[{"type":"thinking","thinking":"Batch 3 compiles — all 8 blocks done with ψ-formulas and signs. Commit, then S3: IIMaj.\n\nIIMaj plan:\n1. Rewrite IIMaj n = 4.56·(√n+1)·(e^{−a√n} + e^{−b√n}) with a = 2·log2 − 1/2, b = 2·log2. Constants: IIMaj n = 2*(√n+1)*(exp(√n/2)+1)*0.57*2^(−(2√n−2)). \n\n2·0.57 = 1.14. 2^(−(2√n−2)) = 2^(2−2√n) = 4·2^(−2√n). So IIMaj = 4.56·(√n+1)·(e^{√n/2}+1)·2^{−2√n}. And 2^{−2√n} = e^{−2(log 2)·√n}: (2:ℝ)^(−2√n) = Real.exp (log 2 · (−2√n)) via Real.rpow_def_of_pos (2 > 0): 2^x = exp (log 2 * x) for real x ✓ (`Real.rpow_def_of_pos : 0 < x → x ^ y = exp (log x * y)`).\n(e^{√n/2}+1)·e^{−2(log2)√n} = e^{√n/2 − 2(log2)√n} + e^{−2(log2)√n} = e^{−a√n} + e^{−b√n} with a = 2log2 − 1/2, b = 2log2: √n/2 − 2(log2)√n = −(2log2 − 1/2)√n ✓.\n\nSo define the unshifted ψ̃(s) := 4.56·(s+1)·(exp(−a·s) + exp(−b·s)) and IIMaj n = ψ̃(√n) — as an Eq lemma (proved by unfolding IIMaj + rpow/exp algebra + ring on exponents).\n\n2. ψ̃ is globally ContDiff ℝ 2: polynomial (s+1), exp of linear: ContDiff.mul etc: `ContDiff.exp`, `ContDiff.mul`, `contDiff_id`, `contDiff_const`, `ContDiff.add`... (contDiff_id.add contDiff_const).mul ((contDiff_const.mul contDiff_id).exp.add ...) — mechanical.\n\n3. Second derivative via chain_rule_sqrt: deriv(deriv (ψ̃∘√)) n = [ψ̃″(s) − ψ̃′(s)/s]/(4n), s = √n.\n\n4. Compute ψ̃′, ψ̃″ at s (s > 0): via HasDerivAt machinery again (or deriv rules since globally smooth — use deriv2_of_hasDerivAt_nhds with explicit D1):\nD1(s) = 4.56·[(e^{−as}+e^{−bs}) + (s+1)(−a e^{−as} − b e^{−bs})]\nD2(s) = 4.56·[2(−a e^{−as} − b e^{−bs}) + (s+1)(a² e^{−as} + b² e^{−bs})]\n\nPer-point HasDerivAt: u = s+1 (u'=1), v = e^{−as}+e^{−bs} (v' = −a e^{−as} − b e^{−bs} via HasDerivAt.exp of const_mul: (hasDerivAt_const_mul? (hasDerivAt_id s).const_mul (−a)).exp gives HasDerivAt (fun s => exp(−a*s)) (exp(−a*s)·(−a)) s — HasDerivAt.exp : HasDerivAt f f' x → HasDerivAt (fun x => exp (f x)) (exp (f x) * f') x ✓. Massage exp(−a s)*(−a) = −a*exp(−a s) by ring.\n\n5. The bound: for n ≥ 20 (s = √n ≥ √20 ≈ 4.4721):\nIIMaj″(n) = [ψ̃″(s) − ψ̃′(s)/s]/(4s²)... and margiñ″ needs IIMaj″(n) < 1/(2n) = 1/(2s²). So need ψ̃″(s) − ψ̃′(s)/s < 2 for s ≥ √20 — wait: IIMaj″(n) < 1/(2n) ⟺ [ψ̃″−ψ̃′/s]/(4s²) < 1/(2s²) ⟺ ψ̃″(s) − ψ̃′(s)/s < 2.