[{"type":"thinking","thinking":"Continue now. Sub-stage 1: derivative infrastructure + easy blocks.\n\nPlan the architecture. The cleanest decomposition:\n\nmarginTilde F n = S(n) − (2·F(Tn) + G(n) + IIMaj n + const) + 1 where:\n- S(n) = ½n log n + cn − 4\n- G(n) = 4T·((2/3)C0hat T + 10.5 L1 + 48) + 40·1.3352 + 2(e⁴+1)(7 Hhat T + 3.3)\n- T(n) = √n + 1\n\nThe function is a sum; deriv of sums: `deriv_add` needs differentiability. Second deriv: messy but mechanical.\n\nKey structural decision: prove a general \"ψ-block\" lemma:\n\nFor φ : ℝ → ℝ twice differentiable, define P φ n := φ (√n + 1). Then (P φ)″ n = ψ[φ](Tn)/(4n) — this IS chain_rule_sqrt_shift (already proved!). \n\nSo write marginTilde F as: S(n) − 2·(F∘T)(n) − Σ blocks − IIMaj n + const, where each block is φᵢ∘T for explicit φᵢ. Then:\n\nderiv (deriv (marginTilde F)) n = 1/(2n) − 2·ψ[F](Tn)/(4n) − Σ wᵢψ[φᵢ](Tn)/(4n) − (IIMaj)″(n).\n\nTo use deriv_add/deriv_sub at the second level, need each piece's first deriv differentiable — ContDiff 2 for each block. For explicit blocks (polynomials in L1, L2, T), ContDiff ℝ ⊤ on the relevant domain (T+2 > 0, L1 > 0 i.e. T > e−2... T ≥ √20+1 > 0, T+2 > 2, log(T+2) > 0 ✓ so log log fine; and √n needs n > 0 ✓ domain n > 20 works; but deriv is global — need differentiability at the specific point n only, plus eventually for the second deriv... deriv (deriv f) n requires deriv f to be differentiable AT n — pointwise suffices for the VALUE? deriv (deriv f) n: deriv f is a global function (defined via Classical.choice where not differentiable); to compute deriv of it at n, need deriv f equals the \"nice\" expression in a NEIGHBORHOOD of n — the eventuallyEq technique used in chain_rule_sqrt_shift. So per block: ContDiff on a neighborhood of Tn (n > 20 ⟹ Tn > √20+1 > 5.4; T+2 > 7.4, log(T+2) > 2 — all smooth there ✓). Since the domains are open half-lines, use the eventuallyEq approach or ContDiffOn→convert.\n\nSimpler uniform approach: define each block φ on ALL of ℝ via the same formula (log of possibly negative/zero args gives junk values log|x|? Real.log x for x ≤ 0 is log of... Real.log x = log |x| for x ≠ 0? No — Real.log x for x < 0: Real.log is defined as Complex... Real.log x = Real.log |x| for x ≠ 0, and 0 at 0? Real.log (x) for x<0: Real.log x = Real.log |x|?? Actually Real.log x = (Complex.log x).re and Complex.log of negative real = log|x| + iπ, so Real.log x = log |x| for x < 0, and Real.log 0 = 0 (Complex.log 0 = 0 by convention... yes Real.log 0 = 0). So formulas are globally defined but smooth only on domains. The eventuallyEq technique handles this: work in 𝓝 n (n > 20) where everything is smooth.\n\nThis is going to be long. Strategy for sub-stage 1 (this session): \n1. Lemma `marginS_deriv2`: second derivative of S = 1/(2n). \n2. Block-wise ψ lemmas for the easy blocks #2, #3, #4, #10, #11 (pure logs) — actually all blocks are needed eventually; do infrastructure first.