[{"type":"thinking","thinking":"Both chain-rule lemmas compile. Commit this infrastructure piece (crash-safe), then proceed to S2: the block lemmas.\n\nActually — commit cadence: helpers compile, commit. Then S2 blocks.\n\nFor S2, per block type I need:\n(1) ContDiffOn ℝ 2 φ (Ioi (−1)) — via ContDiffOn combinators: `.mul`, `.add`, `.log`(?), const. Check `ContDiffOn.log`: grep. Real.log ContDiff: `Real.contDiff_log : ContDiff ℝ n Real.log`?? No — log isn't C^∞ at 0 globally... but Real.log with junk value 0 at 0: is it ContDiff on all ℝ? At x=0 it's not even continuous (log|x| → −∞)... Real.log x = log|x| for x ≠ 0 and Real.log 0 = 0 — NOT continuous at 0. So need ContDiffOn on {0}ᶜ or Ioi 0: `Real.contDiffOn_log : ContDiffOn ℝ n Real.log {0}ᶜ`? Let me grep. And composition: `ContDiffOn.comp` needs the inner function to map into the domain ✓.\n\nBlocks and their ψ formulas (T variable; L1 := log(T+2), L2 := log(L1)):\n\nBlock id: φ(T) = T. ψ = −1/(T−1). deriv φ = 1, deriv² = 0: ψ = 0 − 1/(T−1) < 0 for T > 1 ✓ trivial.\n\nBlock L1: φ = log(T+2). φ′ = 1/(T+2), φ″ = −1/(T+2)². ψ = −1/(T+2)² − 1/((T+2)(T−1)) < 0 for T > 1 ✓ (both terms negative: T+2 > 0, T−1 > 0 ✓).\n\nBlock L2: φ = log(log(T+2)). φ′ = 1/((T+2)L1). φ″ = −(L1+1)/((T+2)L1)². ψ = φ″ − φ′/(T−1) = −(L1+1)/((T+2)L1)² − 1/((T+2)L1(T−1)) < 0 for T > 1 (L1 > 0 ✓ since T+2 > 3 > e) ✓.\n\nBlock L1²: φ = L1². φ′ = 2L1/(T+2). φ″ = 2(1−L1)/(T+2)². ψ = 2(1−L1)/(T+2)² − 2L1/((T+2)(T−1)) < 0 iff (1−L1)/(T+2) < L1/(T−1) ⟸ 1−L1 < 0 (L1 > 1) and RHS > 0 ✓.\n\nBlock L1·L2: φ = L1L2. φ′ = L2/(T+2) + L1/((T+2)L1) = (L2 + 1)/(T+2). φ″ = L2′/(T+2) − L2/(T+2)² + [1/((T+2)L1)]′ = 1/((T+2)²L1) − L2/(T+2)² − 1/((T+2)L1)²·(1+L1)?? let me not hand-derive here — better to let Lean compute and match D-057's table: ψ = [1/L1 − L2 − 1]/(T+2)² − (L2+1)/((T+2)(T−1)). Sign: 1/L1 ≤ 1/2 < 1 + L2 (L2 ≥ 0.69) → bracket < 0 ✓; second term negative ✓.\n\nBlock L2²: φ = L2². ψ = 2[1 − L2(L1+1)]/((T+2)L1)² − 2L2/((T+2)L1(T−1)). Sign: L2(L1+1) ≥ 0.69·3 = 2.07 > 1 ✓ both negative ✓.\n\nBlock T·L1: φ = T·L1. ψ = (T+4)/(T+2)² − L1/(T−1) − T/((T+2)(T−1)) (#5). Sign per D-057: L1 ≥ 2 → ψ ≤ [(T+4)(T−1) − (3T+4)(T+2)]/((T+2)²(T−1)) — hmm their reduction: ψ = (T+4)/(T+2)² − [L1(T+2) + T]/((T+2)(T−1)) — with L1 ≥ 2: L1(T+2)+T ≥ 2(T+2)+T = 3T+4 → the subtracted fraction ≥ (3T+4)/((T+2)(T−1)) → ψ ≤ (T+4)/(T+2)² − (3T+4)/((T+2)(T−1)) = [(T+4)(T−1) − (3T+4)(T+2)]/((T+2)²(T−1)) and numerator = T²+3T−4 − (3T²+10T+8) = −2T²−7T−12 < 0 ✓.\n\nBlock T·L2: φ = T·L2. ψ = [T(L1−1)+4L1]/((T+2)L1)² − L2/(T−1) − T/((T+2)L1(T−1)) (#6). Sign per D-057: suffices 2 ≤ L1·L2·(T+2)... their chain: ψ < 0 ⟺ [T(L1−1)+4L1]/((T+2)L1)² < L2/(T−1) + T/((T+2)L1(T−1)). LHS < L1(T+2)²?? — their note: \"LHS < L1(T+2)², RHS > T(T+2)L1 + L2(T+2)²L1²; divide L1(T+2): condition (T+2) ≤ T + L2(T+2)L1\" — hmm the units look off (LHS is a derivative-value not scaled)... I'll redo the sign proof myself at write time: \n\nψ₆ = A/(D²) − L2/(T−1) − T/((T+2)L1(T−1)) where A = T(L1−1)+4L1, D = (T+2)L1.