\n\nBound: ψ̃″(s) − ψ̃′(s)/s = 4.56·[2(−ae^{−as}−be^{−bs}) + (s+1)(a²e^{−as}+b²e^{−bs}) − (e^{−as}+e^{−bs})/s − (s+1)(−ae^{−as}−be^{−bs})/s]\n≤ 4.56·[(s+1)(a²e^{−as}+b²e^{−bs}) + (s+1)(ae^{−as}+be^{−bs})/s]  (dropping negative terms)\n= 4.56(s+1)[(a²+a/s)e^{−as} + (b²+b/s)e^{−bs}]\n≤ 4.56(s+1)(b²+b/s)(e^{−as}+e^{−bs})   [since b > a > 0: b²+b/s ≥ a²+a/s, termwise]\n\nNeed < 2 ⟺ 2.28(s+1)(b²+b/s)(e^{−as}+e^{−bs}) < 1.\n\nMonotone-factor bound: write (s+1)(b²+b/s) = (s+1)(b²s+b)/s = [(s+1)/s]·(b²s+b). So\nLHS = 2.28·[(s+1)/s]·[(b²s+b)e^{−as}]·[1 + e^{−(b−a)s}] with b−a = 1/2.\n\nThree factors, each positive and (strictly) decreasing on [√20, ∞):\n- u₁(s) = (s+1)/s = 1+1/s: decreasing ✓ (antitone: 1/s antitone for s>0).\n- u₂(s) = (b²s+b)e^{−as}: derivative-negative argument: deriv u₂ = b²e^{−as} + (b²s+b)(−a)e^{−as} = e^{−as}[b² − a(b²s+b)] = e^{−as}[b²(1−as) − ab]: for s ≥ √20 ≈ 4.47, a ≈ 0.8863: 1−as ≤ 1−3.96 < 0 → bracket < 0 ✓ negative. Prove via `strictAntiOn_of_deriv_neg`?? For ANTITONE (non-strict) on [√20,∞): use MonotoneOn/AntitoneOn from derivative ≤ 0: `antitoneOn_of_deriv_nonpos`? There's `antitoneOn_of_deriv_nonpos (hD : Convex ℝ D) (hf : ContinuousOn f D) (hf' : DifferentiableOn ℝ f (interior D)) (h : ∀ x ∈ interior D, deriv f x ≤ 0) : AntitoneOn f D` ✓ exists I believe.\n- u₃(s) = 1+e^{−s/2}: decreasing ✓ (exp of −s/2 antitone: exp monotone, −s/2 antitone → composition antitone; +1 preserves).\n\nSo LHS(s) ≤ 2.28·u₁(√20)·u₂(√20)·u₃(√20), and the product bound: u₁·u₂·u₃ with each positive: u₁(s) ≤ u₁(√20), u₂(s) ≤ u₂(√20), u₃(s) ≤ u₃(√20) → product ≤ product (mul_le_mul with nonneg chains).\n\nThen numeric: u₁(√20) = (√20+1)/√20 ≤ (4.4722+1)/4.4721: need √20 bounds BOTH ways: 4.4721 ≤ √20 ≤ 4.4722: u₁ = 1 + 1/s decreasing in s → u₁(s) ≤ u₁(√20) ≤ 1 + 1/4.4721 ≈ 1.2236.\nu₂(√20) = (b²√20+b)e^{−a√20}: b = 2log2 ≤ 2·0.6932 = 1.3864 (log 2 ≤ 0.6932: from log_two_lt_d9? exists? `Real.log_two_lt_d9 : log 2 < 0.6931471806`? I recall log_two_gt_d9 used earlier; lt version probably exists too — check. Otherwise prove: log 2 < 0.69315 via exp: 2 < exp(0.69315)?? harder. grep needed.)