\n\nHmm — actually, let me reconsider the architecture to minimize deriv-chasing. Instead of computing the second derivative of the WHOLE marginTilde via deriv_add chains (needs each piece's deriv-differentiability + formula assembly), do:\n\nFor each component fᵢ (i = 1..k) with marginTilde F = Σ cᵢ fᵢ (cᵢ constants incl. negative), prove `deriv (deriv fᵢ) n = exprᵢ n` separately, then combine via deriv_add/deriv_sub/deriv_const_mul — each combination step needs Differentiability of the first derivs... `deriv_add (hf : DifferentiableAt ℝ f x) (hg : DifferentiableAt ℝ g x) : deriv (f+g) x = deriv f x + deriv g x` — for SECOND deriv: deriv (deriv (f+g)) n: need deriv (f+g) = deriv f + deriv g in a neighborhood (eventuallyEq) then deriv of sum of derivs: deriv (fun x => deriv f x + deriv g x) n = deriv (deriv f) n + deriv (deriv g) n via deriv_add with DifferentiableAt of (deriv f) and (deriv g) at n. So per piece: need DifferentiableAt ℝ (deriv fᵢ) n. Fine.\n\nPlan the component list for marginTilde F (n : ℝ), T = √n+1:\n\nmarginTilde F n = S(n) − 2·F(Tn) − B1(n) − B2(n) − B3(n) − IIMaj n + C\nwhere:\n- S(n) = (1/2)n log n + (C1−γ/2)n − 4 [no wait, the −4 is inside S in the def: ((1/2)*n*log n + (lagariasC1 − γ/2)*n − 4); constants don't matter for 2nd deriv]\n- B1(n) = 4·Tn·((2/3)·C0hat Tn + 10.5·L1(Tn) + 48) where C0hat T = 0.433 L1 + 0.886 L2 + 2.9, L1 = log(T+2), L2 = log(L1)\n- B2(n) = 2(e⁴+1)·(7·Hhat Tn + 3.3), Hhat T = 3·C0hat T² + 10.5 L1 + 48\n- B3(n) = 40·1.3352 + ... constants\n- IIMaj n\n\nDecompose into φᵢ∘T blocks (for chain_rule_sqrt_shift) — everything except S and IIMaj:\n- F-block: φ = F (weight 2)\n- 4T·(2/3)·0.433·L1 = (8/3)(0.433)·(T·L1) → block φ = T↦T·log(T+2), weight 4·(2/3)(0.433) = 1.154666...\n- 4T·(2/3)·0.886·L2 → block T·log log(T+2), weight 4·(2/3)(0.886) = 2.362666...\n- 4T·(2/3)·2.9 + 4T·48 → block T (identity), weight 4·((2/3)(2.9) + 48) = 4·49.9333 = 199.7333\n- 4T·10.5·L1 → block T·L1, weight 42\n- 2(e⁴+1)·7·3·C0hat² = 42(e⁴+1)·(0.433L1 + 0.886L2 + 2.9)²: expand: (0.433L1+0.886L2+2.9)² = 0.433²L1² + 0.886²L2² + 2.9² + 2·0.433·0.886 L1L2 + 2·0.433·2.9 L1 + 2·0.886·2.9 L2 → blocks L1², L2², const, L1·L2, L1, L2 with weights 42(e⁴+1)·{0.187489, 0.784996, 8.41, 0.767276, 2.5114, 5.1388}\n- 2(e⁴+1)·7·10.5·L1 = 147(e⁴+1)·L1 → block L1\n- constants: 2(e⁴+1)·(7·48 + 3.3), 40·1.3352, −4, +1\n\nSo block types: F, T·L1, T·L2, id(T), L1², L2², L1·L2, L1, L2 — 8 block types + S + IIMaj + F.\n\nψ-blocks (with T+2 substitution: note D-057's table uses L1 := log(T+2), L2 := log log(T+2) and the ψ's in the table are computed w.r.t. T):\n- id: ψ = −1/(T−1) < 0 ✓ (#2)\n- L1 = log(T+2): ψ = −1/(T+2)² − 1/((T+2)(T−1)) (#10) ✓\n- L2: ψ = −(L1+1)/((T+2)L1)² − 1/((T+2)L1(T−1)) (#11) ✓\n- L1²: ψ = 2(1−L1)/(T+2)² − 2L1/((T+2)(T−1)) (#7) ✓ (L1 > 1)\n- L1·L2: ψ = [1/L1 − L2 − 1]/(T+2)² − (L2+1)/((T+2)(T−1)) (#8) ✓\n- L2²: ψ = 2[1 − L2(L1+1)]/((T+2)L1)² − 2L2/((T+2)L1(T−1)) (#9) ✓\n- T·L1: ψ = (T+4)/(T+2)² − L1/(T−1) − T/((T+2)(T−1)) (#5)\n- T·L2: ψ = [T(L1−1)+4L1]/((T+2)L1)² − L2/(T−1) − T/((T+2)L1(T−1)) (#6)\n\nFor the second-derivative assembly, margiñ″(n) = 1/(2n) − [2ψ_F + Σ wᵢψᵢ](Tn)/(4n) − IIMaj″(n). Since ψ_F < 0 (hψ) and each ψᵢ < 0 with wᵢ > 0, the bracket is negative, so −[...]/(4n) > 0, giving margiñ″(n) ≥ 1/(2n) − IIMaj″(n) — need IIMaj″(n) ≤ 0?? No wait — D-057 §3: B̃₄ not concave; |B̃₄″| bounded: margiñ″ ≥ 1/(2n) − |B̃₄″| > 0.\n\nIIMaj(n) = 2(√n+1)(e^{√n/2}+1)·0.57·2^{−(2√n−2)}.\n\nSimplify: 2^{−(2√n−2)} = 4·2^{−2√n} = 4·e^{−2(ln2)√n}. So IIMaj(n) = 2·0.57·4·(√n+1)(e^{√n/2}+1)e^{−2(ln2)√n} = 4.56·(√n+1)·(e^{√n(1/2−2ln2)} + e^{−2(ln2)√n}).\n\nLet a := 2ln2 − 1/2 ≈ 0.8863 (>0), b := 2ln2 ≈ 1.3863. Then IIMaj(n) = 4.56·(√n+1)·(e^{−a√n} + e^{−b√n}).