[{"type":"thinking","thinking":"The linarith failure at 2392 — marginTilde_convex's final `linarith [hIIM, hnn1..hnn9]`. Let me look at the goal + hIIM. The hIIM : IIMaj″ < 1/(3n) and the goal 0 ≤ 1/(2n) + [9 nonneg terms] + (−1)·IIMaj″: 1/(2n) − IIMaj″ > 1/(2n) − 1/(3n) = 1/(6n) > 0 ✓ should close... — linarith's issue: the `1/(2*n)` and `1/(3*n)` reciprocal atoms (the banked gotcha: \"linarith can't relate distinct reciprocal atoms\"!). The old proof had hIIM : IIMaj″ < 1/(2n) with the goal's 1/(2n) — SAME atom, linarith worked. Now with 1/(3n) — DIFFERENT reciprocal atoms — linarith can't relate 1/(2n) and 1/(3n)!! (The banked gotcha from session-032.) Fix: bridge with the explicit equation 1.5·(1/(3n)) = 1/(2n) — wait, the consumer is the OLD marginTilde_convex with mIIM = (−1)·IIMaj: the goal is 0 ≤ 1/(2n) + nonnegs + (−1)·IIMaj″: with hIIM : IIMaj″ < 1/(3n): (−1)·IIMaj″ > −1/(3n): 1/(2n) − 1/(3n) > 0 ⟺ need linarith to see 1/(3n) < 1/(2n)?? — linarith can't (reciprocal atoms). Bridge: `have hbridge : (1 : ℝ) / (3 * n) < 1 / (2 * n) := by ...` — prove via div_lt_div_of_pos_right? — 1/(3n) < 1/(2n) ⟺ 1/(3n) < 1/(2n) — hmm — one_div_lt_one_div_of_le? — for 0 < 2n < 3n: 1/(3n) < 1/(2n) ✓ via `one_div_lt_one_div_of_le (by positivity : 0 < 2n) (by linarith : 2n ≤ 3n)`?? — hmm — one_div_lt_one_div_of_le (hb : 0 < b) (h : b ≤ a) : 1/a < 1/b?? no that's ≤. For strict: `one_div_lt_one_div_of_lt (hb : 0 < b) (h : b < a) : 1/a < 1/b`?? hmm — let me use: `have hbridge : (1 : ℝ) / (3 * n) < 1 / (2 * n) := by rw [show (2 : ℝ) * n = 3 * n - n by ring]; ...` — overcomplicating. Simplest: `have hbridge : (1 : ℝ) / (3 * n) < 1 / (2 * n) := by apply one_div_lt_one_div_of_lt (by positivity : (0:ℝ) < 2 * n) (by linarith [hn0])` — one_div_lt_one_div_of_lt (hb : 0 < b) (hab : b < a) : 1/a < 1/b — with a := 3n, b := 2n: needs 0 < 2n ✓ positivity (n > 0) and 2n < 3n ✓ linarith [hn0] ✓ gives 1/(3n) < 1/(2n) ✓. Hmm — does one_div_lt_one_div_of_lt exist? — there should be (parallel to one_div_le_one_div_of_le). Then `linarith [hIIM, hbridge, hnn1..hnn9]`.\n\nWait — but actually, hold on: the old consumer had hIIM := IIMaj_deriv2_lt (giving < 1/(2n)) and the SAME goal — it worked because the atoms matched. With lt_three (1/(3n)) the atoms differ. But actually — I should think about what the a2 re-proof needs ANYWAY: the dog-leg's marginTilde_convex will have mIIM = (−1.5)·IIMaj and the goal 0 ≤ 1/(2n) + nonnegs + (−1.5)·IIMaj″: with lt_three: 1.5·IIMaj″ < 1.5/(3n) = 1/(2n) — the reciprocal-atom bridge `1.5·(1/(3n)) = 1/(2n)` — via field_simp! — 1.5/(3n) = 1/(2n) ⟺ 1.5·2n = 3n ⟺ 3n = 3n ✓ field_simp [hn0.ne'] closes ✓ (the same bridge I planned for the c = 1.5 work). So for the CURRENT consumer: add the bridge `have hbridge : (1.5 : ℝ) * (1 / (3 * n)) = 1 / (2 * n) := by field_simp [hn0.ne']` — then linarith [hIIM, hbridge, hnn*] : 1.5·IIMaj″ < 1.5·(1/(3n)) = 1/(2n) ✓ — hmm — but the current mIIM is (−1)·IIMaj (coefficient 1, not 1.5) — the goal is 0 ≤ 1/(2n) + nonnegs + (−1)·IIMaj″: 1/(2n) − IIMaj″ > 1/(2n) − 1/(3n) — needs 1/(3n) < 1/(2n) — hbridge as one_div_lt: 1/(3n) < 1/(2n) ✓ linarith handles the atoms via the equation/inequality ✓.\n\nHmm — actually simpler for the CURRENT consumer: since the mIIM is still (−1)·IIMaj (the a2 redefinition changes it to 1.5 LATER), the current proof needs: (−1)·IIMaj″ > −1/(3n) ⟹ 1/(2n) + (−1)·IIMaj″ > 1/(2n) − 1/(3n) > 0: bridge `1/(3n) < 1/(2n)` — via one_div_lt_one_div_of_lt or by field_simp/norm: `have hbridge : (1 : ℝ) / (3 * n) < 1 / (2 * n) := by field_simp [hn0.ne']; linarith`?? — 1/(3n) < 1/(2n) ⟺ 2n < 3n ⟺ 0 < n ✓ field_simp [hn0.ne'] turns it into 2n < 3n ✓ linarith [hn0] closes ✓. Let me use:\n```\n    have hbridge : (1 : ℝ) / (3 * n) < 1 / (2 * n) := by\n      field_simp [hn0.ne']\n      linarith [hn0]\n```\nThen `linarith [hIIM, hbridge, hnn1..hnn9]`.","signature":""},{"type":"text","text":"The reciprocal-atom gotcha — adding the `1/(3n) < 1/(2n)` bridge for linarith."},{"type":"tool_use","id":"Read_554","name":"Read","input":{"file_path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiSkeleton.lean","offset":2388,"limit":10},"extra_content":null}]