[{"type":"thinking","thinking":"Errors:\n\n1. Line 3636 (h1s1'): my convoluted sub-add proof left a goal `1 + s - (1 + s) = 1 + s + s - (1 + s)`. The h1s1' proof: (1−s) = 1 → s = 0 — my chain: h5 : 1−s+s = 1+s (rw [h4]... wait h4 : 1−s = 1: h5 : 1 − s + s = 1 + s := by rw [h4] — rw [h4] at the TYPE: goal `1 − s + s = 1 + s`: rw [h4] rewrites `1 − s` to `1` in the goal: `1 + s = 1 + s` — closed by rfl automatically?? then h6 : 1 = 1+s := h5 — but h5 was `by rw [h4]` — if rw closed it, h5 : 1−s+s = 1+s?? no — if the rw closed the goal by rfl, h5 has the type of the ORIGINAL stated equation ✓. Then h7 : 1 − 1 = (1 + s) − 1 := by rw [h6] — goal `1 − 1 = (1+s) − 1`: rw [h6] rewrites occurrences of `1` — rewrites EVERYTHING: `1+s − (1+s) = 1+s+s − (1+s)`?? that's the garbage shown. rw [h6] with h6 : 1 = 1+s rewrites `1` everywhere — including in already-fine spots. Instead: h7 : (1 : ℂ) = 1 + s → s = 0: from h6: s = (1+s) − 1 = 1 − 1 = 0: compute directly:\n```\nhave h7 : s = 0 := by\n  have h8 : s = (1 + s) - 1 := by ring\n  rw [h8, h6]\n  ring  -- (1) - 1 = 0\n```\nHmm: rw [h8, h6]: goal s = 0: rw [h8]: (1+s)−1 = 0: rw [h6] : rewrites `1+s`?? h6 : 1 = 1+s: rw [h6] rewrites 1 → 1+s everywhere: (1+s)−1 has `1` and `1+s`: rewrites the `1`s: (1+s) − (1+s) = 0 ✓ hmm — rw [h6] rewrites occurrences of 1 — both the standalone 1 and the 1 inside (1+s) — chaos again. Better: linear: from h6 : 1 = 1+s: s = 0 by `linear_combination h6`!! — linear_combination handles complex linear ✓: `exact h6 ▸ by ring`?? Simplest: `have h7 : s = 0 := by linear_combination -h6` — linear_combination h6 : goal s = 0: h6 : 1 = 1+s ⟹ s = 0: `linear_combination -h6`?? the goal s = 0: -h6 : −1 = −1−s: ⟹ s−1 = −1−s?? no: linear_combination c·h6 with c : s = 0 from 1 = 1+s: subtract: 1−(1+s) = 0 ⟹ −s = 0: goal s = 0 ⟸ 0 = −s·... linear_combination -h6: LHS: s − (−(1 − (1+s))) hmm — linear_combination works over commutative rings ✓ ℂ ✓. Use `linear_combination -h6`. Or: have h7 : s = (1 : ℂ) - 1 := by linear_combination h6 — no: `s = 0`: linear_combination h6 * (-1)?? just: `have h7 : s = 0 := by linear_combination -h6` hmm — let me verify: linear_combination e proves LHS_goal − RHS_goal = e_assembled: goal s − 0 = −(1 − (1+s)) = −1+1+s = s ✓ so linear_combination -h6 closes ✓.\n\n   Actually — h1s1' is `(1:ℂ) − s ≠ 1` — MUCH simpler direct: (1−s) = 1 ⟺ s = 0 via `sub_eq_iff_eq_add`: 1 − s = 1 ↔ 1 = 1 + s ↔ s = 0:\n   ```\n   have h1s1' : (1 : ℂ) - s ≠ 1 := fun h => by\n     have h2 : s = 0 := by\n       have h3 : 1 - s = 1 := h\n       linear_combination -h3\n     rw [h2, norm_zero] at hs\n     linarith [hs, hR]\n   ```\n   Similarly h1s0' : (1:ℂ) − s ≠ 0: 1 − s = 0 ⟺ s = 1: linear_combination -h:\n   ```\n   have h1s0' : (1 : ℂ) - s ≠ 0 := fun h => by\n     have h2 : s = 1 := by linear_combination -h\n     exact absurd h2 hs1'\n   ```\n   And hζ1's `intro h` branch: (1−s) = 1 → s = 0: same simplification: linear_combination -h.