[{"type":"thinking","thinking":"All confirmed:\n- `HolomorphicOn_riemannZeta : HolomorphicOn ζ {s | s ≠ 1}` (Zeta23) ✓\n- `integrable_log_over_pow (rneg : r < 0) (Npos) : IntegrableOn (fun x => ‖x^(r−1)‖·‖log x‖) (Ioi N)` ✓\n- `MeasureTheory.intervalIntegral_tendsto_integral_Ioi` ✓ (the N→∞ limit)\n- Namespaces: the grep didn't show namespace lines — ZetaBounds.lean has no namespace? The lemmas are used unqualified within Zeta23 files with opens... `ZetaUpperBnd` was referenced as `Zeta23.FromPNTPlus...`? Earlier grep showed usage `Zeta0EqZeta Npos ...` inside ZetaBounds.lean itself. The namespace question: let me just check what namespace these live in — actually from the earlier grep of MediumPNT.lean using `‖ζ'...‖` and `ZetaUpperBnd` — the files might share `namespace Zeta23` hmm. Quick check needed at build — I'll reference them and qualify as needed.\n\nNow let me write LiRBound.lean §1 (ii′-5b). This is a big file; let me write it in pieces and compile incrementally. Start with the b2Fun machinery (self-contained).\n\n```lean\n/-\nNguyenRH1/LiRBound.lean — tier ii′ sub-stage ii′-5 (D-067): the R-bound\n...\n-/\nimport Mathlib\nimport Zeta23.FromPNTPlus.ZetaBounds   -- for Zeta0EqZeta etc.\nimport NguyenRH1.LiContour\n\nopen Zeta23 Complex Filter Topology\nopen scoped Nat\n\nnoncomputable section\n\nnamespace NguyenRH1\n\n/-! ## 1. The periodic Bernoulli primitive B̃₂ -/\n\n/-- The second periodic Bernoulli function: B̃₂(x) = (u² − u)/2 with\nu = Int.fract x ∈ [0, 1). Continuous (endpoint-matched), 1-periodic,\ntaking values in [−1/8, 0]. (The antiderivative of the sawtooth\nx − ⌊x⌋ − 1/2 off the integers.) -/\ndef b2Fun (x : ℝ) : ℝ := (Int.fract x ^ 2 - Int.fract x) / 2\n```\n\nWait — Int.fract for real x: `Int.fract : α → α` for α with floor... `Int.fract (x : ℝ) : ℝ = x − ⌊x⌋` ✓.\n\nLemmas:\n\n```lean\ntheorem b2Fun_add_one (x : ℝ) : b2Fun (x + 1) = b2Fun x := by\n  unfold b2Fun\n  rw [Int.fract_add_one]\n\ntheorem b2Fun_nonneg_range (x : ℝ) : -1/8 ≤ b2Fun x ∧ b2Fun x ≤ 0 := by\n  have hu0 : 0 ≤ Int.fract x := Int.fract_nonneg x\n  have hu1 : Int.fract x < 1 := Int.fract_lt_one x\n  unfold b2Fun\n  -- (u²−u)/2 = −u(1−u)/2: u(1−u) ∈ [0, 1/4]\n  constructor\n  · -- ≥ −1/8: u(1−u) ≤ 1/4\n    have : Int.fract x * (1 - Int.fract x) ≤ 1/4 := by\n      nlinarith [sq_nonneg (Int.fract x - 1/2)]\n    nlinarith\n  · nlinarith\n```\n\nHmm — (u²−u)/2 ≥ −1/8 ⟺ u²−u ≥ −1/4 ⟺ (u−1/2)² ≥ 0 ✓ nlinarith [sq_nonneg (u − 1/2)].\n(u²−u)/2 ≤ 0 ⟺ u(u−1) ≤ 0 ✓ from u ≥ 0, u ≤ 1: nlinarith.