\nb² ≤ 1.3864² = 1.92211; b²√20+b ≤ 1.92211·4.4722 + 1.3864 ≈ 8.5968 + 1.3864 = 9.9832. e^{−a√20} ≤ e^{−3.96} (a√20 ≥ (2·0.6931−0.5)·4.4721 = 0.8862·4.4721 ≈ 3.9632 ≥ 3.96) ≤ 1/Σ₆(3.96): Σ₆(3.96) = 1+3.96+3.96²/2+3.96³/6+3.96⁴/24+3.96⁵/120+3.96⁶/720: compute: 3.96²=15.6816/2=7.8408; 3.96³=62.0991/6=10.3499; 3.96⁴=245.912/24=10.2463; 3.96⁵=973.813/120=8.1151; 3.96⁶=3856.3/720=5.3560. Sum: 1+3.96+7.8408+10.3499+10.2463+8.1151+5.356 = 46.868. e^{−a√20} ≤ 1/46.868 ≈ 0.021336 (via e^{a√20} ≥ Σ₆(a√20) ≥ Σ₆(3.96) — wait: e^t ≥ Σ₆(t) and t ≥ 3.96 → e^t ≥ Σ₆(t) ≥ Σ₆(3.96) (each term increasing in t) ✓ then e^{−t} ≤ 1/Σ₆(3.96) ✓.\nu₃(√20) = 1+e^{−√20/2} ≤ 1+e^{−2.23605} (√20/2 ≥ 2.23605) ≤ 1+1/Σ₆(2.23605)?? e^{−2.236} ≤ 1/(1+2.236+2.5+...) — e^{2.236} ≥ 1+2.236+2.5+1.863+1.0417+0.4659 ≈ 9.1 hmm let me: 2.236²=4.9997/2=2.4998; 2.236³=11.179/6=1.8632; 2.236⁴=24.997/24=1.0415; 2.236⁵=55.89/120=0.4658. Σ₅ ≈ 1+2.236+2.4998+1.8632+1.0415+0.4658 = 9.1063 → e^{−2.236} ≤ 0.1098. u₃ ≤ 1.1098.\n\nProduct: 2.28·1.2236·(9.9832·0.021336)·1.1098 = 2.28·1.2236·0.21301·1.1098 ≈ 0.6596 < 1 ✓ comfortable.\n\nFor Lean, I'll simplify the numeric tail: use cruder constants with margin: need final < 1 with provable rational arithmetic. Let me define the bound chain with decimals provable by norm_num. The exp lower bound: Real.sum_le_exp_of_nonneg (hx : 0 ≤ x) (n : ℕ) : ∑ i ∈ Finset.range n, x ^ i / i ! ≤ Real.exp x — used in EulerMascheroni file as `Real.sum_le_exp_of_nonneg (by norm_num) 7` ✓ signature confirmed. With n = 7 (range 7 = 0..6 ✓).\n\nt ≥ 3.96: a·√20 ≥ (2·0.6931471803−0.5)·4.4721 = 0.8862943606·4.4721: compute: 0.8862943606·4.4721 ≈ 3.96330...: 0.8862943606·4 = 3.54518, ·0.4721 = 0.418423: total ≈ 3.9636 ≥ 3.96 ✓. In Lean: a = 2·log2 − 1/2 ≥ 2·0.6931471803 − 0.5 = 0.8862943606 (log_two_gt_d9 ✓). √20 ≥ 4.4721 (le_sqrt, 4.4721² = 19.99968... norm_num ✓). a·√20 ≥ 0.8862943606·4.4721 ≥ 3.96 (norm_num: 0.8862943606·4.4721 = 3.96360... ✓).\n\nAlso need a ≤ b: a = 2log2−1/2 < 2log2 = b ✓ linarith (log 2 > 0: Real.log_pos... log 2 > 0 via log_two_gt_d9 ✓).\nb bounds: b = 2log2: upper b ≤ 2·0.6931471806 = 1.3862943612 (need log_two_lt_d9 — check existence; if missing, prove log 2 ≤ 0.7 via: log 2 = −log(1/2)... or Real.log_two_lt_d9 — grep).\nb²: b ≤ 1.3863 → b² ≤ 1.92183 (mul_le_mul with nonneg: b ≤ 1.3862943612: b² ≤ 1.3862943612² = 1.92181... norm_num).\ne^{−as} vs e^{−bs}: e^{−bs} ≤ e^{−as} since −bs ≤ −as (b ≥ a, s ≥ 0) → exp_le_exp.mpr ✓.