\nWant ψ₆ < 0 ⟺ A(T−1)(T+2)L1 < L2((T+2)L1)²(T−1)... let me just do: ψ₆ < 0 ⟺ A/((T+2)L1)² < L2/(T−1) + T/((T+2)L1(T−1)) = [L2(T+2)L1 + T]/((T+2)L1(T−1)).\n⟺ A(T−1) < [L2(T+2)L1 + T](T+2)L1/((T+2)L1)² ·... multiply both sides by ((T+2)L1)²(T−1) > 0:\nA(T−1)(T+2)L1... no: LHS·mult = A(T−1); RHS·mult = [L2(T+2)L1 + T](T+2)L1.\nSo ψ₆ < 0 ⟺ A(T−1) < (T+2)L1[L2(T+2)L1 + T] = L2(T+2)²L1² + T(T+2)L1.\nA(T−1) = [T(L1−1)+4L1](T−1) = (T−1)(TL1 − T + 4L1) = (T−1)(L1(T+4) − T).\nSuffices: (T−1)(L1(T+4) − T) ≤ T(T+2)L1 (dropping the L2 term ≥ 0) ⟺ expand: (T−1)L1(T+4) − T(T−1) ≤ T(T+2)L1 ⟺ L1[(T−1)(T+4) − T(T+2)] ≤ T(T−1) ⟺ L1[T²+3T−4 − T²−2T] = L1(T−4) ≤ T(T−1) ⟺ L1 ≤ T(T−1)/(T−4) (for T > 4). At T ≥ 5.4721: T(T−1)/(T−4) ≥ 5.4721·4.4721/1.4721 ≈ 24.47/1.4721 ≈ 16.62, and L1 = log(T+2): is log(T+2) ≤ T(T−1)/(T−4)?? For all T ≥ 5.47: log(T+2) ≤ (T+2)−1 = T+1 and T+1 ≤ 16.62 at T=5.47?? 6.47 ≤ 16.62 ✓ and T(T−1)/(T−4) grows like T while log grows slower ✓. So suffices: log(T+2) ≤ T(T−1)/(T−4) — provable: log(T+2) ≤ T+1 (log x ≤ x−1) and T+1 ≤ T(T−1)/(T−4) ⟺ (T+1)(T−4) ≤ T(T−1) ⟺ T²−3T−4 ≤ T²−T ⟺ −3T−4 ≤ −T ⟺ 2T ≥ −4 ✓ always. So even simpler: L1 ≤ T+1 ≤ T(T−1)/(T−4) ✓✓. Wait need T > 4 for the division direction: T ≥ 5.47 ✓.\n\nSo ψ₆ < 0 via: A(T−1) ≤ T(T+2)L1 ≤ T(T+2)L1 + L2(T+2)²L1² — hmm need A(T−1) ≤ T(T+2)L1: A(T−1) = (T−1)(L1(T+4)−T); is (T−1)(L1(T+4)−T) ≤ T(T+2)L1 ⟺ L1[(T−1)(T+4) − T(T+2)] ≤ T(T−1) ⟺ L1(T−4) ≤ T(T−1): since T−4 > 0: ⟺ L1 ≤ T(T−1)/(T−4) ✓ shown. But sign of (T−4): at T ≥ 5.47 positive ✓. Also need L1(T+4) − T ≥ 0? Not needed — if A(T−1) ≤ 0 even better; the inequality L1(T−4) ≤ T(T−1) is all we need (both sides ≥ 0 anyway).\n\nHmm wait, one more check: A = T(L1−1)+4L1 — is A > 0? L1 ≥ 2 → T(L1−1) ≥ T > 0 ✓.\n\nOK — all 8 signs are provable with: L1 ≥ 2 (via e² ≤ 7.389 ≤ T+2), L2 ≥ log 2 ≥ 0.6931, T ≥ √20+1, and elementary algebra. The block-by-block Lean work: derive φ′, φ″ formulas (deriv computations), combine into ψ, sign via nlinarith/field manipulations.\n\nThis is the long grind. To keep pieces committable, do blocks in batches: batch 1 = id, L1, L2 (simplest, products none); batch 2 = L1², L1L2, L2²; batch 3 = T·L1, T·L2.\n\nFor each block, the statement shape:\n\ntheorem blockL1_psi (n : ℝ) (hn : 20 ≤ n) :\n    deriv (deriv fun n => Real.log (Real.sqrt n + 1 + 2)) n = ψ-form := ...\n\nHmm — actually better structure: per block, two lemmas:\n(i) `blockX_deriv2`: the second-derivative FORMULA at T (T > 1 say): deriv (deriv φ) T = ... and deriv φ T = ... — or directly the ψ value: (deriv (deriv φ) T − deriv φ T/(T−1)) = closed form.\n(ii) `blockX_sign`: closed form < 0 for T ≥ T₀.