\n\n```lean\ntheorem abs_b2Fun_le (x : ℝ) : |b2Fun x| ≤ 1/8 := by\n  have ⟨h1, h2⟩ := b2Fun_nonneg_range x\n  rw [abs_of_nonpos h2]\n  linarith\n\ntheorem continuous_b2Fun : Continuous b2Fun := by\n  unfold b2Fun\n  -- b2Fun = g ∘ Int.fract with g u = (u²−u)/2, g 0 = 0 = g 1\n  exact ContinuousOn.comp_fract'' (f := fun u : ℝ => (u^2 - u)/2) (by fun_prop) (by norm_num)\n```\nCheck comp_fract'' signature: `theorem ContinuousOn.comp_fract'' {f : α → β} (h : ContinuousOn f I) (hf : f 0 = f 1) : Continuous fun x => f (fract x)` — I = Icc 0 1. So `(ContinuousOn.comp_fract'' (h) (hf))` with h : ContinuousOn (fun u => (u^2−u)/2) (Icc 0 1) — `by fun_prop`? fun_prop proves continuity of polynomial fun on a set: `by continuity`? `by fun_prop` should give ContinuousOn... maybe need `Continuous.continuousOn (by fun_prop)` or continuity tactic. hf : (0²−0)/2 = (1²−1)/2 ⟹ 0 = 0 ✓ norm_num.\nThe resulting Continuous fun x => f (fract x) — our b2Fun x = (fract x ^ 2 − fract x)/2 = f (fract x) ✓ defeq after unfold ✓.\n\nHasDerivAt:\n```lean\ntheorem hasDerivAt_b2Fun {x : ℝ} (hx : Int.fract x ≠ 0) :\n    HasDerivAt b2Fun (Int.fract x - 1/2) x := by\n  have hf : HasDerivAt (Int.fract : ℝ → ℝ) 1 x := by\n    have hIco : Set.Ico (⌊x⌋ : ℝ) (⌊x⌋ + 1) ∈ 𝓝 x :=\n      Set.Ico_mem_nhds ((Int.floor_le x).lt_of_ne (fun h => hx (by rw [Int.fract]; rw [h]; simp))) (Int.lt_floor_add_one x)\n```\nHmm — Int.fract x ≠ 0 ⟺ x ≠ ⌊x⌋: Int.fract x = x − ⌊x⌋: fract = 0 ⟺ x = ⌊x⌋ ✓. The Ico_mem_nhds needs ⌊x⌋ ≤ x (floor_le ✓) but STRICT for the nhds: Ico_mem_nhds (h1 : a < x) (h2 : x < b) : Ico a b ∈ 𝓝 x ✓ so need ⌊x⌋ < x from fract ≠ 0: (Int.floor_le x).lt_of_ne (Ne.symm ...) — floor_le : ⌊x⌋ ≤ x; lt_of_ne needs ⌊x⌋ ≠ x: from hx: fract x ≠ 0: x − ⌊x⌋ ≠ 0 ⟺ x ≠ ⌊x⌋ ✓.\n    Then on Ico, floor y = ⌊x⌋: `Int.floor_eq_on_Ico (n : ℤ) (x : α) (h : x ∈ Set.Ico n (n+1)) : ⌊x⌋ = n`? exists ✓ I think.\n    HasDerivAt fract 1 x: fract y = y − ⌊y⌋; near x: ⌊y⌋ = ⌊x⌋: fract =ᶠ[𝓝 x] fun y => y − ↑⌊x⌋: HasDerivAt (fun y => y − c) 1: hasDerivAt_id x |>.sub_const — then congr_of_eventuallyEq:\n```lean\n    have hEq : (Int.fract : ℝ → ℝ) =ᶠ[𝓝 x] fun y => y - (⌊x⌋ : ℝ) := by\n      filter_upwards [hIco] with y hy\n      rw [Int.fract, Int.floor_eq_on_Ico _ _ hy]\n    exact (hasDerivAt_id x |>.sub_const _).congr_of_eventuallyEq hEq  -- hmm congr direction + the value: (1 : ℝ) − 0 = 1\n```\nHasDerivAt.congr_of_eventuallyEq (h : f =ᶠ[𝓝 x] g)... `HasDerivAt.congr_of_eventuallyEq (hf : HasDerivAt f f' x) (h : f =ᶠ[𝓝 x] g) : HasDerivAt g f' x` ✓. (hasDerivAt_id x).sub_const (c) : HasDerivAt (fun y => y − c) (1 − 0) x — 1−0 needs simp to 1: `.sub_const` gives HasDerivAt (fun y => y − c) (1 − 0)? Hmm: (hasDerivAt_id x).sub_const c : HasDerivAt (fun y => y − c) (1 - 0) x — the derivative value 1−0 — simp only [sub_zero] to get 1. Fine — iterate at compile.