\n\n2. Line 3815 (h5 in Fexpr's h1): `le_trans hĈ2 (mul_le_mul_of_nonneg_left ?m hA0)` — I want `2 ≤ A0count * C0hat (R+2)`: my chain: hĈ2 : 2 ≤ Ĉ₀(4k+9/2), hĈR2 : Ĉ₀(4k+9/2) ≤ Ĉ₀(R+2), and Ĉ₀(R+2) ≤ A₀·Ĉ₀(R+2) (1 ≤ A₀): le_trans hĈ2 (le_trans hĈR2 (the A₀-mult)): the A₀-mult: `mul_le_mul_of_nonneg_left (h : 1 ≤ C0hat (R+2)) hA0 : A₀·1 ≤ A₀·Ĉ₀(R+2)` — hĈpos... let me just write:\n```\nhave h2 : (2 : ℝ) ≤ A0count * C0hat (R + 2) := by\n  have h1 : (1 : ℝ) ≤ C0hat (R + 2) := by linarith [hĈ2, hĈR2]\n  have h2 : C0hat (R + 2) ≤ A0count * C0hat (R + 2) := by\n    have h3 : C0hat (R + 2) = 1 * C0hat (R + 2) := by ring\n    rw [h3]\n    exact mul_le_mul_of_nonneg_right hA0count_ge_one hĈR2pos.le\n  linarith [hĈ2, hĈR2, h2]\n```\n   A0count_ge_one : 1 ≤ A0count ✓ exists. mul_le_mul_of_nonneg_right (h : a ≤ b) (hc : 0 ≤ c) : a*c ≤ b*c: with a := 1, b := A0count: 1·Ĉ ≤ A₀·Ĉ ✓.\n\nLet me fix all three spots (h1s0', h1s1', hζ1's intro branch) + h2.","signature":""},{"type":"text","text":"Fixing h1s0'/h1s1'/hζ1 with `linear_combination` and the h2 chain with `A0count_ge_one`."},{"type":"tool_use","id":"Edit_229","name":"Edit","input":{"replace_all":false,"file_path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiCircleBound.lean","old_string":"      have hζ1 : riemannZeta (1 - s) ≠ 0 := by\n        apply riemannZeta_ne_zero_of_one_le_re\n        · have h1 : (1 - s).re = 1 - s.re := by simp\n          rw [h1]\n          linarith [hreg2]\n        · intro h\n          have h2 : (1 - s) = 1 → s = 0 := by\n            intro h3\n            have h4 : 1 - s = 1 := h3\n            have h5 : 1 - s + s = 1 + s := by rw [h4]\n            rw [sub_add_cancel] at h5\n            have h6 : 1 = 1 + s := h5\n            have h7 : 1 - 1 = (1 + s) - 1 := by rw [h6]\n            rw [sub_self, add_sub_cancel_left] at h7\n            exact h7.symm\n          exact absurd (h2 h) (fun h6 => by rw [h6, norm_zero] at hs; linarith [hs, hR])\n      have hs0' : s ≠ 0 := fun h => by rw [h, norm_zero] at hs; linarith [hs, hR]\n      have hs1' : s ≠ 1 := fun h => by\n        have h1 : s.re = 1 := by rw [h]; simp\n        linarith [h1, hreg2]\n      have h1s0' : (1 : ℂ) - s ≠ 0 := fun h => by\n        have h2 : (1 - s) = 0 → s = 1 := by\n          intro h3\n          have h4 : 1 - s = 0 := h3\n          have h5 : 1 - s + s = 0 + s := by rw [h4]\n          rw [sub_add_cancel, zero_add] at h5\n          exact h5.symm\n        exact absurd (h2 h) hs1'\n      have h1s1' : (1 : ℂ) - s ≠ 1 := fun h => by\n        have h2 : (1 - s) = 1 → s = 0 := by\n          intro h3\n          have h4 : 1 - s = 1 := h3\n          have h5 : 1 - s + s = 1 + s := by rw [h4]\n          rw [sub_add_cancel] at h5\n          have h6 : 1 = 1 + s := h5\n          have h7 : 1 - 1 = (1 + s) - 1 := by rw [h6]\n          rw [sub_self, add_sub_cancel_left] at h7\n          exact h7.symm\n        exact absurd (h2 h) hs0'","new_string":"      have hζ1 : riemannZeta (1 - s) ≠ 0 := by\n        apply riemannZeta_ne_zero_of_one_le_re\n        · have h1 : (1 - s).re = 1 - s.re := by simp\n          rw [h1]\n          linarith [hreg2]\n        · intro h\n          have h2 : s = 0 := by linear_combination -h\n          rw [h2, norm_zero] at hs\n          linarith [hs, hR]\n      have hs0' : s ≠ 0 := fun h => by rw [h, norm_zero] at hs; linarith [hs, hR]\n      have hs1' : s ≠ 1 := fun h => by\n        have h1 : s.re = 1 := by rw [h]; simp\n        linarith [h1, hreg2]\n      have h1s0' : (1 : ℂ) - s ≠ 0 := fun h => by\n        have h2 : s = 1 := by linear_combination -h\n        exact absurd h2 hs1'\n      have h1s1' : (1 : ℂ) - s ≠ 1 := fun h => by\n        have h2 : s = 0 := by linear_combination -h\n        exact absurd h2 hs0'"},"extra_content":null}]