\n\nHmm wait: (e^{√n/2}+1)·e^{−2(ln2)√n} = e^{√n/2 − 2(ln2)√n} + e^{−2(ln2)√n} = e^{−a√n} + e^{−b√n} ✓.\n\nSo IIMaj(n) = 4.56·(√n+1)·(e^{−a√n} + e^{−b√n}). Much nicer! Second derivative: let s = √n. d/dn = (1/(2s))·d/ds. IIMaj = 4.56·(s+1)·(e^{−as}+e^{−bs}).\n\nd/ds: 4.56·[(e^{−as}+e^{−bs}) + (s+1)(−a e^{−as} − b e^{−bs})].\nd²/ds²: 4.56·[2(−a e^{−as} − b e^{−bs}) + (s+1)(a² e^{−as} + b² e^{−bs})].\n\nIIMaj″(n) = (1/(2s))²·(d²/ds²) + (d/ds)·(d/dn(1/(2s)))·... careful: d/dn f(s(n)) = f'(s)/(2s); d²/dn² = f″(s)/(4s²) − f'(s)/(4s³) = [f″(s) − f'(s)/s]/(4s²) — the same ψ-form with s = √n = T−1 ✓ consistent.\n\n|IIMaj″(n)| bound: with s ≥ √20 ≈ 4.472:\nf(s) = (s+1)(e^{−as}+e^{−bs}) ≤ (s+1)·2e^{−as} (a < b)\nf'(s) = (e^{−as}+e^{−bs}) − (s+1)(a e^{−as}+b e^{−bs}): |f'| ≤ 2e^{−as} + (s+1)·2b e^{−as} = 2e^{−as}(1 + b(s+1)).\nf″(s) = −2(a e^{−as}+b e^{−bs}) + (s+1)(a²e^{−as}+b²e^{−bs}): |f″| ≤ 4b e^{−as} + (s+1)·2b²e^{−as} = 2b e^{−as}(2 + b(s+1)).\n\nIIMaj″(n) = [f″(s) − f'(s)/s]/(4s²)·4.56. \n\n|IIMaj″(n)| ≤ 4.56/(4s²)·[2b e^{−as}(2 + b(s+1)) + 2e^{−as}(1+b(s+1))/s]\n= 4.56·2e^{−as}/(4s²)·[b(2+b(s+1)) + (1+b(s+1))/s].\n\nNeed: 1/(2n) = 1/(2s²) > |IIMaj″| ⟺ 1 > 2s²|IIMaj″|:\n2s²|IIMaj″| ≤ 4.56·e^{−as}·[b(2+b(s+1)) + (1+b(s+1))/s].\n\nAt s = √20 ≈ 4.4721: a s = 3.9637; e^{−3.9637} ≈ 0.01902; b(s+1) ≈ 1.3863·5.4721 ≈ 7.586; bracket ≈ 1.3863·(2+7.586) + (1+7.586)/4.4721 ≈ 13.28 + 1.92 ≈ 15.2. So 2s²|IIMaj″| ≤ 4.56·0.01902·15.2 ≈ 1.318 > 1. TOO LOOSE vs the measured 0.189. My crude bounds lose too much (the e^{−as}+e^{−bs} both counted at e^{−as}, signs ignored in f' — the actual f' is negative-dominated... hmm the measured max of 2n|B̃₄″| = 0.189 < 1 but my bound gives 1.32).\n\nBetter: keep e^{−as} + e^{−bs} structure: factor e^{−as}: e^{−as}(1 + e^{−(b−a)s}) with b−a = 1/2: 1 + e^{−s/2}. At s=4.47: 1 + 0.107 = 1.107. Sharper: replace 2 by 1.107. And in f', f″: the b-weighted term gets e^{−bs} ≈ 0.00203 vs e^{−as} ≈ 0.019 — ratio 0.107. Tighten:\n|f′| ≤ e^{−as}[(1 + a(s+1)) + 0.107·(1 + b(s+1))] — hmm wait f'(s) = (e^{−as}+e^{−bs}) − (s+1)(a e^{−as}+b e^{−bs}): the first group positive ~e^{−as}(1.107), second group negative — |f'| ≤ max? |f'| ≤ (s+1)(a e^{−as}+b e^{−bs}) dominates: ≈ (s+1)·e^{−as}(a + b·0.107) ≈ 5.4721·0.01902·(0.8863+0.1484) ≈ 0.1076. And f″(s) = −2(a e^{−as}+b e^{−bs}) + (s+1)(a²e^{−as}+b²e^{−bs}): the second term ≈ 5.4721·0.01902·(0.7855+0.2057) ≈ 0.1031; first term ≈ −2·0.01902·(0.8863+0.1484) ≈ −0.0393. f″ ≈ 0.0638 (positive). IIMaj″(n) = 4.56·[f″ − f'/s]/(4s²): f'/s: f' ≈ ... f' is negative? f'(s) = (e^{−as}+e^{−bs}) − (s+1)(a e^{−as} + b e^{−bs}) ≈ 0.0210 − 0.1076 ≈ −0.0866. So −f'/s ≈ +0.0194. f″ − f'/s ≈ 0.0638 + 0.0194 ≈ 0.0832. IIMaj″ ≈ 4.56·0.0832/(4·20) ≈ 0.00474. 2n·IIMaj″ ≈ 0.19 ✓ matches the measured 0.189. So IIMaj″ > 0 actually (it's CONVEX here?) — D-057 says \"not concave\" — right, B̃₄″ > 0 means margiñ″ = 1/(2n) − ψ-sum/(4n) − B̃₄″ — with B̃₄″ ≈ +0.0047 < 1/(2n) = 0.025 ✓ fine.\n\nFor the Lean proof: bound IIMaj″(n) ≤ 4.56·[f″_upper − f'_lower/s]/(4s²)... since we need an UPPER bound on IIMaj″ (the margin has −IIMaj″ and we need margiñ″ ≥ 1/(2n) − IIMaj″ — if IIMaj″ could be negative, even better; but need the inequality 1/(2n) − IIMaj″ > 0 ⟺ IIMaj″ < 1/(2n). So an upper bound suffices:\n\nIIMaj″(n) = 4.56[f″(s) − f'(s)/s]/(4s²).