\n\nThen in assembly: chain_rule_sqrt_shift_on φ hC n hn gives deriv(deriv (φ∘T)) n = ψ_raw/(4n); rw [block formula] → closed form/(4n) < 0... Actually for the margin, blocks appear with NEGATIVE weights (margin = S − B̃), and −wᵢ·(φᵢ∘T) has second deriv −wᵢψᵢ/(4n) > 0. \n\nderiv computations: to get deriv (deriv φ) T for φ = log(·+2): deriv φ = fun T => (T+2)⁻¹ (via deriv_of... `Real.deriv_log`? For composition: deriv (fun T => Real.log (T + 2)) = fun T => (T+2)⁻¹: HasDerivAt (log (T+2)) (1/(T+2)) via Real.hasDerivAt_log (T+2 ≠ 0) composed with +2. Then deriv (fun T => (T+2)⁻¹) = −(T+2)⁻²: HasDerivAt (fun T => (T+2)⁻¹) (−(T+2)⁻¹²...) — HasDerivAt.inv of (T+2) ✓. Use the eventuallyEq + deriv technique or `deriv (deriv ...)` via HasDerivAt.deriv at the point — but deriv (deriv φ) T requires deriv φ as a FUNCTION to be nice near T: prove `deriv φ =ᶠ[𝓝 T] fun x => (x+2)⁻¹` then deriv of that. Standard pattern per block ~15-25 lines.\n\nAlternatively — a slicker approach: prove each block's ψ-statement via `HasDerivAt` twice + the `deriv` extraction: define f0 = φ, prove HasDerivAt φ (D1 T) T with D1 explicit AND D1 differentiable with HasDerivAt D1 (D2 T) T — then deriv φ T = D1 T (h.deriv), and deriv (deriv φ) T = D2 T requires deriv φ =ᶠ D1 near T — from HasDerivAt at ALL nearby points (same proof parameterized) → eventuallyEq. Same work.\n\nLet me define a tiny helper to reduce boilerplate? \"Twice-differentiable with given derivatives\": \n\n```lean\n/-- Package: φ has first derivative D1 near T and D1 has derivative D2 at T. -/\nstructure TwiceAt (φ D1 : ℝ → ℝ) (T D2 : ℝ) : Prop where ...\n```\nHmm — simpler: lemma form: if (∀ x ∈ 𝓝 T, HasDerivAt φ (D1 x) x) and HasDerivAt D1 D2 T, then deriv (deriv φ) T = D2. Proof: deriv φ =ᶠ D1 (each point: h.deriv), so deriv (deriv φ) T = deriv D1 T = D2 ✓. \n\n```lean\ntheorem deriv2_of_hasDerivAt_nhds {φ D1 : ℝ → ℝ} {T D2 : ℝ}\n    (h1 : ∀ᶠ x in 𝓝 T, HasDerivAt φ (D1 x) x) (h2 : HasDerivAt D1 D2 T) :\n    deriv (deriv φ) T = D2 := by\n  have he : deriv φ =ᶠ[𝓝 T] D1 := by\n    filter_upwards [h1] with x hx using hx.deriv\n  rw [h2.deriv_congr_of_eventuallyEq?? ]\n```\n`HasDerivAt.deriv : HasDerivAt f f' x → deriv f x = f'`; and `Filter.EventuallyEq.deriv : f =ᶠ[𝓝 x] g → deriv f x = deriv g x`? The chain_rule proof used `hmul.congr_of_eventuallyEq hg1` — HasDerivAt.congr_of_eventuallyEq : HasDerivAt f f' x → f =ᶠ[𝓝 x] g → HasDerivAt g f' x — direction matters: hg1 : deriv(...) =ᶠ explicit; they proved hmul : HasDerivAt explicit ... then congr to get HasDerivAt (deriv ...) ✓. So here: h2 : HasDerivAt D1 D2 T; he : deriv φ =ᶠ D1; congr_of_eventuallyEq needs f =ᶠ g where f is the PROVED one (D1): he.symm : D1 =ᶠ deriv φ → HasDerivAt (deriv φ) D2 T → .deriv gives deriv (deriv φ) T = D2 ✓.\n\n```lean\ntheorem deriv2_of_hasDerivAt_nhds {φ D1 : ℝ → ℝ} {T D2 : ℝ}\n    (h1 : ∀ᶠ x in 𝓝 T, HasDerivAt φ (D1 x) x) (h2 : HasDerivAt D1 D2 T) :\n    deriv (deriv φ) T = D2 :=\n  (h2.congr_of_eventuallyEq (by filter_upwards [h1] with x hx using hx.deriv.symm)).deriv\n```\nHmm — EventuallyEq.symm: he : deriv φ =ᶠ D1 → he.symm : D1 =ᶠ deriv φ ✓. congr_of_eventuallyEq : HasDerivAt f f' x → f =ᶠ[𝓝 x] g → HasDerivAt g f' x — so h2.congr_of_eventuallyEq he.symm : HasDerivAt (deriv φ) D2 T ✓ .deriv ✓.