\n\nHmm — actually, for the \"dropping negative terms\" step I need care with signs: ψ̃″ − ψ̃′/s = 4.56[−2(ae^{−as}+be^{−bs}) + (s+1)(a²e^{−as}+b²e^{−bs}) − (e^{−as}+e^{−bs})/s + (s+1)(ae^{−as}+be^{−bs})/s].\n\nSince ae^{−as}+be^{−bs} > 0 and e^{−as}+e^{−bs} > 0: −2(...) ≤ 0 and −(...)/s ≤ 0, so\nψ̃″ − ψ̃′/s ≤ 4.56[(s+1)(a²e^{−as}+b²e^{−bs}) + (s+1)(ae^{−as}+be^{−bs})/s] ✓ (drop two negative terms — rigorous via linarith).\n\nThen factor: (s+1)(a²+a/s) ≤ (s+1)(b²+b/s) and same for the b-term — actually: a²e^{−as} + b²e^{−bs} ≤ (b²)(e^{−as}+e^{−bs}) and (ae^{−as}+be^{−bs})/s ≤ (b/s)(e^{−as}+e^{−bs}): so sum ≤ (s+1)(b²+b/s)(e^{−as}+e^{−bs}) ✓ (a² ≤ b² from 0 ≤ a ≤ b; a ≤ b; e^{−as}, e^{−bs} ≥ 0; mul_le_mul chains).\n\nThen: (s+1)(b²+b/s) = (s+1)(b²s+b)/s ✓ (field: b²+b/s = (b²s+b)/s, s ≠ 0).\ne^{−as}+e^{−bs} ≤ e^{−as} + e^{−as} = 2e^{−as}?? or keep (1+e^{−(b−a)s})e^{−as}: e^{−bs} = e^{−as}·e^{−(b−a)s}: exp_add ✓. The 2e^{−as} route is looser but simpler! Check: 2.28·1.2236·(9.9832·0.021336)·2 = 2.28·1.2236·0.21301·2 ≈ 1.189 > 1 ✗ TOO LOOSE. So need the (1+e^{−s/2}) factorization (b−a = 1/2 exactly). OK keep u₃.\n\nSo final chain for the IIMaj bound lemma (n ≥ 20):\nderiv (deriv IIMaj... wait — need the identity IIMaj n = ψ̃(√n) first, and the theorem we need for assembly is:\n  deriv (deriv IIMaj) n < 1/(2n)  for n ≥ 20.\nHmm — actually for assembly: margiñ″(n) = 1/(2n) − Σ-stuff/(4n) − IIMaj″(n) ≥ 1/(2n) − IIMaj″(n) > 0. So need IIMaj″(n) < 1/(2n) ✓.\n\nLemma IIMaj_deriv2_lt (n) (hn : 20 ≤ n) : deriv (deriv IIMaj) n < 1/(2*n).\n\nProof structure:\n- have hId : IIMaj = fun n => psiTilde (Real.sqrt n) (function-level eq, by funext + algebra)\n- rw → deriv (deriv (fun n => psiTilde (√n))) n = [ψ̃″(√n) − ψ̃′(√n)/√n]/(4n) via chain_rule_sqrt psiTilde hcont n (by linarith)\n- rw [psiTilde formulas at s := √n] → explicit\n- bound chain → < 2/(4n)?? wait: need [ψ̃″−ψ̃′/s]/(4n) < 1/(2n) ⟺ ψ̃″ − ψ̃′/s < 2 (n > 0). Provide hbound : ψ̃″(s) − ψ̃′(s)/s ≤ 4.56(s+1)(b²+b/s)(e^{−as}+e^{−bs}) and hfinal : 4.56(s+1)(b²+b/s)(e^{−as}+e^{−bs}) < 2 via the factor bound 2.28·u₁·u₂·u₃ ≤ 0.6596 < 1: 4.56(...)= 2·2.28(...)< 2 ✓.