\nUpper bound f″: f″(s) = −2(a e^{−as} + b e^{−bs}) + (s+1)(a² e^{−as} + b² e^{−bs}) ≤ (s+1)(a² e^{−as} + b² e^{−bs}).\nUpper bound −f'/s: −f'(s) = −(e^{−as}+e^{−bs}) + (s+1)(a e^{−as}+b e^{−bs}) ≤ (s+1)(a e^{−as}+b e^{−bs}).\nSo IIMaj″(n) ≤ 4.56/(4s²)·[(s+1)(a²e^{−as}+b²e^{−bs}) + (s+1)(a e^{−as}+b e^{−bs})/s]\n= 4.56(s+1)/(4s²)·[(a² + a/s)e^{−as} + (b² + b/s)e^{−bs}]\n≤ 4.56(s+1)/(4s²)·(a² + a/s)·(e^{−as} + e^{−bs})·(b²+b/s)/(a²+a/s) hmm — since b > a: (b²+b/s) ≥ (a²+a/s), factor:\n≤ 4.56(s+1)(b²+b/s)/(4s²)·(e^{−as}+e^{−bs}).\n\nNeed < 1/(2s²) ⟺ 4.56(s+1)(b²+b/s)(e^{−as}+e^{−bs})/2 < 1 ⟺ 2.28(s+1)(b²s+b)/s·(e^{−as}+e^{−bs}) < 1.\n\nAt s=4.4721: 2.28·5.4721·(1.9218·4.4721+1.3863)/4.4721·0.02103 = 2.28·5.4721·(9.9796)/4.4721·0.02103 ≈ 2.28·5.4721·2.2315·0.02103 ≈ 0.5857 < 1 ✓. And the LHS is decreasing in s eventually (exponential beats polynomial) — need a rigorous \"decreasing on [√20,∞)\" or bound pieces: (s+1)/s ≤ 1.225 (decreasing ✓), (b²s+b) increasing but e^{−as} decays: show h(s) := (b²s+b)e^{−as} decreasing for s ≥ √20: log-derivative: b²/(b²s+b) − a ≤ b²/(b²s) − a = 1/s − a < 0 for s > 1/a ≈ 1.128 ✓ decreasing. And e^{−(b−a)s} decreasing → e^{−as}+e^{−bs} = e^{−as}(1+e^{−s/2}) ≤ e^{−as}·(1+e^{−√20/2}). So the whole LHS ≤ 2.28·[(s+1)/s]·[(b²s+b)e^{−as}]·[1+e^{−√20/2}] ≤ (evaluate each at s=√20 via monotonicity) — all three factors positive and decreasing ⟹ LHS(s) ≤ LHS(√20) ≈ 0.586 < 1 for all s ≥ √20 ✓ rigorous with three monotonicity lemmas!\n\nFactor monotonicity:\n- u₁(s) = (s+1)/s = 1 + 1/s: decreasing ✓ trivial.\n- u₂(s) = (b²s+b)e^{−as}: log-deriv argument — needs derivatives; or ratio argument: u₂(s+t)/u₂(s) ≤ ... for Lean, differentiable decreasing proofs via deriv < 0: `deriv_neg_...`? There are lemmas: `strictAntiOn_of_deriv_neg` (for intervals) ✓.\n- u₃(s) = 1 + e^{−s/2}: decreasing ✓.\n\nBut hold on — u₂ ≤ u₂(√20): (b²s+b)e^{−as} at √20: (1.9218·4.4721+1.3863)·0.01902 ≈ 9.9796·0.01902 ≈ 0.1898. u₁ ≤ 1.2236; u₃ ≤ 1.1068. Product: 2.28·1.2236·0.1898·1.1068 ≈ 0.586 ✓.\n\nOK so the B̃₄ part is doable analytically with clean monotonicity pieces. The numerics: e^{−a√20} — need an UPPER bound on e^{−3.9637}: e^{−3.9637} ≤ 0.0191? For Lean, proving e^{−a√20} ≤ r for explicit rational r: a√20 = (2ln2 − 1/2)√20 = 2√20 ln2 − √20/2 = √20(2ln2 − 1/2)... need bounds: ln 2 ≤ 0.6932 (provable: Real.log 2 ≤ 0.6932 via log_le_sub_one or exp bounds — Mathlib has `Real.log_two_lt_d9`? There's `Real.log_two_gt_d9 : 0.6931471803 < log 2` and lt version? I recall log_two_gt_d9 exists; upper: from exp: 2 < e^{0.6932}? provable via sum_le_exp... hmm — alternatively bound e^{−a√20} using exp_add_one_le... The standard trick: e^x ≤ 1/(1−x) for x<1? or e^{−t} ≤ 1/(1+t) ... For t ≥ 0: e^t ≥ 1 + t so e^{−t} ≤ 1/(1+t). Even sharper: e^t ≥ 1 + t + t²/2 → e^{−t} ≤ 1/(1+t+t²/2). With t = a√20 ≈ 3.9637: 1/(1+3.9637+7.855) ≈ 1/12.82 ≈ 0.078 — too loose vs 0.019. Use higher truncations: e^t ≥ Σ_{k≤6} tᵏ/k! → e^{−t} ≤ 1/Σ. t³/6≈10.38, t⁴/24≈10.29, t⁵/120≈8.16, t⁶/720≈5.39: Σ ≈ 1+3.9637+7.855+10.38+10.29+8.16+5.39 ≈ 47.03 → e^{−t} ≤ 0.02127 ✓ close enough? We need total ≤ 1: product with 0.02127: 2.28·1.2236·(9.9796·0.02127)·1.1068 ≈ 0.655 < 1 ✓ still fine.