\n\nWith that helper, per block: h1 via neighborhood HasDerivAt constructions, h2 via HasDerivAt of D1. Both from HasDerivAt combinators (`.log`, `.mul`, `.add`, `.inv`, `.const_mul`, `.div`...). Real.hasDerivAt_log : x ≠ 0 → HasDerivAt Real.log x⁻¹ x ✓.\n\nBlock list with D1, D2 (T variable, L1 := log(T+2), L2 := log(L1)):\n\nid: φ = id. D1 = 1. D2 = 0. ψ = 0 − 1/(T−1).\nL1: φ = log(T+2). D1 = (T+2)⁻¹. D2 = −(T+2)⁻².\nL2: φ = log(L1). D1 = (T+2)⁻¹·L1⁻¹ = 1/((T+2)L1). D2 = −[(L1+1)/((T+2)L1)²].\n   D2 computation: d/dT [(T+2)⁻¹ · L1⁻¹] = −(T+2)⁻²L1⁻¹ + (T+2)⁻¹·(−L1⁻²·(T+2)⁻¹) = −1/((T+2)²L1) − 1/((T+2)²L1²) = −(L1+1)/((T+2)²L1²) ✓ matches #11's first term (−(L1+1)/((T+2)L1)²) ✓.\nL1²: D1 = 2L1(T+2)⁻¹. D2 = 2(1−L1)/(T+2)².\n   d/dT [2L1·(T+2)⁻¹] = 2(T+2)⁻¹·(T+2)⁻¹ + 2L1·(−(T+2)⁻²) = 2/(T+2)² − 2L1/(T+2)² = 2(1−L1)/(T+2)² ✓.\nL1L2: D1 = L2(T+2)⁻¹ + L1·((T+2)L1)⁻¹ = L2/(T+2) + 1/(T+2) = (L2+1)/(T+2). D2: d/dT[(L2+1)·(T+2)⁻¹] = L2′·(T+2)⁻¹ + (L2+1)(−(T+2)⁻²) = 1/((T+2)L1)·(T+2)⁻¹ − (L2+1)/(T+2)² = 1/((T+2)²L1) − (L2+1)/(T+2)² = [1/L1 − L2 − 1]/(T+2)² ✓ matches #8.\nL2²: D1 = 2L2·((T+2)L1)⁻¹. D2 = d/dT[2L2·((T+2)L1)⁻¹]: = 2·L2′·((T+2)L1)⁻¹ + 2L2·(−((T+2)L1)⁻²·((T+2)L1)′); L2′ = 1/((T+2)L1); ((T+2)L1)′ = L1 + 1. So D2 = 2/((T+2)L1)² − 2L2(L1+1)/((T+2)L1)² = 2[1 − L2(L1+1)]/((T+2)L1)² ✓ #9.\nT·L1: φ = T·L1. D1 = L1 + T/(T+2). D2 = (T+2)⁻¹ + [(T+2) − T]/(T+2)² = 1/(T+2) + 2/(T+2)² = (T+4)/(T+2)² ✓ #5 first term.\nT·L2: φ = T·L2. D1 = L2 + T/((T+2)L1). D2 = L2′ + [((T+2)L1) − T(L1+1)]/((T+2)L1)² = 1/((T+2)L1) + [(T+2)L1 − TL1 − T]/((T+2)L1)² = 1/((T+2)L1) + [2L1 − T]/((T+2)L1)² = [(T+2)L1 + 2L1 − T]/((T+2)L1)² = [TL1 + 2L1 + 2L1 − T]/... = [T(L1−1) + 4L1]/((T+2)L1)² ✓ #6 first term.\n\nAll check out against D-057's table. \n\nNow — the second term of each ψ is −D1/(T−1) ✓ handled in assembly.\n\nContDiffOn ℝ 2 for each block on Ioi(−1): \n- id: contDiffOn_id ✓ (ContDiff).\n- T·L1, T·L2 etc: products. ContDiffOn.mul. L1 = log(T+2): ContDiffOn.log: `ContDiffOn.log (hf : ContDiffOn ℝ n f s) (hs : ∀ x ∈ s, f x ≠ 0) : ContDiffOn ℝ n (fun x => Real.log (f x)) s`? Let me check name — likely `ContDiffOn.log` in Mathlib/Analysis/SpecialFunctions/Log/Deriv.lean or ContDiff section. For T ∈ Ioi(−1): T+2 > 1 > 0 ≠ 0 ✓. L2 = log(L1): need L1 ≠ 0 on Ioi(−1): T+2 > 1 → L1 > 0 ✓.\n\nAssembly of the margin's second derivative: marginTilde F n — the def uses `let T := √n+1`. When differentiating, unfold marginTilde: write it as an explicit sum of functions of n:\n\nmarginTilde F = Sfun − [2·(F∘T) + 4·(fun n => Tn·((2/3)C0hat Tn + 10.5 log(Tn+2) + 48)) + const + 2(e⁴+1)·(7·(fun n => Hhat Tn) + 3.3) + IIMaj + 1] + 1\n\nThis is painful to decompose at the deriv level directly from the def. Better: prove an eventuallyEq (or plain eq) lemma rewriting marginTilde F n into a sum of block-composed functions with explicit weights:\n\nmarginTilde F n = S(n) − 2 F(Tn) − Σᵢ wᵢ φᵢ(Tn) − IIMaj n + C\n\nwith:\n- w·φ list:\n  φ₁ = F, w₁ = 2\n  φ₂ = T·L1 (T*(log(T+2))), w₂ = 4·((2/3)·0.433 + 10.5) = 4·10.788666... = 43.154666...: exact: (2/3)·0.433 = 0.288666..., +10.5 = 10.788666..., ×4 = 43.154666...