\n```lean\n  unfold b2Fun\n  -- ((fract)² − fract)/2: deriv = (2·fract·1 − 1)/2 = fract − 1/2\n  have := (hf.pow 2).sub hf |>.div_const 2  -- HasDerivAt (fun x => (fract x^2 − fract x)/2) ((2*fract x*1 − 1)/2)\n  convert this using 1  -- align derivative value: (2u·1 − u... hmm (hf.pow 2) : HasDerivAt fract² (2·fract x^(2-1)·1) = (2·fract x)·1... \n  ring  -- (2u−1)/2 = u − 1/2 ✓\n```\nCareful with the exact forms; will iterate.\n\nThen §2 — the IBP identity. Structure:\n\n```lean\n/-! ## 2. The level-2 Euler–Maclaurin identity for ζ -/\n\n-- B1 (Zeta23) equals fract − 1/2 on [0, ∞):\ntheorem b1_eq_fract_sub_half {x : ℝ} (hx : 0 ≤ x) : EulerMaclaurin.B1 x = Int.fract x - 1/2 := ...\n```\nHmm — B1's full name/namespace: from EulerMaclaurin.lean — the grep didn't show a namespace line! Let me check if EulerMaclaurin.lean has `namespace` — the earlier grep \"^namespace\" found nothing? The grep output only showed... it printed nothing for namespace (the command `grep -n \"^namespace\\|^end\" ... | head -4` output was empty in the earlier call for both files? Looking back: the command output showed only the sed outputs — no namespace lines printed ⟹ NO namespace in those files — so B1, Zeta0EqZeta etc. are ROOT-level (or Zeta23 root?). They might be just plain root declarations. ZetaUpperBnd referenced plainly. OK — so `B1`, `Zeta0EqZeta` (wait — `riemannZeta0`, `Zeta0EqZeta`) at root namespace. Hmm — but there could be name collisions... fine.\n\nB1 def: `B1 (x : ℝ) : ℝ := x - ⌊x⌋₊ - 1/2`. Nat floor. For 0 ≤ x: ⌊x⌋₊ = ⌊x⌋ (Int): lemma: `Nat.floor_eq_int_floor`? Search: `Nat.floor_eq_iff`, `Int.floor_natCast`... The right conversion: `Nat.floor_int (hx : 0 ≤ x) : ⌊x⌋₊ = ⌊x⌋.toNat`? hmm. There's `Nat.floor_eq_intFloor`? Let me just grep for the Nat-Int floor conversion at build. Common: `Nat.floor_eq_iff`, `Int.floor_natCast`, `Nat.floor_ofNat`... The cleanest: `Int.natCast_floor_eq_natFloor`? I'll grep in a minute.\n\nThe IBP identity:\n\n```lean\ntheorem integral_b1_cpow_eq_integral_b2 {s : ℂ} (hs : 0 < s.re) :\n    ∫ x in Set.Ioi (1:ℝ), (B1 x : ℂ) * (x : ℂ) ^ (-s - 1)\n      = (s + 1) * ∫ x in Set.Ioi (1:ℝ), (b2Fun x : ℂ) * (x : ℂ) ^ (-s - 2) := by\n```\nProof skeleton:\n1. For each integer N ≥ 1: ∫ x in (1:ℝ)..N, ↑B̃₁·x^{−s−1} = (s+1)·∫ x in 1..N, ↑B̃₂·x^{−s−2}:\n   - Split into unit intervals [k, k+1] (k = 1..N−1): `intervalIntegral.integral_add_adjacent_intervals`? hmm — summing ∫ₖ^{k+1} over k∈Finset.range... there's a lemma for ∫ₐ^b splitting at integers? Simplest: induction on N using ∫₁^(N+1) = ∫₁^N + ∫_N^(N+1) (`intervalIntegral.integral_add_adjacent_intervals`? that's for three points: ∫ₐ^b + ∫_b^c = ∫ₐ^c — `intervalIntegral.integral_add_adjacent_intervals (hab : IntervalIntegrable f volume a b) (hbc : ...)