\n\npsiTilde ContDiff ℝ 2: build:\n  contDiff_const.mul ((contDiff_id.add contDiff_const).mul (((contDiff_const.mul contDiff_id).exp).add (contDiff_const.mul contDiff_id).exp))\nWait psiTilde s = 4.56 * ((s+1) * (exp(-a*s) + exp(-b*s))): structure: const_mul of product: (contDiff_id.add contDiff_const) : C^∞ (s+1); (contDiff_const.mul contDiff_id).exp : exp(-a·s) — contDiff_const.mul contDiff_id : fun s => −a*s: ContDiff.const_mul : ContDiff ℝ n f → (c) → ContDiff ℝ n (fun x => c * f x) ✓ (c := −a, f := id). .exp ✓ ContDiff.exp. .add for the sum; .mul for (s+1)*(sum); .const_mul 4.56. Types: a, b as real literals — define a, b as local lets or inline (2*Real.log 2 − 1/2). Inline is fine but verbose; define private defs:\n  private def iimA : ℝ := 2 * Real.log 2 - 1 / 2\n  private def iimB : ℝ := 2 * Real.log 2\n  private def psiTilde (s : ℝ) : ℝ := 4.56 * ((s + 1) * (Real.exp (-iimA * s) + Real.exp (-iimB * s)))\n\nThe ψ̃′/ψ̃″ formulas via deriv2_of_hasDerivAt_nhds with:\n  D1 s = 4.56 * ((Real.exp (-a*s) + Real.exp (-b*s)) + (s+1) * (-a * Real.exp (-a*s) - b * Real.exp (-b*s)))\n  D2 s = 4.56 * (2 * (-a * Real.exp (-a*s) - b * Real.exp (-b*s)) + (s+1) * (a^2 * Real.exp (-a*s) + b^2 * Real.exp (-b*s)))\n\nh1: ∀ᶠ x in 𝓝 s, HasDerivAt psiTilde (D1 x) x — actually global: eventually_of_forall (all x: everything smooth) ✓.\nPer-point: u := (·+1): (hasDerivAt_id' x).add_const 1 : HasDerivAt (fun t => t+1) 1 x; va := fun t => exp(-a*t): (hasDerivAt_id' x).const_mul (-a) : HasDerivAt (fun t => -a*t) (-a) x; .exp : HasDerivAt (fun t => exp(-a*t)) (exp(-a*x)*(-a)) x; massage exp(-a*x)*(-a) = -a*exp(-a*x) (mul_comm/ring). v := va.add vb; uv := u.mul v; D = const_mul 4.56.\nD1's derivative: w := fun t => exp(-a*t)+exp(-b*t) (w' = -a e^{−ax} − b e^{−bx}), z := fun t => (t+1)*(−a e^{−at} − b e^{−bt}): z' = 1·(−a e−b e) + (s+1)·(a² e + b² e): (−a e^{−at} − b e^{−bt})' = a²e^{−at} + b²e^{−bt}: from va': HasDerivAt (fun t => −a*exp(−a*t)) (a² exp(−a*x))?? d/dt[−a e^{−at}] = −a·(−a e^{−at}) = a² e^{−at}: (−a e^{−at})' : (va').const_mul (−a)?? va' as function: fun t => −a*exp(−a*t) = const_mul (−a) of va: HasDerivAt (−a * va) (−a * va'(x)) = (−a)*(exp(−a*x)*(−a)) = a²exp(−a*x) ✓. Then sum, product, const — mechanical.\n\nThe D2 statement: d/ds D1 = 4.56·[w' + z']: w' + z' = (−a e − b e) + (−a e − b e) + (s+1)(a²e + b²e) = 2(−a e^{−as} − b e^{−bs}) + (s+1)(a²e^{−as}+b²e^{−bs}) ✓.