[{"type":"thinking","thinking":"Six issues:\n\n1. **1338**: `rw [div_le_iff₀ (Real.exp_pos 1), one_mul]` — goal `1 / Real.exp 1 ≤ 0.36789` — div_le_iff₀ : a/b ≤ c ↔ a ≤ c·b — with a := 1, b := e, c := 0.36789: goal becomes `1 ≤ 0.36789·e` — then `one_mul` — pattern `1 * ?a` — NOT found (the goal is `1 ≤ 0.36789·e` — the LHS `1` is not `1 * something`). My mistake — div_le_iff₀ gives `a ≤ c * b` = `1 ≤ 0.36789 * e` — no `1 *` anywhere. Remove `one_mul`: `rw [div_le_iff₀ (Real.exp_pos 1)]` then `nlinarith [Real.exp_one_gt_d9]`. The goal: `1 ≤ 0.36789·e` — from e > 2.7182818283: 0.36789·e > 0.36789·2.7182818283 = 1.000030... ≥ 1 ✓ nlinarith (linear: 0.36789·e atom ✓ norm_num computes 0.36789·2.7182818283 ≈ 1.00003 ✓).\n\nWait — nlinarith [Real.exp_one_gt_d9]: Real.exp_one_gt_d9 : 2.7182818283 < exp 1. Goal: 1 ≤ 0.36789 * exp 1. From 0.36789·2.7182818283 ≥ 1: 0.36789 × 2.7182818283 — compute precisely: 0.36789 × 2.7182818283 = 0.36789×2 = 0.73578; ×0.7 = 0.257523; ×0.0182818283 = 0.0067271... — total 0.73578+0.257523+0.006727 = 1.00003 ✓ — nlinarith verifies 0.36789·2.7182818283 ≥ 1 as a decimal identity ✓ should close.\n\n2. **1363 warning**: pinch_horizontal's hTn1 unused?? — wait — hTn1 IS used in (a) (h4 : log(√n+2) ≤ log(T+2)... no wait — (a) uses `by linarith [hTn]` for √n+2 ≤ T+2 — hmm — and hTn1 : T ≤ √n+1 — do I use it in pinch_horizontal? — the pinch is entirely about T (Z2hat T, C0hat T, Hhat T) and the log-arg bound uses √n+2 ≤ T+2 (hTn, not hTn1!). So hTn1 unused in pinch_horizontal ✓ — rename to _hTn1? — but the interface (6-d calls pinch_horizontal hn hTn hTn1 positionally) — renaming to _hTn1 keeps positional compat ✓ do it.\n\n3. **1552**: `Real.exp_one_lt_d9.le : exp 1 ≤ 2.7182818286` vs expected `exp 1 ≤ 2.7183` — my h4 : Real.exp 1 ≤ 2.7183 := Real.exp_one_lt_d9.le — MISMATCH (2.7182818286 ≠ 2.7183). Fix: `have h4 : Real.exp 1 ≤ 2.7182818286 := Real.exp_one_lt_d9.le` and adjust the following norm_num/linarith: goal after rw [h3] was `Real.exp 1 ≤ (1.4:ℝ)^3` — with norm_num computing 1.4^3 = 2.744: goal `exp 1 ≤ 2.744` — from h4 : e ≤ 2.7182818286 ≤ 2.744 ✓ linarith. My code: `norm_num at h4 ⊢` — norm_num at ⊢ would compute 1.4^3 → 2.744 ✓ then `linarith [h4]` — with h4 : e ≤ 2.7182818286 ✓ — but the ERROR was the ascription mismatch itself. Fix the type ascription: `have h4 : Real.exp 1 ≤ 2.7182818286 := Real.exp_one_lt_d9.le`.\n\nHmm wait — actually — `norm_num at h4 ⊢` — norm_num at h4 might rewrite 2.7182818286 — leave as is. Just fix the ascription.\n\n4. **1565**: he13's `rw [div_le_iff₀ h2πpos, one_mul]` — goal `1/(2π)·exp(1/3) ≤ 1` — div_le_iff₀ : a/b ≤ c ↔ a ≤ c·b — hmm — the LHS is `1/(2π) * exp(1/3)` — a PRODUCT, not a single division!! div_le_iff₀'s pattern `?a / ?b ≤ ?c` — my LHS: `(1 / (2π)) * exp(1/3)` — the outer op is * — no match ✓ as reported. Fix: the goal `1/(2π)·exp(1/3) ≤ 1` — rewrite as exp(1/3)/(2π) ≤ 1 first: `rw [mul_comm, ← div_eq_mul_one_div]`?? — hmm — (1/(2π))·e13 = e13/(2π) via: `show Real.exp (1/3) / (2 * Real.pi) ≤ 1`?? — (1/(2π))·e13 vs e13/(2π) = e13·(1/(2π)) ✓ mul_comm-ish: `rw [mul_comm, div_eq_mul_one_div]`?? — div_eq_mul_one_div : a/b = a·(1/b) — e13/(2π) = e13·(1/(2π)) — so e13·(1/(2π)) = e13/(2π) — ← direction. Let me: `rw [show (1 / (2 * Real.pi)) * Real.exp (1/3 : ℝ) = Real.exp (1/3 : ℝ) / (2 * Real.pi) by rw [div_eq_mul_one_div, mul_comm], div_le_iff₀ h2πpos, one_mul]; exact he13` — hmm — div_eq_mul_one_div : a/b = a*(1/b) — rw [div_eq_mul_one_div] on goal `(1/(2π))·e13 = e13/(2π)` — pattern ?a/?b — matches