\n  φ₃ = T·L2, w₃ = 4·(2/3)·0.886 = 2.362666...\n  φ₄ = id, w₄ = 4·((2/3)·2.9 + 48) = 4·49.9333... = 199.7333...\n  φ₅ = L1², w₅ = 2(e⁴+1)·7·3·0.433² = 42(e⁴+1)·0.187489 = 7.874538(e⁴+1)\n  φ₆ = L1·L2, w₆ = 42(e⁴+1)·2·0.433·0.886 = 42(e⁴+1)·0.767276 = 32.225592(e⁴+1)\n  φ₇ = L2², w₇ = 42(e⁴+1)·0.886² = 42(e⁴+1)·0.784996 = 32.969832(e⁴+1)\n  φ₈ = L1, w₈ = 42(e⁴+1)·2·0.433·2.9 + 2(e⁴+1)·7·10.5 = 42(e⁴+1)·2.5114 + 147(e⁴+1) = (105.4788 + 147)(e⁴+1) = 252.4788(e⁴+1)\n  φ₉ = L2, w₉ = 42(e⁴+1)·2·0.886·2.9 = 42(e⁴+1)·5.1388 = 215.8296(e⁴+1)\n  C (constants): −[40·1.3352 + 2(e⁴+1)·(7·(3·2.9² + 48) + 3.3) + 1] − 4 + 1 hmm wait: margin = S − B̃ + 1, S includes −4. Constants inside Hhat: 3·(2.9² from C0hat² expansion) + 48: C0hat² = (0.433L1+0.886L2+2.9)² → const part 2.9² = 8.41; Hhat const = 3·8.41 + 48 = 73.23. 2(e⁴+1)(7·73.23 + 3.3) = 2(e⁴+1)(512.61+3.3) = 2(e⁴+1)·515.91. Plus 40·1.3352 = 53.408. Plus B̃'s trailing +1 and outer +1 and S's −4. Total const = −4 − 53.408 − 1031.82(e⁴+1) − 1 + 1 — whatever, it's a constant K; second deriv kills it.\n\nFor the deriv assembly, the constant just needs to be identified as constant — don't compute its value! Structure the eventuallyEq as: marginTilde F =ᶠ (or =) fun n => S n − 2 F(Tn) − Σ wᵢφᵢ(Tn) − IIMaj n + K for SOME K — prove via ring after unfolding C0hat/Hhat/IIMaj defs... ring with log/exp atoms ✓ (it's pure algebra in atoms log(Tn+2), log log, exp, etc. — the weights are decimal literals; ring normalizes ✓). Let K be the existential: state as `∃ K, marginTilde F = fun n => ... + K`? Or just compute K explicitly as the same expression — simpler: state equality with the constant written out. Actually simplest: the assembly lemma states marginTilde F n = (explicit expression with constant inline) — just unfold everything and ring.\n\nThen: ContinuousOn on Ici 20: each piece continuous on n ≥ 20 (or even n > 0): S: n log n — continuous on Ici 20 ✓ (log continuous at positive: n ≥ 20 > 0 — ContinuousOn.log with ≠ 0... n ↦ log n on Ici 20: continuousOn of fun n => n * log n: ContinuousOn.mul continuousOn_id (continuousOn_log? Real.continuousOn_log : ContinuousOn Real.log {0}ᶜ ✓ comp)). F∘T: F ContDiff 2 → continuous; T continuous ✓. Blocks: products/sums of continuous. IIMaj: exp/sqrt/rpow: 2^(−(2√n−2)) : Real.rpow 2 x — continuous in x ✓ (Real.continuous_const_rpow? `Real.continuous_const_rpow : 0 < a → Continuous (a ^ ·)`? or rpow of nat? `2 ^ (-(2*√n−2))` with real exponent — `Real.rpow`: continuous when base > 0 ✓). Assemble ~40 lines.\n\nDifferentiableOn f and deriv f on interior (Ioi 20): each piece differentiable on Ioi 20 (needs n ≠ 0 ✓, log args ✓); deriv f differentiableOn: from the second-derivative FORMULAS (deriv f = explicit combination of differentiable functions on Ioi 20 ✓ — via the eventuallyEq-deriv technique per piece, or: since we compute deriv (deriv piece) n for every n > 20 via chain rules, we get DifferentiableAt (deriv piece) n as a byproduct? The chain-rule lemmas give deriv (deriv piece) n — which REQUIRES deriv piece differentiable at n implicitly? No — deriv (deriv f) n is defined via junk if not differentiable. Hmm — for hf''_nonneg we need 0 ≤ deriv (deriv f) n which we compute explicitly — the chain-rule applications already give the VALUES. For hf' and hf'' (DifferentiableOn of f and deriv f): \n- f differentiableOn Ioi 20: piecewise ✓.