\n\nBut t itself = a·√20 with a = 2ln2 − 1/2: need t ≥ t₀ for the inequality e^{−t} ≤ 1/Σ(t)... careful: e^{−t} ≤ 1/Σ₆(t) requires t ≥ 0 — and to get a numeric bound, need t ≥ t₀ ⟹ e^{−t} ≤ e^{−t₀} ≤ 1/Σ₆(t₀)?? e^{−t} ≤ e^{−t₀} ✓ (t ≥ t₀). And t ≥ t₀: t = √20·(2ln2−1/2): need ln2 ≥ 0.6931 (Real.log_two_gt_d9 ✓ exists) and √20 ≥ 4.4721 (Real.sqrt bounds: 4.4721² ≤ 20 — norm_num can verify 4.4721² = 19.999... let me check: 4.4721² = 19.99968... ✓ ≤ 20). So a ≥ 2(0.6931) − 0.5 = 0.8862; t ≥ 4.4721·0.8862 ≈ 3.9632. Use t₀ = 3.96. Σ₆(3.96): 1+3.96+7.8408+10.3499+10.2419+8.1125+5.3566 ≈ 46.86 → e^{−t} ≤ e^{−3.96} ≤ 1/46.86 ≈ 0.021341 ✓.\n\nHmm wait — direction: e^{−t} ≤ e^{−t₀} since −t ≤ −t₀ ✓; e^{−t₀} ≤ 1/Σ₆(t₀) since e^{t₀} ≥ Σ₆(t₀) (Real.sum_le_exp_of_nonneg or `Real.add_one_le_exp`... Mathlib: `Real.sum_le_exp_of_nonneg (hx : 0 ≤ x) (n : ℕ) : ∑ i ∈ range n, x^i / i! ≤ exp x`? The EulerMascheroni file used `Real.sum_le_exp_of_nonneg (by norm_num) 7` ✓ exists!). Then e^{−t₀} = 1/e^{t₀} ≤ 1/Σ₆(t₀) ✓.\n\nThis is all doable but it's a LOT. Total realistic estimate stands at ~800-1200 lines. The session continues; work in committable pieces. Since pieces only become valuable when the main theorem is done, but committing helper lemmas progressively is still crash-safe (they're proved, tagged, don't break the build).\n\nARCHITECTURE — do it as one new file? LiSkeleton is getting big. Create `NguyenRH1/MarginConvexity.lean`? The package lib glob includes all lean files under NguyenRH1/ presumably (Lean lib roots: `NguyenRH1` lib with globs — default glob is the lib name module + submodules? lakefile [[lean_lib]] name = \"NguyenRH1\" — default globs: `NguyenRH1` and `NguyenRH1.*`? Default is `.andSubmodules`. I think default glob for lean_lib is the module itself only... Actually Lake default: `globs := #[Glob.one name]`?? Hmm — but the package built NguyenRH1.LiSkeleton etc, so submodules ARE included (default is .andSubmodules for lean_lib? evidence: build compiled LiSkeleton/SuzukiSkeleton/ValleySkeleton ✓). So I can add MarginConvexity.lean and it will be built. Need also to import it in NguyenRH1.lean (the root) for PrintAxioms visibility — check root file.\n\nActually simpler: keep everything in LiSkeleton.lean (it's the same paper §3.1 content). File length is fine (~600 lines now). I'll add to LiSkeleton.lean.\n\nOrder of work for sub-stage 1 (now):\n1. ψ-block SECOND-DERIVATIVE lemmas for the 8 explicit block types: each: `deriv (deriv (fun n => φᵢ (√n+1))) n = ψᵢ(Tn)/(4n)` — via chain_rule_sqrt_shift with φᵢ proved ContDiff ℝ 2... chain_rule_sqrt_shift needs ContDiff ℝ 2 φ GLOBALLY — problem: log(T+2) etc are not globally smooth (log at negative/zero args is junk log|x| — actually Real.log IS smooth away from 0; log(T+2) as a function on ℝ is smooth except T = −2; L2 = log(log(T+2)) smooth except T+2 ≤ 1 i.e. T ≤ e−2... singularities at T = −2 (log 0 = 0 junk, not differentiable) and where log(T+2) = 0 (T = −1). So global ContDiff 2 fails. chain_rule_sqrt_shift as stated needs global ContDiff.