` ✓). Induction: base N = 1: ∫₁^1 = 0 both sides ✓. Step: add the IBP'd unit piece.\n   - Unit piece: ∫ x in k..(k+1), ↑B̃₁ x * x^(−s−1) = (s+1)·∫ x in k..(k+1), ↑B̃₂ x * x^(−s−2):\n     via integral_mul_deriv_eq_deriv_mul_of_hasDerivAt with u = ↑b2Fun, u' = ↑(fract−1/2), v = x^(−s−1), v' = −(s+1)x^(−s−2):\n     ∫ u' v + u v' = u v |: the lemma form: `∫ x in a..b, u' x * v x + u x * v' x = u b * v b - u a * v a` (integral_deriv_mul_eq_sub_of_hasDerivAt ✓). At k+1, k (integers ≥ 1): b2Fun = 0 (fract = 0) ⟹ boundary 0. So ∫ u'v + ∫ uv' = 0 ⟹ ∫ u'v = −∫ uv' = −∫ b2Fun·(−(s+1)x^{−s−2}) = (s+1)∫b2Fun x^{−s−2} ✓.\n     And u' = ↑(fract x − 1/2) vs the B1 form: ↑(B1 x) = ↑(x − ⌊x⌋₊ − 1/2) = ↑(fract x − 1/2) for x ≥ 0 ✓ (B1-eq lemma).\n     Integrability: u' = fract−1/2 bounded measurable ✓ on the finite interval; v continuous ✓. `IntervalIntegrable` of each + products.\n     HasDerivAt u on Ioo(k, k+1): fract x ≠ 0 there ✓ (fract x = 0 ⟺ x integer).\n     v's HasDerivAt: x^(−s−1) at x > 0: Zeta23 has `ZetaSum_aux1φderiv`-style or Mathlib `HasDerivAt.cpow_const`? `hasDerivAt_cpow_const`? For (x:ℂ)^c with real x > 0: `Complex.hasDerivAt_cpow_const`? There's `hasDerivAt_cpow_const`? Hmm — Zeta23's `Real.differentiableAt_cpow_const_of_ne` (seen at ZetaBounds.lean:533!) — `Real.differentiableAt_cpow_const_of_ne (s : ℂ) {x : ℝ} (xpos : 0 < x) : DifferentiableAt ℝ (fun x => (x:ℂ)^s) x` — differentiableAt but I need the derivative VALUE: HasDerivAt (fun x => (x:ℂ)^c) (c·x^(c−1)) x — the formula: d/dx x^c = c·x^{c−1} for real x > 0: Mathlib: `hasDerivAt_cpow_const`? there's `Real.hasDerivAt_cpow_const`? or `Complex.hasDerivAt_cpow_const (x) (hx : 0 < x)?`... Hmm — there's `hasDerivAt_cpow_const (c : ℂ) {x : ℝ} (hx : 0 < x) : HasDerivAt (fun x => (x:ℂ)^c) (c * x^(c−1)) x`? I think `Complex.hasDerivAt_cpow_const` exists with something like that. Check at build.\n2. N → ∞ both sides:\n   - LHS: ∫ x in 1..N, ↑B̃₁ x^(−s−1) → ∫ x in Ioi 1, ... via intervalIntegral_tendsto_integral_Ioi (needs IntegrableOn on Ioi 1: |B̃₁ x^{−s−1}| ≤ (1/2)x^{−σ−1} integrable for σ > 0 ✓ — `integrableOn_of_Zeta0_fun`-style; Zeta23 has `integrableOn_of_Zeta0_fun` for THEIR integrand (⌊x⌋+1/2−x)x^{−s−1} = −B̃₁·x^{−s−1} ✓ EXACTLY our integrand up to sign!! `integrableOn_of_Zeta0_fun {N} (N_pos) {s} (s_re_gt : 0 < s.re) : IntegrableOn ... (Ioi N)` ✓).\n   - RHS: B̃₂ version: IntegrableOn of ↑b2Fun·x^{−s−2} on Ioi 1 for σ > −1: |·| ≤ (1/8)x^{−σ−2} ✓ build a small integrability lemma (mirror integrableOn_of_Zeta0_fun's proof or bound directly: Continuous(B̃₂)·measurable × power integrable: `integrableOn_Ioi_rpow_of_lt`? (x^a integrable on Ioi 1 ⟺ a < −1 ✓ Mathlib has `integrableOn_Ioi_rpow_of_lt` ✓).