\n- deriv f differentiableOn Ioi 20: show deriv f =ᶠ/= explicit smooth function g on Ioi 20, and g differentiableOn: deriv f = g on Ioi 20 via deriv_add/deriv_sub at each point (needs each piece differentiable at each point of Ioi 20 ✓); then DifferentiableOn g via the explicit form ✓ and `DifferentiableOn.congr`.\n\nOK. The total assembly is substantial but mechanical. Given the volume, organize the remaining work as:\n\nS2a: helper `deriv2_of_hasDerivAt_nhds` + block derivative/ψ lemmas (8 blocks × (formula + sign)) — big; split into 2-3 commits.\nS2b: IIMaj (identity + chain_rule_sqrt application + bound).\nS2c: assembly + convexOn_of_deriv2_nonneg.\n\nStart S2a with the helper + block id/L1/L2 (prove formulas AND signs). \n\nFor block id: φ = id: chain_rule_sqrt_shift_on needs ContDiffOn ℝ 2 id (Ioi (−1)): `contDiffOn_id`? or (contDiff_id).contDiffOn ✓.\nψ-form: deriv (deriv id) T − deriv id T/(T−1) = 0 − 1/(T−1): deriv id T = 1 (deriv_id''), deriv (deriv id) T: deriv (fun _ => 1)?? deriv id = fun _ => 1 (deriv_id' : deriv id = 1? `deriv_id'' : deriv id x = 1`; and deriv (deriv id) T = deriv (fun _ => 1) T?? — need deriv id =ᶠ const 1: `deriv_id` : deriv id = fun x => 1 (as a function, `deriv_id'`?). There's `deriv_id'' (x) : deriv id x = 1` and `deriv_const`. Simplest: deriv (deriv id) T = 0 via: deriv id =ᶠ[𝓝 T] fun _ => 1 (eventually_of_forall fun _ => deriv_id'' _), then .deriv → deriv (fun _ => 1) T = deriv_const T 1 = 0... `Filter.EventuallyEq.deriv : f =ᶠ[𝓝 x] g → deriv f x = deriv g x` ✓ exists I believe.\n\nBlock L1 (φ T = log (T+2)):\nh1 : ∀ᶠ x in 𝓝 T, HasDerivAt (fun t => Real.log (t+2)) ((x+2)⁻¹) x: from Real.hasDerivAt_log (x+2 ≠ 0) .add_const?? HasDerivAt Real.log (x+2)⁻¹ (x+2) → comp with fun t => t+2: (Real.hasDerivAt_log h).add_const 2?? HasDerivAt.add_const : HasDerivAt f f' x → HasDerivAt (fun y => f (y + c)) f' x ✓ gives HasDerivAt (fun t => log (t+2)) (x+2)⁻¹ x ✓. Neighborhood: eventually (x+2 ≠ 0) near T (T > −1 → T+2 > 1 > 0: eventually_pos → ne) ✓.\nh2 : HasDerivAt (fun t => (t+2)⁻¹) (−(T+2)⁻¹^2) T: (hasDerivAt_id' T).add_const 2 → HasDerivAt (fun t => t+2) 1 T → .inv (T+2 ≠ 0) : HasDerivAt (fun t => (t+2)⁻¹) (−1/(T+2)²)... HasDerivAt.inv : HasDerivAt f f' x → f x ≠ 0 → HasDerivAt (fun x => (f x)⁻¹) (−f'/f x ^2) x ✓ gives −1/(T+2)² ✓ then massage = −(T+2)⁻¹^2 by ring.\n\nThen deriv2_of_hasDerivAt_nhds gives deriv (deriv φ) T = −(T+2)⁻². And deriv φ T = (T+2)⁻¹ from the h1's point-instance .deriv. ψ formula: −(T+2)⁻² − (T+2)⁻¹/(T−1) = −1/(T+2)² − 1/((T+2)(T−1)) — as a rw-able equation. Sign: for T ≥ T₀: both terms < 0: 1/(T+2)² > 0, 1/((T+2)(T−1)) > 0 ✓ nlinarith or direct.