RHS's e13/(2π)?? — rewrites it to e13·(1/(2π)) — then goal (1/(2π))·e13 = e13·(1/(2π)) — mul_comm closes?? — rw [mul_comm] — rewrites the FIRST product — (1/(2π))·e13 → e13·(1/(2π)) — goal becomes e13·(1/(2π)) = e13·(1/(2π)) ✓ rfl. Messy but works. SIMPLER: state he13' differently — as `Real.exp (1/3) / (2 * Real.pi) ≤ 1`?? — but I use it in hIIMajAbs as ((1/2π)·e13)·IIMaj ≤ 1·IIMaj — mul_le_mul_of_nonneg_right (he13' : (1/2π)·e13 ≤ 1) — needs the (1/2π)·e13 form. Alternative: prove he13' via `div_le_one_of_le (h : a ≤ b) (hb : 0 < b)?` — hmm — or `div_le_iff₀` with the division form: prove he13'/aux: exp(1/3)/(2π) ≤ 1 by rw [div_le_iff₀ h2πpos, one_mul]; exact he13 — then convert to the product form in hIIMajAbs:\n```lean\n  have he13' : Real.exp (1 / 3 : ℝ) / (2 * Real.pi) ≤ 1 := by\n    rw [div_le_iff₀ h2πpos, one_mul]\n    exact he13\n```\nand in hIIMajAbs: `have h1 := mul_le_mul_of_nonneg_right he13' hIIMaj0` — h1 : (e13/(2π))·IIMaj ≤ 1·IIMaj — and the calc's step `_ ≤ 1 * IIMaj := h1` — the calc has `((1/(2π))·e13)·IIMaj` — vs h1's `(e13/(2π))·IIMaj` — mismatch — fix the calc's middle form to (e13/(2π))·IIMaj: `(1/(2π))·(e13·IIMaj) = (e13/(2π))·IIMaj := by ring`?? — (1/(2π))·(e13·IIMaj) vs (e13/(2π))·IIMaj — e13/(2π) = e13·(1/(2π)) — (1/(2π))·e13·IIMaj vs e13·(1/(2π))·IIMaj — commuted ✓ ring ✓.\n\nSo:\n```lean\n  have hIIMajAbs : (1 / (2 * Real.pi)) * (Real.exp (1 / 3 : ℝ) * IIMaj (n : ℝ)) ≤\n      IIMaj (n : ℝ) := by\n    have h1 := mul_le_mul_of_nonneg_right he13' hIIMaj0\n    calc (1 / (2 * Real.pi)) * (Real.exp (1 / 3 : ℝ) * IIMaj (n : ℝ))\n        = (Real.exp (1 / 3 : ℝ) / (2 * Real.pi)) * IIMaj (n : ℝ) := by ring\n      _ ≤ 1 * IIMaj (n : ℝ) := h1\n      _ = IIMaj (n : ℝ) := by ring\n```\n\n5. **1603**: `Real.log_nonneg hlog1` — hlog1 : 1 ≤ Real.log T — log_nonneg expects 1 ≤ T. My hA used `(Real.log_nonneg hlog1)` intending `0 ≤ log T` — hlog1 : 1 ≤ log T — so log_nonneg needs 1 ≤ T (not 1 ≤ log T). Fix: `Real.log_nonneg (by linarith [hTn, hsqrt316] : (1:ℝ) ≤ T)` — T ≥ 316 ≥ 1 ✓. In hA: `mul_le_mul hTn1 h1 (Real.log_nonneg ...) (by linarith [hsqrt0])` — the third arg: 0 ≤ log T := Real.log_nonneg hT1ge — provide hT1ge : 1 ≤ T := by linarith [hTn, hsqrt316].\n\nHmm — hA is in hFAlt: `exact mul_le_mul hTn1 h1 (Real.log_nonneg hlog1) (by linarith [hsqrt0])` — change to `(Real.log_nonneg (by linarith [hTn, hsqrt316] : (1 : ℝ) ≤ T))`.\n\n6. **1612/1614/1616 timeouts** — deterministic timeout at isDefEq / tactic execution — at lines 1612-1616 — what's there? Probably the hX/hHhat or the final assembly region... 1612 — let me figure out: the file's §6 starts ~1330; 1612 is around hHhat or hBn. The isDefEq timeout at 1612:48 — col 48 — hmm — likely `unfold Bn; rfl` or one of the `unfold` calls exploding (unfold IIMaj positivity?). Line 1612 — let me count: 1330 (C0hat_le_self) + pinch ~80 + sfBoundProp ~... — 1612 is likely `have hBn : Bn FAlt (n : ℝ) = ... := by unfold Bn; rfl` region or hHhat. The timeout at isDefEq — the `rfl` after unfold Bn?? — the let-expression's defeq check exploding — OR `unfold DogLegRight IIMaj; positivity` in hDLR0 — unfold of a def with rpow — positivity traversing — hmm.\n\nLet me get the line contents at 1610-1620.","signature":""},{"type":"text","text":"Six issues. Checking what's at lines 1610–1620 (the timeout region) before fixing."},{"type":"tool_use","id":"Read_173","name":"Read","input":{"file_path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiSideBounds.lean","offset":1606,"limit":24},"extra_content":null}]