 = 106698.5 → /10000 = 10.66985. So 11.6318·0.9173 = 10.66985. f(2.5024) = 10.66985 − 11.63215616 = −0.96230616 ≤ −0.96203 ✓.\n   In pinchF_at_critical: h4 : (11.6318 : ℝ) * 0.9173 = 10.66985 := by norm_num — hmm — norm_num verifies 11.6318·0.9173 = 10.66985 EXACTLY?? 11.6318·0.9173 = 10.66985014·... (not exactly 10.66985!). norm_num would compute the exact product 10.66985014 — my stated 10.66985 is not the exact product (10.66985014 ≠ 10.66985)!! — state h4 as (11.6318)·0.9173 ≤ 10.6699?? — 10.66985014 ≤ 10.6699 ✓ or keep the exact 10.66985014: f ≤ 10.66985014 − 11.63215616 = −0.96230602 ≤ −0.96203 ✓. State h4 : (11.6318:ℝ)·0.9173 = 10.66985014 := by norm_num (norm_num verifies the exact product ✓ — might be slow on big decimals but fine). Actually simpler: h3 : 11.6318·log 2.5024 ≤ 11.6318·0.9173 (mul_le_mul_of_nonneg_left ✓ as before) then linarith [h3, h2] with h2 : 4.6484·2.5024 ≥ 11.63215616 — hmm — f = 11.6318·log − 4.6484·2.5024 ≤ 11.6318·0.9173 − 11.63215616·... — and 11.6318·0.9173 = 10.66985014 (norm_num-computed inside linarith? — linarith doesn't compute products of literals... it does norm_num-style evaluation of literal products? linarith handles literal arithmetic ✓ — linarith [h3, h2] with h3 : 11.6318·log 2.5024 ≤ 11.6318·0.9173 and h2 : 4.6484·2.5024 ≥ 11.63215616: goal f ≤ −0.96203: f = 11.6318·log − 4.6484·2.5024 ≤ 11.6318·0.9173 − 11.63215616 = −0.96230602 ≤ −0.96203: linarith must evaluate 11.6318·0.9173 and 4.6484·2.5024 numerically — linarith's preprocessing does compute literal products ✓ (norm_num-level). Give h4 : (11.6318 : ℝ) * 0.9173 = 10.66985014 := by norm_num and h2 : (4.6484 : ℝ) * 2.5024 ≥ 11.63215616 := by norm_num — wait 4.6484·2.5024 = 11.63215616 EXACTLY (computed earlier: 4.6484×2.5024 = 11.63215616·... let me verify: 4.6484×2.5024 = 4.6484×2.5 = 11.621; 4.6484×0.0024 = 0.01115616; total 11.63215616 ✓ exact ✓).\n   So: f(2.5024) = 11.6318·log(2.5024) − 4.6484·2.5024 ≤ 10.66985014 − 11.63215616 = −0.96230602 ≤ −0.96203 ✓ linarith [h3, h2, h4] where h3 : 11.6318·log 2.5024 ≤ 11.6318·0.9173, h4 : 11.6318·0.9173 = 10.66985014, h2 : 4.6484·2.5024 = 11.63215616 (norm_num computes ✓).\n3. pinchF'_at_critical: 11.6318 ≤ 4.6484·2.5024 = 11.63215616 ✓ (margin 3.6e-4 ✓).\n4. hcorr's h6 : 4.6474 ≤ 11.6318/2.5024: 4.6474·2.5024 = 11.62965376 ≤ 11.6318 ✓.\n5. The final fold with the rounding haves.\n\nLet me also double-check the hcorr's 0.001 budget: correction ≤ (−f′(2.5024))·(2.5024−log 12) ≤ 0.0000254·0.0175 = 4.4e-7 ≤ 0.001 ✓ trivially.\n\nAnd the pinch's final bound: state pinch_tight_R with ≤ 0.397094 — the proof: branch L ≥ 2.5024: f ≤ f(2.5024) ≤ −0.96231 ≤ 0.397094 ✓; branch L < 2.5024: f ≤ −0.96231+0.001 = −0.96131 ≤ 0.397094 ✓. Hmm wait — pinchF_at_critical gives f(2.5024) ≤ −0.96203 (the stated bound) — with the new coefficients it's −0.962306 — state pinchF_at_critical : pinchF 2.5024 ≤ -0.96203 ✓ (−0.962306 ≤ −0.96203 ✓ fine).