\n\nRather than a uniform ψ-lemma shape, per block state:\n\ntheorem blockXX_psi {T : ℝ} (hT : 1 < T) : deriv (deriv φ) T − deriv φ T / (T−1) = [closed form]\ntheorem blockXX_neg {T : ℝ} (hT : T₀ ≤ T) : [closed form] < 0 — or fold sign into one lemma: ... < 0 directly with the closed form.\n\nThen assembly uses chain_rule_sqrt_shift_on + block psi to evaluate.\n\nT₀ := √20 + 1. From hT : Real.sqrt 20 + 1 ≤ T derive: T ≥ 5.4721...? Need numeric: √20 ≥ 4.4721 (since 4.4721² = 19.99968... ≤ 20 ✓ norm_num-able? 4.4721² = 19.999675... as decimal: 44721² = 1,999,967,  hmm compute: 44721² = 44721·44721: 44000²=1.936e9... let me just trust: 4.4721² = 19.99968... < 20 ✓; so (4.4721 : ℝ)² ≤ 20 → 4.4721 ≤ √20 via Real.le_sqrt (h : 0 ≤ y... `Real.le_sqrt (hy : 0 ≤ y) : x ≤ sqrt y ↔ x² ≤ y`? or Real.le_sqrt' — check at use). Also L1 ≥ 2: T+2 ≥ √20+3 ≥ 7.4721 ≥ e²: e² = exp 2 ≤ 2.71829² (exp_one_lt_d9 : exp 1 < 2.7182818286? — the EulerMascheroni file used exp_one_lt_d9 : Real.exp 1 < 2.7182818286 ✓). 2.7182818286² = 7.389057... ≤ 7.4721 ✓ norm_num. So log(T+2) ≥ log(e²) = 2 via Real.log_le_log?? : 0 < x → x ≤ y → log x ≤ log y: name? `Real.log_le_log_iff`? or `Real.log_le_log (h₀ : 0 < x) (h : x ≤ y) : log x ≤ log y` ✓ I think exists. log (exp 2) = 2: Real.log_exp ✓.\nL2 ≥ log 2: L1 ≥ 2 → log L1 ≥ log 2 ✓ log_le_log. log 2 ≥ 0.6931: Real.log_two_gt_d9 : 0.6931471803 < log 2 ✓.\n\nOK. Time to write. Batch 1: helper + block id + block L1 + block L2 (formulas + signs). Given message size limits, write in one edit, then build, fix, commit.\n\nBlock sign lemma statements — use the margin's needed range: T ≥ √20+1. Statement shapes:\n\n```lean\ntheorem blockTL1_psi {T : ℝ} (hT : -1 < T) :\n    deriv (deriv fun t => t * Real.log (t + 2)) T -\n      deriv (fun t => t * Real.log (t + 2)) T / (T - 1) =\n      (T + 4) / (T + 2) ^ 2 - Real.log (T + 2) / (T - 1) - T / ((T + 2) * (T - 1)) := ...\n```\nHmm wait — the ψ in chain-rule form is φ″(T) − φ′(T)/√n where √n = T−1 — the formalized chain_rule_sqrt_shift writes `deriv φ (√n+1)/Real.sqrt n` — with T = √n+1, that's φ′(T)/(T−1) — but stated with Real.sqrt n not (T−1)! In assembly I'll have the ψ-expression as φ″(Tn) − φ′(Tn)/√n — and block psi lemmas stated with (T−1) need √n = T−1 bridge: rw [show Real.sqrt n = (Real.sqrt n + 1) − 1 by ring]. OK handle in assembly.\n\nPer-block statement — I'll parametrize by T and give the ψ-function value directly:\n\nblock_id_psi (T) (hT : 1 < T) : deriv (deriv id) T − deriv id T/(T−1) = −1/(T−1)\nblock_id_neg (T) (hT : 1 < T) : deriv (deriv id) T − deriv id T/(T−1) < 0 — combine? Just prove _psi then sign separately or inline. To save space: per block ONE lemma: `deriv (deriv φ) T − deriv φ T/(T−1) = closed ∧ ...`? Statements getting complicated. Alternative uniform packaging:\n\ndef psiOf (φ : ℝ → ℝ) (T : ℝ) : ℝ := deriv (deriv φ) T − deriv φ T / (T − 1)\n\nThen per block: `psiOf φX T = closedX T` (hT : −1 < T) and `closedX T < 0` (hT : √20+1 ≤ T). In assembly: chain_rule gives deriv(deriv (φ∘T)) n = (φ″(Tn) − φ′(Tn)/√n)/(4n); rewrite √n = Tn−1 → psiOf φ Tn / (4n); rw [block_psi] → closedX(Tn)/(4n).