\n   - Tendsto along NATURALS N → ∞ (the boundary-vanishing needs integer N ✓ that's why atTop ℕ): intervalIntegral_tendsto_integral_Ioi with b := fun n : ℕ => (n:ℝ) ✓ (as in their proof: (b := fun n ↦ (n:ℝ)) tendsto_natCast_atTop_atTop ✓).\n\n3. §3 — hasDerivAt of the B₂ parameter integral for re s > −1. Clone of hasDerivAt_Zeta0Integral:\n```lean\ntheorem hasDerivAt_b2Integral {s : ℂ} (hs : -1 < s.re) :\n    HasDerivAt (fun z => ∫ x in Set.Ioi (1:ℝ), (b2Fun x : ℂ) * (x : ℂ) ^ (-z - 2))\n      (∫ x in Set.Ioi (1:ℝ), (b2Fun x : ℂ) * (x : ℂ) ^ (-s - 2) * (-Real.log x)) s\n```\nvia hasDerivAt_integral_of_dominated_loc_of_deriv_le with ε = (s.re + 1)/2, bound x^{−(s.re+3)/2}·|log x| — integrable via integrable_log_over_pow (r := −(s.re+1)/2 < 0 ✓: their form: ‖x^(r−1)‖·‖log x‖ with r−1 = −(s.re+3)/2 ⟹ r = −(s.re+1)/2 ✓ < 0).\n\n4. §4 — the extended identity via identity principle:\n```lean\ntheorem riemannZeta_eq_level2 {s : ℂ} (hs : -1 < s.re) (hs1 : s ≠ 1) :\n    riemannZeta s = 1/2 + 1/(s - 1) - s * (s + 1) * ∫ x in Set.Ioi (1:ℝ), (b2Fun x : ℂ) * (x : ℂ)^(-s - 2) := by\n```\nProof: \n- On V := {re > 0, ≠ 1}: ζ(s) = ζ₀ 1 s (Zeta0EqZeta) = 1/2 + 1/(s−1) − s∫B̃₁... wait ζ₀ 1 s = Σ_{range 2} + (−1^(1−s))/(1−s) + (−1^(−s))/2 + s∫(⌊x⌋+1/2−x)x^{−s−1}: simplify: sum = 1 (0^s = 0 junk + 1); (−1^(1−s))/(1−s): (1:ℂ)^(1−s) = 1 (one_cpow) ✓: = −1/(1−s) = 1/(s−1); (−1^(−s))/2 = −1/2. And (⌊x⌋+1/2−x) = −B̃₁: s∫(−B̃₁)x^{−s−1} = −s∫B̃₁x^{−s−1}. So ζ₀ 1 s = 1/2 + 1/(s−1) − s∫B̃₁x^{−s−1} = 1/2 + 1/(s−1) − s(s+1)∫B̃₂x^{−s−2} (by §2) ✓ on V.\n- Both sides analytic on U := {s | −1 < re s ∧ s ≠ 1}: \n  - LHS: HolomorphicOn_riemannZeta.mono ✓.\n  - RHS: the B₂ integral hasDerivAt everywhere on U (§3) ⟹ DifferentiableOn ⟹ analyticOnNhd (DifferentiableOn.analyticOnNhd, U open? {−1 < re} open ✓ minus {1} open ✓).\n- Preconnected: U = {−1<re} ∩ {≠1}: half-plane minus point — path-connected: mimic Zeta23's isPathConnected_aux (they proved it for {z ≠ 1 ∧ 0 < re}!). Generalize: their proof is for 0 < re; mine needs −1 < re — copy their proof with the shift (or find a slicker way: {−1<re} ∖ {1}: IsConnected via... the Zeta23 proof uses explicit paths — copy & adapt).\n- Agree on V ⊆ U with accumulation: eventuallyEq at 2? V is open nonempty ⊂ U; the eqOn_of_preconnected_of_eventuallyEq needs agreement in a NEIGHBORHOOD of a point: at s = 2: V ∈ 𝓝 2 ✓ (V open, 2 ∈ V) so agreement on V gives eventuallyEq at 2 ✓. Use AnalyticOnNhd.eqOn_of_preconnected_of_eventuallyEq (as in Zeta0EqZeta proof) — signature: (hf : AnalyticOnNhd f U) (hg : AnalyticOnNhd g U) (hU : IsPreconnected U) (hx : x ∈ U) (hfg : f =ᶠ[𝓝 x] g)?? their usage: `AnalyticOnNhd.eqOn_of_preconnected_of_eventuallyEq f_an g_an preconU h2 ?