\n\nLet me apply all the coefficient updates.","signature":""},{"type":"text","text":"The final fold's coefficients need exact matching with the pinch form (mixed-sign rounding issue). Updating the pinch to the exact rounded coefficients (11.6318, 4.6484)."},{"type":"tool_use","id":"Bash_316","name":"Bash","input":{"command":"cd D:/Python/Quantum/research/lean/NguyenRH1 && python - <<'EOF'\np = 'NguyenRH1/LiTightR.lean'\nsrc = open(p, encoding='utf-8').read()\n\n# pinchF/pinchF' coefficients: 11.6321 -> 11.6318\nsrc = src.replace(\"\"\"def pinchF (L : ℝ) : ℝ := 11.6321 * Real.log L - 4.6484 * L\"\"\",\n                  \"\"\"def pinchF (L : ℝ) : ℝ := 11.6318 * Real.log L - 4.6484 * L\"\"\")\nsrc = src.replace(\"\"\"def pinchF' (L : ℝ) : ℝ := 11.6321 / L - 4.6484\"\"\",\n                  \"\"\"def pinchF' (L : ℝ) : ℝ := 11.6318 / L - 4.6484\"\"\")\n\n# hasDerivAt_pinchF: coefficients\nsrc = src.replace(\"\"\"  have h1 := (Real.hasDerivAt_log hx0).const_mul 11.6321\n  have h2 := (hasDerivAt_id' x).const_mul 4.6484\n  have h4 := h1.sub h2\n  unfold pinchF pinchF'\n  rw [show (11.6321 : ℝ) / x - 4.6484 = 11.6321 * x⁻¹ - 4.6484 * 1 from by\n    rw [div_eq_mul_inv, mul_one]]\n  rw [show (fun L : ℝ => 11.6321 * Real.log L - 4.6484 * L)\n      = (fun L : ℝ => 11.6321 * Real.log L) - (fun L : ℝ => 4.6484 * L) from by\n    ext y; simp only [Pi.sub_apply]]\n  simpa only [Pi.sub_apply] using h4\"\"\", \"\"\"  have h1 := (Real.hasDerivAt_log hx0).const_mul 11.6318\n  have h2 := (hasDerivAt_id' x).const_mul 4.6484\n  have h4 := h1.sub h2\n  unfold pinchF pinchF'\n  rw [show (11.6318 : ℝ) / x - 4.6484 = 11.6318 * x⁻¹ - 4.6484 * 1 from by\n    rw [div_eq_mul_inv, mul_one]]\n  rw [show (fun L : ℝ => 11.6318 * Real.log L - 4.6484 * L)\n      = (fun L : ℝ => 11.6318 * Real.log L) - (fun L : ℝ => 4.6484 * L) from by\n    ext y; simp only [Pi.sub_apply]]\n  simpa only [Pi.sub_apply] using h4\"\"\")\n\n# antitoneOn_pinchF': coefficient\nsrc = src.replace(\"\"\"  have h1 : 11.6321 / b ≤ 11.6321 / a := by\n    apply div_le_div_of_nonneg_left (by norm_num : (0 : ℝ) ≤ 11.6321) ha0 hab\"\"\", \"\"\"  have h1 : 11.6318 / b ≤ 11.6318 / a := by\n    apply div_le_div_of_nonneg_left (by norm_num : (0 : ℝ) ≤ 11.6318) ha0 hab\"\"\")\n\n# pinchF_at_critical: coefficients + exact constants\nsrc = src.replace(\"\"\"lemma pinchF_at_critical : pinchF 2.5024 ≤ -0.96203 := by\n  rw [show pinchF 2.5024 = 11.6321 * Real.log 2.5024 - 4.6484 * 2.5024 from rfl]\n  have h1 := log_two_point_five_zero_two_four_le\n  have h2 : (4.6484 : ℝ) * 2.5024 ≥ 11.6321 := by norm_num\n  have h3 : 11.6321 * Real.log 2.5024 ≤ 11.6321 * 0.9173 :=\n    mul_le_mul_of_nonneg_left h1 (by norm_num)\n  have h4 : (11.6321 : ℝ) * 0.9173 = 10.67012533 := by norm_num\n  linarith [h3, h2, h4]\"\"\", \"\"\"lemma pinchF_at_critical : pinchF 2.5024 ≤ -0.96203 := by\n  rw [show pinchF 2.5024 = 11.6318 * Real.log 2.5024 - 4.6484 * 2.5024 from rfl]\n  have h1 := log_two_point_five_zero_two_four_le\n  have h2 : (4.6484 : ℝ) * 2.5024 = 11.63215616 := by norm_num\n  have h3 : 11.6318 * Real.log 2.5024 ≤ 11.6318 * 0.9173 :=\n    mul_le_mul_of_nonneg_left h1 (by norm_num)\n  have h4 : (11.6318 : ℝ) * 0.9173 = 10.66985014 := by norm_num\n  linarith [h3, h2, h4]\"\"\")\n\n# pinchF'_at_critical: coefficients\nsrc = src.replace(\"\"\"lemma pinchF'_at_critical : pinchF' 2.5024 ≤ 0 := by\n  rw [show pinchF' 2.5024 = 11.6321 / 2.5024 - 4.6484 from rfl]\n  have h1 : (11.6321 : ℝ) ≤ 4.6484 * 2.5024 := by norm_num\n  have h2 : (11.6321 : ℝ) / 2.5024 ≤ 4.6484 :=\n    (div_le_iff₀ (show (0 : ℝ) < 2.5024 by norm_num)).mpr h1\n  linarith [h2]\"\"\", \"\"\"lemma pinchF'_at_critical : pinchF' 2.5024 ≤ 0 := by\n  rw [show pinchF' 2.5024 = 11.6318 / 2.5024 - 4.6484 from rfl]\n  have h1 : (11.6318 : ℝ) ≤ 4.6484 * 2.5024 := by norm_num\n  have h2 : (11.6318 : ℝ) / 2.5024 ≤ 4.6484 :=\n    (div_le_iff₀ (show (0 : ℝ) < 2.5024 by norm_num)).mpr h1\n  linarith [h2]\"\"\")\n\n# pinch_tight_R statement: coefficients + final bound\nsrc = src.replace(\"\"\"lemma pinch_tight_R {L : ℝ} (hL : Real.log 12 ≤ L) :\n    11.6321 * Real.log L - 4.6484 * L ≤ 0.3975 := by\"\"\", \"\"\"lemma pinch_tight_R {L : ℝ} (hL : Real.log 12 ≤ L) :\n    11.6318 * Real.log L - 4.6484 * L ≤ 0.397094 := by\"\"\")\n\n# the h6 : pinchF L = ... rfl occurrences\nsrc = src.replace(\"\"\"    have h6 : pinchF L = 11.6321 * Real.log L - 4.6484 * L := rfl\n    linarith [hval', h6]\"\"\", \"\"\"    have h6 : pinchF L = 11.6318 * Real.log L - 4.6484 * L := rfl\n    linarith [hval', h6]\"\"\")\n\n# hcorr's h2/h6 coefficients\nsrc = src.replace(\"\"\"      have h2 : -pinchF' 2.5024 = (4.6484 : ℝ) - (11.6321 : ℝ) / 2.5024 := by\n        unfold pinchF'\n        ring\"\"\", \"\"\"      have h2 : -pinchF' 2.5024 = (4.6484 : ℝ) - (11.6318 : ℝ) / 2.5024 := by\n        unfold pinchF'\n        ring\"\"\")\nsrc = src.replace(\"\"\"        have h6 : (4.6474 : ℝ) ≤ (11.6321 : ℝ) / 2.5024 := by\n          rw [le_div_iff₀ (show (0 : ℝ) < 2.5024 by norm_num)]\n          have h7 : (4.6474 : ℝ) * 2.5024 ≤ 11.6321 := by norm_num\n          exact h7\n        linarith [h6]\"\"\", \"\"\"        have h6 : (4.6474 : ℝ) ≤ (11.6318 : ℝ) / 2.5024 := by\n          rw [le_div_iff₀ (show (0 : ℝ) < 2.5024 by norm_num)]\n          have h7 : (4.6474 : ℝ) * 2.5024 ≤ 11.6318 := by norm_num\n          exact h7\n        linarith [h6]\"\"\")\n\nopen(p, 'w', encoding='utf-8').write(src)\nprint(\"done\")\nEOF\nexport PATH=\"$HOME/.elan/bin:$PATH\" && lake env lean NguyenRH1/LiTightR.lean 2>&1 | head -25","description":"Update pinch coefficients, recompile","timeout":420000},"extra_content":null}]