\n\nHmm, actually the chain-rule conclusion IS (φ″(Tn) − φ′(Tn)/√n)/(4n) — psiOf with (T−1) replaced by √n: equal after the ring rewrite ✓.\n\nDefine psiOf near the blocks. Sign lemmas per block with explicit closed forms. Let me write batch 1 now (helper + psiOf def + id + L1 + L2). \n\nFor L2 (log (log (t+2))):\nh1 : ∀ᶠ x in 𝓝 T, HasDerivAt (fun t => Real.log (Real.log (t+2))) (((x+2)*Real.log (x+2))⁻¹) x:\n Real.hasDerivAt_log (h : log(x+2) ≠ 0) : HasDerivAt Real.log (log(x+2))⁻¹ (log (x+2)) → comp with (fun t => log (t+2)) whose HasDerivAt is ((x+2)⁻¹) (from Real.hasDerivAt_log (x+2≠0)).add_const 2: HasDerivAt.comp x h_outer h_inner : HasDerivAt (g₂ ∘ g₁) (g₂' * g₁') x — gives HasDerivAt (fun t => log (log (t+2))) ((log(x+2))⁻¹ * (x+2)⁻¹) x — massage to (((x+2)*log(x+2))⁻¹): mul_inv_rev/ring.\n Neighborhood conditions: x near T > −1: x+2 > 1 (eventually, since T+2 > 1: continuous → eventually (x+2) > 1): so log(x+2) > 0 ≠ 0 ✓ and x+2 ≠ 0 ✓.\nh2 : HasDerivAt (fun t => ((t+2)*Real.log (t+2))⁻¹) (−((T+2)*L1)⁻¹² * ((T+2)*L1)′...) T: inner g(t) = (t+2)·log(t+2): HasDerivAt g (L1 + 1) T: (id+2).mul (log(t+2)): HasDerivAt.mul : HasDerivAt (f*g) (f'*g x + f x*g') — f = t+2: f' = 1; gives (1·L1 + (T+2)·(T+2)⁻¹) = L1 + 1 (field: (T+2)·(T+2)⁻¹ = 1 since ≠0 — need the simplification; HasDerivAt.mul gives literally 1 * L1 + (T+2) * (T+2)⁻¹; massage with have : (T+2)*(T+2)⁻¹ = 1 := mul_inv_cancel₀ h — then the stated D1g := L1+1 via convert/ring_nf). Then .inv (g T ≠ 0: (T+2)>0, L1>0 ✓): HasDerivAt g⁻¹ (−D1g/(g T)²) = −(L1+1)/((T+2)L1)² ✓.\n\nThen ψ_L2 = −(L1+1)/((T+2)L1)² − (((T+2)L1)⁻¹)/(T−1) = −(L1+1)/((T+2)L1)² − 1/((T+2)L1(T−1)) < 0: both terms negative (L1+1 > 0, denominators positive) ✓ nlinarith with pos facts or div_neg... straightforward: term1 < 0: −(L1+1)/pos² < 0 ✓; term2 < 0 ✓; sum < 0: linarith [h1neg, h2neg] where each via div_pos/neg_div... fine.\n\nWriting now. Keep h1/h2 as `have`s inside each psi lemma. Note deriv2_of_hasDerivAt_nhds helper first.\n\nFor id block: simpler: psiOf id T = 0 − 1/(T−1): deriv id T = 1: `deriv_id''`? and deriv (deriv id) T = 0: eventuallyEq (fun _ => 1): have : deriv id =ᶠ[𝓝 T] fun _ => (1:ℝ) := eventually_of_forall (fun _ => deriv_id'' _)... `deriv_id'' : deriv id x = 1`? name check: `deriv_id'` : deriv id = 1? In Mathlib: `deriv_id' : deriv id = fun x => 1`? and `deriv_id'' (x : 𝕜) : deriv id x = 1` ✓ exists I believe.\n\nOK write batch 1.","signature":""},{"type":"text","text":"Both compile. Committing infrastructure, then starting the block table."},{"type":"tool_use","id":"Bash_97","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: chain-rule infrastructure for marginTilde_convex — chain_rule_sqrt_shift_on (localized to Ioi (-1), for L1/L2-singular blocks) + chain_rule_sqrt (unshifted, for IIMaj)\" && git log --oneline -1","description":"Commit chain-rule infrastructure"},"extra_content":null}]