\n\nThen chain_rule_sqrt with hψ : ContDiff ℝ 2 psiTilde → deriv(deriv (psiTilde∘√)) n = (D2(s) − D1(s)/s)/(4n) — need deriv psiTilde s = D1 s and deriv (deriv psiTilde) s = D2 s — from deriv2_of_hasDerivAt_nhds + h1.self_of_nhds.deriv ✓.\n\nIdentity IIMaj n = psiTilde (√n):\nunfold IIMaj psiTilde iimA iimB: goal: 2*(√n+1)*(exp(√n/2)+1)*0.57*2^(-(2√n−2)) = 4.56*((√n+1)*(exp(−a√n)+exp(−b√n))).\nSteps: 2^(-(2√n−2)) = exp(log 2 * (-(2√n−2))) (Real.rpow_def_of_pos (by norm_num : (0:ℝ)<2)); then split exp: exp(log2·(2−2√n)) = exp(2log2)·exp(−2(log2)√n)... = 4·exp(−b√n)?? exp(2 log 2) = (exp(log 2))² = 2² = 4: Real.exp_log (0<2) ✓.\nHmm — cleaner: 2^(-(2√n−2)) = 2^(2−2√n) = 2^2 * 2^(−2√n) = 4·2^(−2√n) (rpow_add: 2^(x+y) = 2^x·2^y: Real.rpow_add (hx : 0 < x)? `Real.rpow_add (hx : 0 < x) (y z : ℝ) : x ^ (y + z) = x^y * x^z` ✓). And 2^(−2√n) = exp(−b√n): rpow_def_of_pos: 2^(−2√n) = exp(log 2 * (−2√n)) = exp(−(2log2)√n) = exp(−b√n) (ring on exponent). And exp(√n/2)·exp(−b√n) = exp(√n/2 − b√n) = exp(−a√n): exp_add ✓ exponent ring: √n/2 − 2log2·√n = −(2log2−1/2)√n ✓.\nSo: (e^{√n/2}+1)·4·2^{−2√n} = 4·(e^{√n/2}·e^{−b√n} + e^{−b√n}) = 4·(e^{−a√n}+e^{−b√n}) ✓.\nLHS constants: 2·0.57·4 = 4.56 ✓.\nLean proof: sequence of rws + ring_nf on exponent equalities. The exponent equalities like log 2 * (2 − 2*√n) = 2*log2 − (2*log2)*√n — ring ✓ (atoms log 2, √n).\n\nThen the final numeric bound with monotonicity — actually wait. Let me reconsider whether I even NEED monotonicity over [√20,∞). Alternative: the bound must hold for ALL n ≥ 20 — yes monotonicity needed (can't check pointwise). OK as planned.\n\nFor u₂(s) = (b²s+b)e^{−as} antitone on [√20,∞): deriv u₂ = e^{−as}(b² − a(b²s+b)): need ≤ 0 on (interior) (s > √20): b² ≤ a(b²s+b) ⟺ b²(1−as) ≤ ab: since 1−as < 0 (as ≥ 3.96 > 1) LHS < 0 < ab (a,b > 0)... wait a > 0: a = 2log2−1/2 ≥ 0.886 > 0 ✓ b > 0 ✓ ab > 0; LHS b²(1−as): b² > 0, (1−as) < 0 → negative ≤ ab ✓ linarith-ish (needs b² > 0, 1−as < 0, ab > 0, and the product sign: nlinarith [hb2pos, hasneg, habpos]).\n\nantitoneOn_of_deriv_nonpos: check exact name — `antitoneOn_of_deriv_nonpos (hD : Convex ℝ D) (hf : ContinuousOn f D) (hf' : DifferentiableOn ℝ f (interior D)) (h : ∀ x ∈ interior D, deriv f x ≤ 0) : AntitoneOn f D` ✓ I'm fairly confident this exists (convex/Deriv.lean).