[{"type":"thinking","thinking":"`HasDerivAt.congr_of_eventuallyEq (h : HasDerivAt f f' x) (h₁ : f₁ =ᶠ[𝓝 x] f) : HasDerivAt f₁ f' x` — method form: `hmain.congr_of_eventuallyEq hg1` where hg1 : deriv g =ᶠ nice (f₁ = deriv g, f = nice) ✓. So: `hmul.congr_of_eventuallyEq hg1` — NOT hg1.symm and NOT hg1.hasDerivAt hmul.\n\nAlso `deriv_add_const (c : F) : deriv (fun y ↦ f y + c) x = deriv f x` — unconditional (no differentiability hypothesis! because deriv of non-differentiable is 0 both sides). \n\nFixes:\n- h1 in hg1: `rw [deriv_add_const]; exact hsm.deriv` — wait hsm.deriv : deriv sqrt m = 1/(2*√m) ✓.\n- hA's h1: `hsqrt.add_const 1` — HasDerivAt.add_const (h : HasDerivAt f f' x) (c) : HasDerivAt (fun x => f x + c) f' x ✓.\n- hder: `(hmul.congr_of_eventuallyEq hg1).deriv`. But hmul's function is `(fun n => deriv φ (√n+1)) * fun n => (2*√n)⁻¹` (Pi.mul form) while hg1's RHS is `fun n => deriv φ (√n+1) * (2*√n)⁻¹` — defeq but congr_of_eventuallyEq needs syntactic match on f? The EventuallyEq hg1 : deriv g =ᶠ (fun n => ... * ...) — when applying, Lean unifies f₁ := deriv g, f := nice lambda; h : HasDerivAt f ... — hmul's function is Pi.mul form; unification up to reducible defeq: Pi.mul application `(fun a => ...) * (fun a => ...)` vs `fun n => ... * ...` — Pi.instMul apply is reducible? Pi.mul_apply is a simp lemma; instMul pi instance... f * g unfolds via Pi.mul_def? It's `Mul.mul f g = fun i => f i * g i` through instance — usually reducible enough for unification? Risky. Mitigate: state hmul with explicit ascribed type using the lambda form:\n```lean\nhave hmul : HasDerivAt (fun n => deriv φ (√n + 1) * (2 * √n)⁻¹)\n    (deriv (deriv φ) (√n + 1) * (1 / (2 * √n)) * (2 * √n)⁻¹ +\n     deriv φ (√n + 1) * (-(2 * (1 / (2 * √n))) / (2 * √n) ^ 2)) n := hA.mul hB\n```\nType ascription forces unification of Pi.mul with lambda — defeq check at term level: hA.mul hB has type HasDerivAt (f * g) ... with (f * g) Pi.mul; ascription wants fun n => f n * g n. Pi.mul f g is DEFINED as fun n => f n * g n (Pi.instMul: mul f g i = f i * g i) — defeq ✓ should typecheck.\n\nThen `(hmul.congr_of_eventuallyEq hg1).deriv` gives deriv (deriv g) n = big expr; rw into goal; field_simp; ring.\n\nApply fixes.","signature":""},{"type":"text","text":"`HasDerivAt.congr_of_eventuallyEq` is the method form, and `deriv_add_const` is unconditional. Applying fixes."},{"type":"tool_use","id":"Edit_177","name":"Edit","input":{"file_path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\Scratch.lean","old_string":"    have h1 : deriv (fun x => √x + 1) m = 1 / (2 * √m) := by\n      simpa using (hsm.add (hasDerivAt_const m 1)).deriv\n    rw [show (fun x => φ (√x + 1)) = φ ∘ (fun x => √x + 1) from rfl,\n      deriv_comp m (hφ1 (√m + 1)) hhm, h1]\n    ring","new_string":"    have h1 : deriv (fun x => √x + 1) m = 1 / (2 * √m) := by\n      rw [deriv_add_const]\n      exact hsm.deriv\n    rw [show (fun x => φ (√x + 1)) = φ ∘ (fun x => √x + 1) from rfl,\n      deriv_comp m (hφ1 (√m + 1)) hhm, h1]\n    ring"},"extra_content":null},{"type":"tool_use","id":"Edit_178","name":"Edit","input":{"file_path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\Scratch.lean","old_string":"    have h1 : HasDerivAt (fun x => √x + 1) (1 / (2 * √n)) n := by\n      simpa using hsqrt.add (hasDerivAt_const n 1)","new_string":"    have h1 : HasDerivAt (fun x => √x + 1) (1 / (2 * √n)) n :=\n      hsqrt.add_const 1"},"extra_content":null},{"type":"tool_use","id":"Edit_179","name":"Edit","input":{"file_path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\Scratch.lean","old_string":"  have hmul := hA.mul hB\n  have hder := (hg1.hasDerivAt hmul).deriv\n  rw [hder]\n  have hn0 : (√n) ≠ 0 := Real.sqrt_ne_zero'.2 hn\n  have hnsq : (√n)^2 = n := Real.sq_sqrt hn.le\n  field_simp [hn0, hnsq]\n  ring","new_string":"  have hmul : HasDerivAt (fun n => deriv φ (√n + 1) * (2 * √n)⁻¹)\n      (deriv (deriv φ) (√n + 1) * (1 / (2 * √n)) * (2 * √n)⁻¹ +\n        deriv φ (√n + 1) * (-(2 * (1 / (2 * √n))) / (2 * √n) ^ 2)) n :=\n    hA.mul hB\n  have hder := (hmul.congr_of_eventuallyEq hg1).deriv\n  rw [hder]\n  have hn0 : (√n) ≠ 0 := Real.sqrt_ne_zero'.2 hn\n  have hnsq : (√n)^2 = n := Real.sq_sqrt hn.le\n  field_simp [hn0, hnsq]\n  ring"},"extra_content":null}]