_ s_mem` — h2 : (2:ℂ) ∈ U, then ?_ is the eventuallyEq at 2, then s_mem — returns eqOn. ✓ copy that.\n\n5. §5 — the disc bound:\n```lean\ntheorem norm_riemannZeta_le_of_mem_closedBall {t : ℝ} (ht : 10 ≤ |t|) {s : ℂ}\n    (hs : s ∈ Metric.closedBall (2 + t * Complex.I) (5/2)) :\n    ‖riemannZeta s‖ ≤ 3/4 * |t| ^ 2 := by\n```\n- re s ≥ 2 − 5/2 = −1/2 > −1 ✓ (disc); s ≠ 1 (|im s − t| ≤ 5/2, |t| ≥ 10 ⟹ |im s| ≥ 7.5 ≠ 0) ✓.\n- Apply the rep: ‖ζ s‖ ≤ 1/2 + 1/‖s−1‖ + ‖s‖‖s+1‖·(1/8)/(re s + 1):\n  - the integral: ‖∫ ↑b2Fun x^{−s−2}‖ ≤ ∫ (1/8)x^{−σ−2} = (1/8)/(σ+1) ✓ (norm_integral_le_integral_norm + the rpow integral: ∫₁^∞ x^{−σ−2}dx = 1/(σ+1): `integral_Ioi_rpow_of_lt`? name: `integral_rpow_of_lt`? For Ioi 1: ∫ x^c on Ioi 1 = −1/(c+1) for c < −1: `integrableOn_Ioi_rpow_of_lt` + `integral_Ioi_rpow_of_lt`? I'll find. Or bound directly via intervalIntegral limit.)\n  - ‖s‖ ≤ |t| + 9/2 (‖s‖ ≤ ‖2+it‖ + 5/2 ≤ 2+|t|+5/2 ✓ via norm triangle + ‖2+it‖ ≤ 2+|t|: norm_le_abs_re_add_abs_im).\n  - ‖s+1‖ ≤ |t| + 11/2.\n  - ‖s−1‖ ≥ |t| ≥ 10: 1/‖s−1‖ ≤ 1/10. Hmm: ‖s−1‖ ≥ |im(s−1)| = |t| ✓ abs_im_le_norm.\n  - re s + 1 ≥ 1/2 ⟹ 1/(re s+1) ≤ 2.\n  So ‖ζ‖ ≤ 1/2 + 1/10 + (|t|+9/2)(|t|+11/2)/4 — then ≤ 3/4·|t|² for |t| ≥ 10 by nlinarith-style expansion: (|t|+9/2)(|t|+11/2)/4 = (t² + 10|t| + 99/4)/4 = t²/4 + 2.5|t| + 6.1875; +0.6: total ≤ t²/4 + 2.5|t| + 6.8 ≤ (3/4)t² ⟺ t²/2 ≥ 2.5|t| + 6.8 ⟸ |t| ≥ 10: t²/2 ≥ 50: 2.5·10 + 6.8 = 31.8 ≤ 50 ✓. nlinarith can do with the right hints (|t| ≥ 10, t² ≥ 100...).\n\nThat's the full ii′-5b plan. It's chunky — maybe 400-600 lines of Lean. Let me start writing. First the b2Fun section, compile, then IBP, then hasDerivAt, then identity, then the bound. I'll build incrementally, compiling after each section.\n\nOne check — Nat/Int floor conversion lemma name. grep:","signature":""},{"type":"text","text":"All pieces confirmed. One last name check (Nat↔Int floor conversion), then I start writing LiRBound.lean."},{"type":"tool_use","id":"Bash_189","name":"Bash","input":{"command":"cd D:/Python/Quantum/zeta-23-lean/.lake/packages/mathlib && grep -rn \"theorem floor_natCast\\|natFloor_eq\\|theorem Int.floor_natCast\\|natCast_floor\" Mathlib/Algebra/Order/Floor/Ring.lean Mathlib/Algebra/Order/Floor/Defs.lean 2>/dev/null | head -6; grep -rn \"Nat.floor_eq_int_floor\\|floor_natFloor\\|natFloor\" Mathlib/Algebra/Order/Floor/Ring.lean | head -6","description":"Find Nat/Int floor conversion lemma"},"extra_content":null}]