\n\nOptions: (a) generalize chain_rule_sqrt_shift to a local version (ContDiffOn on a neighborhood of Tn, or twice-differentiable at points near Tn); (b) define blocks globally with junk but prove an eventuallyEq to a globally-smooth function near Tn... (a) is cleaner: prove `chain_rule_sqrt_shift_at` : if φ is C² near Tn = √n+1 (n > 0), then deriv(deriv (φ∘T)) n = ψ/(4n). The existing proof structure mostly localizes (it uses hg1 eventuallyEq already); the global hφ2 usage: Differentiable ℝ (deriv φ) everywhere — replace with DifferentiableAt at points near Tn... Specifically needs: φ differentiable near Tn (for hg1 at m near n: need hφ1 at √m+1 for m near n — from neighborhood differentiability); deriv φ differentiable at Tn (for hA). So hypothesis: ContDiffOn ℝ 2 φ (Set.Ioi t₀) with √n+1 > t₀... or simpler: (hφ : ContDiffOn ℝ 2 φ (Set.Ioi (-2))) — for L1 blocks singular at −2 only... but L2 blocks singular at −1 too, T·L2 etc. Use (Set.Ioi (-1))?? L2 = log(log(T+2)): singular when T+2 ≤ 1 ⟺ T ≤ −1, so smooth on Ioi (−1)?? log(T+2) at T ∈ (−1, ∞): T+2 > 1, log(T+2) > 0 ✓ smooth ✓. L1 blocks smooth on Ioi (−2) ⊃ Ioi (−1) ✓. id/T-blocks smooth everywhere ✓. So uniform domain: Set.Ioi (−1) works for all blocks (Tn ≥ √20+1 > −1 ✓ and a neighborhood: for n > 20, m near n has √m+1 > −1 ✓).\n\nSo prove: `chain_rule_sqrt_shift_on (φ : ℝ → ℝ) (hφ : ContDiffOn ℝ 2 φ (Set.Ioi (-1))) (n : ℝ) (hn : 0 < n) : deriv (deriv fun n => φ (√n+1)) n = (deriv (deriv φ) (√n+1) − deriv φ (√n+1)/√n)/(4n)`. \n\nWait — but for the FINAL theorem, F is globally C² (hF : ContDiff ℝ 2 F) — fine, restrict. \n\nHmm, ContDiffOn for the explicit blocks: proving ContDiffOn ℝ 2 (fun T => T * Real.log (T+2)) (Ioi (−1)): via ContDiffOn.mul, contDiffOn_log? Mathlib: `Real.contDiffOn_log : ContDiffOn ℝ ⊤ Real.log {0}ᶜ`? There's `ContDiffOn.log : ContDiffOn ℝ n f s → (∀ x ∈ s, f x ≠ 0) → ContDiffOn ℝ n (fun x => Real.log (f x)) s` — `ContDiffOn.log` exists I believe. So build each block's ContDiffOn by combinators. ~10-20 lines each.\n\n2. ψ-formula lemmas: deriv φᵢ and deriv (deriv φᵢ) at T (for T > −1) with the closed forms — deriv computations via deriv_mul/deriv_const_mul/log deriv lemmas: `Real.deriv_log : deriv Real.log x = x⁻¹` (x ≠ 0); `Real.deriv_log'`? For compositions: deriv (fun T => Real.log (T+2)) T = 1/(T+2) via deriv_comp/deriv_add_const. Second: −1/(T+2)² etc. Each block ~20-40 lines.\n\n3. ψ-sign lemmas: for T ≥ T₀ = √20+1: ψᵢ(T) < 0. Using bounds: L1 ≥ log(√20+3) ≥ 2.011?? — wait: L1 = log(T+2), T ≥ √20+1 ⟹ T+2 ≥ √20+3 ≈ 7.4721, L1 ≥ log 7.4721 ≈ 2.0110. Need Real.log lower bound at a point: log 7.4721 ≥ 2? ⟺ 7.4721 ≥ e² ⟸ e² ≤ 7.4721: e² ≤ (2.7183)² ≈ 7.389... hmm e ≈ 2.71828, e² ≈ 7.389 ≤ 7.4721 ✓: prove e ≤ 2.71829 (Real.exp_one_lt_d9 ✓ exists — used in EulerMascheroni file!) then e² ≤ 2.71829² = 7.3891... ≤ 7.4721 ≤ T+2 → 2 = log(e²) ≤ log(T+2) via Real.log_le_log / exp_le_exp... `Real.log_le_log (hx : 0 < x) (h : x ≤ y) : log x ≤ log y`? exists I think. And L2 = log L1 ≥ log 2 ≥ 0.6931 (log_two_gt_d9 ✓) — D-057 uses L2 ≥ 0.6983 — we have L2 ≥ log 2.011 ≥ log 2 ≥ 0.693 ✓ (the sign arguments have slack).\n\nEach sign proof: manipulate the ψ closed form into a manifestly-negative combination using L1 ≥ 2, L2 ≥ 0.693, T ≥ 5.47 — nlinarith with provided facts, or explicit algebraic reductions like D-057's. ~20-40 lines each, 8 blocks.\n\n4. IIMaj: rewrite as 4.56(s+1)(e^{−as}+e^{−bs}) — wait, the formalized IIMaj is `2 * (√n + 1) * (exp (√n/2) + 1) * 0.57 * 2^(−(2√n−2))`. Constants: 2·0.57·2^{−(2√n−2)}: 2^{−(2√n−2)} = 2²·2^{−2√n} = 4·2^{−2√n}... 