[{"type":"thinking","thinking":"Batch 2 compiles (L1², L1·L2, L2² with signs + helpers + bounds). Commit, then batch 3: T·L1 and T·L2 (the two product blocks with the more delicate sign arguments).\n\nBlock T·L1: φ = fun t => t * log (t+2).\n- D1 t = log(t+2) + t·(t+2)⁻¹: (hasDerivAt_id' t).mul (log_shift) : 1·L1 + t·(t+2)⁻¹ — massage to L1 + t*(t+2)⁻¹ (one_mul).\n- D2 = d/dT [L1 + T(T+2)⁻¹]: u = log(t+2) (u'=(T+2)⁻¹), v = t·(t+2)⁻¹: v' = (t+2)⁻¹ + t·(-(T+2)⁻²)... wait v = id·inv: v' = 1·(T+2)⁻¹ + T·(-1/(T+2)²) = (T+2)⁻¹ − T/(T+2)² = (T+2−T)/(T+2)² = 2/(T+2)². D2 = (T+2)⁻¹ + 2/(T+2)² = (T+4)/(T+2)² ✓ massage field_simp.\n- ψ = (T+4)/(T+2)² − (L1 + T(T+2)⁻¹)/(T−1) → closed form: (T+4)/(T+2)² − L1/(T−1) − T/((T+2)(T−1)): massage: (L1 + T(T+2)⁻¹)/(T−1) = L1/(T−1) + T/((T+2)(T−1)) ✓ field_simp.\n- Sign (hT : √20+1 ≤ T, L1 ≥ 2): ψ = (T+4)/(T+2)² − L1/(T−1) − T/((T+2)(T−1)). Per D-057: L1 ≥ 2 → L1/(T−1) ≥ 2/(T−1) → ψ ≤ (T+4)/(T+2)² − 2/(T−1) − T/((T+2)(T−1)) = [(T+4)(T−1) − 2(T+2)²... let me recompute with common denom (T+2)²(T−1): (T+4)(T−1) − 2(T+2)² − T(T+2) all over (T+2)²(T−1): numerator: (T²+3T−4) − 2(T²+4T+4) − (T²+2T) = T²+3T−4 −2T²−8T−8 −T²−2T = −2T²−7T−12 < 0 ✓ (D-057's). In Lean:\n  have : psiOf ... = [(T+4)(T−1) − 2(T+2)² − T(T+2)]/((T+2)²(T−1))?? — rather than rewrite ψ to that, do inequality:\n  ψ ≤ (T+4)/(T+2)² − 2/(T−1) − T/((T+2)(T−1)) [from L1 ≥ 2: L1/(T−1) ≥ 2/(T−1), so −L1/(T−1) ≤ −2/(T−1); div_le_div_of_le (T−1 > 0)? need L1/(T−1) ≥ 2/(T−1) ⟸ L1 ≥ 2 via div_le_div_of_nonneg... `div_le_div_right (hc : 0 < c) : a/c ≤ b/c ↔ a ≤ b` ✓ (div_le_div_right).mpr hL1-ish]\n  then show (T+4)/(T+2)² − 2/(T−1) − T/((T+2)(T−1)) < 0: combine: = [−2T²−7T−12]/((T+2)²(T−1)) via field_simp+ring (have hnum : ... := by field_simp; ring), and −2T²−7T−12 < 0 (nlinarith or: −(2T²+7T+12) < 0 since 2T²+7T+12 > 0), denom > 0 ✓ div_neg_of_neg_of_pos.\n  linarith to combine.\n\nBlock T·L2: φ = fun t => t * log (log (t+2)).\n- D1 t = L2 + t·((t+2)·L1)⁻¹... per-point: (id).mul (log_log): 1·L2 + t·((L1)⁻¹*(t+2)⁻¹): massage to L2 + t*((t+2)*L1)⁻¹.\n- D2 = d/dT [L2 + T·((T+2)L1)⁻¹]: u = log_log (u' = (L1)⁻¹(T+2)⁻¹), v = t·z(t) with z = ((t+2)L1)⁻¹: v' = 1·z + T·z' where z' = −(L1+1)/((T+2)L1)² (from hz_inner pattern). D2 = (L1)⁻¹(T+2)⁻¹ + ((T+2)L1)⁻¹ + T·(−(L1+1)/((T+2)L1)²) = [2·L1·... ]: let me verify against #6: D2 should be [T(L1−1)+4L1]/((T+2)L1)².\n  D2 = 1/((T+2)L1) + 1/((T+2)L1) − T(L1+1)/((T+2)L1)² = 2/((T+2)L1) − T(L1+1)/((T+2)L1)² = [2(T+2)L1 − T(L1+1)]/((T+2)²L1²)?? hmm: 2/((T+2)L1) = 2(T+2)L1/((T+2)²L1²). And target [T(L1−1)+4L1]/((T+2)L1)² = [T(L1−1)+4L1](T+2)/((T+2)²... no — target denominator is ((T+2)L1)² = (T+2)²L1² ✓: 2(T+2)L1 − T(L1+1) = 2TL1 + 4L1 − TL1 − T = TL1 + 4L1 − T = T(L1−1) + 4L1 ✓.\n- ψ = D2 − D1/(T−1) = [T(L1−1)+4L1]/((T+2)L1)² − L2/(T−1) − T/((T+2)L1(T−1)) ✓ (D1/(T−1) = L2/(T−1) + T((T+2)L1)⁻¹/(T−1) = L2/(T−1) + T/((T+2)L1(T−1)) ✓).