\n\nu₁(s) = (s+1)/s = 1 + s⁻¹: antitone on Ici √20: s⁻¹ antitone on s > 0: `antitoneOn_inv`? or via deriv too — simplest: u₁ s = 1 + 1/s: AntitoneOn (fun s => 1+1/s) (Ici √20): from one_div antitone: `StrictAntiOn...`? There's `AntitoneOn.inv`? Hmm: prove via deriv_nonpos too (deriv (1+1/s) = −1/s² < 0) — uniform approach: use antitoneOn_of_deriv_nonpos for u₁ and u₃ as well? u₃ = 1+exp(−s/2): deriv = −½e^{−s/2} < 0 ✓. Uniform = less lemma-name hunting. But deriv computations for u₁: (1+s⁻¹)' = −s⁻² — via deriv2-style HasDerivAt... small.\n\nHmm, this IIMaj bound is itself ~200 lines. Fine — it's the last big piece. Write it as:\n- iimA, iimB, psiTilde defs\n- psiTilde_contDiff\n- IIMaj_eq_psiTilde (function eq)\n- psiTilde_D1/D2 formulas (hasDerivAt_psiTilde, deriv2 formula)\n- iimaj_deriv2_lt : deriv (deriv IIMaj) n < 1/(2n) for n ≥ 20 — containing the factor monotonicity + numerics.\n\nLet me also double-check `Real.rpow_add` and `Real.exp_add`, `Real.rpow_def_of_pos`, `Real.exp_log`, `Real.log_exp` names — standard ✓. `Real.sum_le_exp_of_nonneg` ✓ confirmed. `Real.log_two_lt_d9` — check.\n\nActually wait — before diving in, sanity-check the DERIVATIVE of IIMaj via chain_rule_sqrt requires ContDiff ℝ 2 psiTilde GLOBALLY ✓ (exp/polynomial) fine.\n\nAlso note: chain_rule_sqrt's conclusion: deriv(deriv (fun n => ψ(√n))) n = (ψ″(√n) − ψ′(√n)/√n)/(4n) ✓.\n\nAlso — hId as FUNCTION equality IIMaj = psiTilde ∘ √: `funext`; then rw [hId]; the deriv sees the composition — chain_rule_sqrt is stated for `fun n => ψ (√n)` — composition vs explicit lambda: deriv (fun n => ψ (√n)) vs deriv (ψ ∘ √) — defeq ✓ rw/show.\n\nLet me check log_two_lt_d9.","signature":""},{"type":"text","text":"All 8 blocks compile. Committing, then checking a name for the IIMaj numerics."},{"type":"tool_use","id":"Bash_153","name":"Bash","input":{"command":"cd /d/Python/Quantum && git add research/lean/NguyenRH1/NguyenRH1/LiSkeleton.lean && git -c user.name=\"NGUYEN-RH-1\" -c user.email=\"research@local\" commit -m \"Session 011: block table batch 3 (T·L1, T·L2) — all 8 D-057 ψ-blocks now formalized with sign proofs\" && grep -rn \"log_two_lt_d9\\|log_two_gt_d9\" /d/Python/Quantum/zeta-23-lean/.lake/packages/mathlib/Mathlib/Analysis/SpecialFunctions/Log/*.lean | head -4","description":"Commit batch 3; check log_two bounds"},"extra_content":null}]