2^(-(2√n−2)) = 2^(2−2√n) = 4·2^{−2√n} ✓ (rpow of real exponent: 2^(−(2√n−2)) = 2^(2 − 2√n) ✓). So IIMaj = 2·0.57·4·(√n+1)(e^{√n/2}+1)·2^{−2√n} = 4.56·(s+1)(e^{s/2}+1)·e^{−2(ln2)s} = 4.56(s+1)(e^{−as}+e^{−bs}) ✓ with a = 2ln2−1/2, b = 2ln2.\n\nFormalizing this identity + second derivative + bound: the second derivative of e^{−a√n} compositions — deriv of exp ∘ (·) with sqrt — doable via chain_rule_sqrt_shift_on too! e^{−a·T̃} where T̃ = √n... hmm IIMaj as function of s = √n directly (not s+1): IIMaj = 4.56(s+1)(e^{−as}+e^{−bs}) with s = √n: define ψ̃(s) = 4.56(s+1)(e^{−as}+e^{−bs}) and IIMaj n = ψ̃(√n). Chain rule for √n (not √n+1): d²/dn² ψ̃(√n) = [ψ̃″(s) − ψ̃′(s)/s]/(4n). Need a second chain-rule variant (T = √n, no +1): `chain_rule_sqrt (ψ : ℝ→ℝ) (C² near √n) : deriv(deriv (ψ∘√)) n = [ψ″(√n) − ψ′(√n)/√n]/(4n)`. Fine — another infrastructure lemma.\n\nThen bound the bracket: ψ̃″(s) − ψ̃′(s)/s ≤ [computed above] ≤ ... and the three-factor monotonicity bound. ~150-250 lines.\n\n5. Assembly: marginTilde F = S + Σ — deriv rules + the final inequality + convexOn_of_deriv2_nonneg. Check the exact Mathlib lemma name/signature for convexity from nonneg second derivative. There's `convexOn_of_deriv2_nonneg` in Mathlib.Analysis.Convex.Deriv? I recall: `convexOn_of_deriv2_nonneg (hfc : Convex ℝ s) (hf : ContinuousOn f s) (hf' : DifferentiableOn ℝ f (interior s)) (hf'' : ∀ x ∈ interior s, 0 ≤ deriv (deriv f) x) : ConvexOn ℝ s f`? hmm need to check — maybe it's via monotonicity of deriv: `convexOn_of_deriv_mono`... Let me check when I get there.\n\nAlso ContinuousOn (marginTilde F) (Ici 20): marginTilde is continuous on n > 0 domain... at n = 20 endpoint: √20 fine, all functions continuous at 20 ✓ (sqrt continuous at 20 > 0, log fine, exp fine). ContinuousOn on Ici 20 via continuity of each piece (need Real.log arguments nonzero: n ≥ 20, T+2 > 0 ✓; 2^(...) fine). ~30 lines.\n\nOK — this is the plan. Get going. Order:\nS1: chain_rule_sqrt_shift_on (local version) + chain_rule_sqrt_on (√n version).\nS2: block ContDiffOn + ψ formula + sign for the 8 blocks (do #2 id, #10 L1, #11 L2 first — simplest).\nS3: IIMaj identity + bound.\nS4: assembly.\n\nLet me check NguyenRH1.lean root first, then start S1. Actually — keep in LiSkeleton.lean; no root change needed.\n\nS1: prove `chain_rule_sqrt_shift_on`. Model on the existing proof, replacing global differentiability with ContDiffOn on Ioi (−1). Details:\n\n```lean\ntheorem chain_rule_sqrt_shift_on (φ : ℝ → ℝ) (hφ : ContDiffOn ℝ 2 φ (Set.Ioi (-1)))\n    (n : ℝ) (hn : 0 < n) :\n    deriv (deriv fun n => φ (Real.sqrt n + 1)) n =\n      (deriv (deriv φ) (Real.sqrt n + 1) -\n        deriv φ (Real.sqrt n + 1) / Real.sqrt n) / (4 * n) := by\n```\nProof sketch (mirroring the existing one):\n- Tn = √n + 1 > 1 > −1 ✓; for m near n (m > 0), √m+1 > 1 ∈ Ioi (−1) ✓ — need a neighborhood of n mapping into Ioi(−1): m > 0 suffices (T_m > 1 > −1 ✓) ✓ so Ioi 0 neighborhood works.\n- From ContDiffOn: φ differentiable on Ioi(−1): hφ.differentiableOn (by norm_num)?? ContDiffOn.differentiableOn : ContDiffOn ℝ n f s → 1 ≤ n → DifferentiableOn ℝ f s ✓. Then DifferentiableAt at interior points: DifferentiableOn.differentiableAt (h : DifferentiableOn ℝ f s) (hx : s ∈ 𝓝 x) → DifferentiableAt ℝ f x ✓. Ioi(−1) ∈ 𝓝 (Tn) since Tn > −1 (Ioi_mem_nhds).