\n- Sign: my reduction: ψ < 0 ⟸ A(T−1) ≤ T(T+2)L1 (with A = T(L1−1)+4L1) ⟸ L1(T−4) ≤ T(T−1) ⟸ L1 ≤ T+1 ≤ T(T−1)/(T−4). Let me recheck the full chain: ψ = A/D² − L2/(T−1) − T/(D(T−1)) where D = (T+2)L1. Want < 0 ⟺ A/D² < L2/(T−1) + T/(D(T−1)) ⟺ A(T−1) < L2D²(T−1)... multiply by D²(T−1) > 0: A(T−1) < L2D² + TD = L2(T+2)²L1² + T(T+2)L1. Suffices: A(T−1) ≤ T(T+2)L1 (since L2(T+2)²L1² ≥ 0). A(T−1) = [T(L1−1)+4L1](T−1). Claim: [T(L1−1)+4L1](T−1) ≤ T(T+2)L1 ⟺ (T−1)(TL1 + 4L1 − T) ≤ T(T+2)L1 ⟺ (T−1)L1(T+4) − T(T−1) ≤ (T²+2T)L1 ⟺ L1[(T−1)(T+4) − (T²+2T)] ≤ T(T−1) ⟺ L1[T²+3T−4−T²−2T] = L1(T−4) ≤ T(T−1) ✓. Since T ≥ 5.4721: T−4 > 0. L1(T−4) ≤ (T+1)(T−4)?? L1 ≤ T+1: log(T+2) ≤ T+1 ⟸ log x ≤ x−1 at x = T+2 ✓ Real.log_le_sub_one_of_pos. (T+1)(T−4) = T²−3T−4 ≤ T(T−1) = T²−T ⟺ −3T−4 ≤ −T ⟺ 2T ≥ −4 ✓. Chain: L1(T−4) ≤ (T+1)(T−4) ≤ T(T−1) ✓.\n  In Lean: needs (T−4) > 0, mul_le_mul_of_nonneg_right (L1 ≤ T+1) (T−4 ≥ 0) ✓.\n  Then assembly: A(T−1) ≤ T(T+2)L1: from L1(T−4) ≤ T(T−1): nlinarith [hprod]?? The claim A(T−1) ≤ T(T+2)L1 expands: (T(L1−1)+4L1)(T−1) ≤ (T²+2T)L1 — expand both: LHS = (TL1−T+4L1)(T−1) = T²L1−TL1−T²+T+4TL1−4L1 = T²L1+3TL1−T²+T−4L1; RHS = T²L1+2TL1. Difference RHS−LHS = −TL1+T²−T+4L1 = L1(4−T) + T(T−1) = T(T−1) − L1(T−4) ≥ 0 ✓ exactly the proved fact. nlinarith [hstep] with hstep : L1*(T−4) ≤ T*(T−1): expansion is quadratic in atoms {T, L1} — nlinarith should do: goal (T*(L1−1)+4*L1)*(T−1) ≤ T*(T+2)*L1: products of atoms — nlinarith with hstep ✓ (it can expand polynomial identities over atoms).\n  Then: A(T−1) ≤ TD ≤ TD + L2D²: need L2D² ≥ 0 ✓.\n  Then final: ψ = A/D² − L2/(T−1) − T/(D(T−1)): show < 0: common: multiply through: since D²(T−1) > 0: suffices A(T−1) − L2D² − TD < 0: from A(T−1) ≤ TD: A(T−1) − TD ≤ 0, and −L2D² < 0 → sum < 0 ✓. And ψ·[D²(T−1)] = A(T−1) − L2D² − TD (field identity). In Lean: have key : psiOf ... * (((T+2)*L1)^2 * (T−1)) = A*(T−1) − L2*((T+2)*L1)^2 − T*(T+2)*L1 := by rw [psi-formula]; field_simp [...]; ring. Then sign: ψ = key/[positive] < 0: nlinarith? psi = stuff → psi * P = N with P > 0, N < 0 → psi < 0: from N = ψ*P: ψ = N/P < 0: div_neg_of_neg_of_pos ✓.\n\nOK write batch 3. Also the hz_inner-style derivation for z' is reused from L2² block (the (t+2)*L1 inverse). Write both blocks now.","signature":""},{"type":"text","text":"Batch 2 compiles. Committing, then batch 3 (T·L1, T·L2 — the product blocks)."},{"type":"tool_use","id":"Bash_144","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: marginTilde_convex block table batches 1-2 — psiOf machinery (deriv2_of_hasDerivAt_nhds), blocks id/L1/L2/L1²/L1·L2/L2² with ψ-formulas + sign proofs, block_range_bounds (T≥5.4721, L1≥2, L2≥0.6931)\" && git log --oneline -1","description":"Commit block table batches 1-2"},"extra_content":null}]