\n- deriv φ differentiable at Tn: ContDiffOn.deriv'? For ContDiffOn, getting differentiability of deriv φ at an interior point: `ContDiffOn.differentiableOn_deriv`? Hmm. Existing proof used hφ.deriv' (n := 1) : ContDiff ℝ 1 (deriv φ) — global version. For On: there's `ContDiffOn.deriv` (for functions on normed spaces with unique differentiability)... Simpler: use the fact that ContDiffOn ℝ 2 on an OPEN set gives ContDiffOn ℝ 1 of the derivative within the set: `ContDiffOn.deriv (hf : ContDiffOn ℝ n f s) (hs : IsOpen s) : ContDiffOn ℝ (n-1) (deriv f) s`? I believe `ContDiffOn.deriv` exists with IsOpen hypothesis (or uniqueDiffOn). Then DifferentiableOn at level 0... ContDiffOn ℝ 1 (deriv φ) s → differentiableOn (1 ≤ 1) ✓ → differentiableAt at interior pts ✓.\n\nThe rest mirrors: hsqrt at n and at nearby m (m > 0 ✓); hg1 eventuallyEq proof uses hφ1 at √m+1 — need DifferentiableAt φ (√m+1) for m ∈ Ioi 0: from differentiableOn + neighborhood ✓.\n\nhA: HasDerivAt (fun n => deriv φ (√n+1)) ... : needs DifferentiableAt ℝ (deriv φ) (√n+1) ✓ from above.\n\nEverything else identical. OK.\n\nAnd `chain_rule_sqrt_on` for the IIMaj part: same with T = √n (no +1): deriv(deriv (φ∘√)) n = (φ″(√n) − φ′(√n)/√n)/(4n). Domain: φ C² on Ioi 0 (since √n > 0 for n > 0; exp-blocks globally smooth anyway — for IIMaj blocks (polynomial × exp of linear in s) are globally C^∞ — just use global ContDiff version! ψ̃(s) = 4.56(s+1)(e^{−as}+e^{−bs}) is smooth on ALL of ℝ (polynomial + exp) ✓ so chain_rule_sqrt needs only a GLOBAL version with T = √n:\n\n`chain_rule_sqrt (ψ : ℝ → ℝ) (hψ : ContDiff ℝ 2 ψ) (n : ℝ) (hn : 0 < n) : deriv (deriv fun n => ψ (Real.sqrt n)) n = (deriv (deriv ψ) (√n) − deriv ψ (√n)/√n)/(4n)`.\n\nNearly identical proof to chain_rule_sqrt_shift (copy, drop the +1). \n\nLet me write S1 now: two lemmas. First check ContDiffOn API names:\n- `ContDiffOn.differentiableOn : ContDiffOn ℝ n f s → 1 ≤ n → DifferentiableOn ℝ f s`? Signature check needed.\n- ContDiffOn.deriv name — check.\n- `ContDiff.contDiffOn : ContDiff ℝ n f → ContDiffOn ℝ n f s` ✓ exists (for F).\n\nLet me grep.","signature":""},{"type":"text","text":"Continuing. Sub-stage 1 architecture (from D-057):\n\n- **8 ψ-block types** (id, L1, L2, L1², L1·L2, L2², T·L1, T·L2) — each needs: ContDiffOn on `Ioi (−1)`, a second-derivative formula via a **localized** chain-rule lemma, and a sign proof\n- **IIMaj** rewrites as `4.56(s+1)(e^{−as}+e^{−bs})` with `s = √n`, `a = 2ln2 − 1/2`, `b = 2ln2` — globally smooth, needs a `T = √n` chain-rule variant + a 3-factor monotonicity bound (all factors decreasing, value at √20 ≈ 0.586 < 1)\n- Assembly via `convexOn_of_deriv2_nonneg`\n\nFirst, pin down the ContDiffOn API names."},{"type":"tool_use","id":"Bash_86","name":"Bash","input":{"command":"grep -rn \"theorem ContDiffOn.differentiableOn\\|protected theorem ContDiffOn.differentiableOn\" /d/Python/Quantum/zeta-23-lean/.lake/packages/mathlib/Mathlib/Analysis/Calculus/ContDiff/Defs.lean | head -3; grep -rn \"theorem ContDiffOn.deriv\\b\\|ContDiffOn.deriv \" /d/Python/Quantum/zeta-23-lean/.lake/packages/mathlib/Mathlib/Analysis/Calculus/ContDiff/*.lean | head -5; grep -rn \"convexOn_of_deriv2_nonneg\" /d/Python/Quantum/zeta-23-lean/.lake/packages/mathlib/Mathlib/Analysis/Convex/Deriv.lean | head -5","description":"Find ContDiffOn API and convexity lemma"},"extra_content":null}]