[{"type":"thinking","thinking":"I now have the full picture of the machinery. Now let me think carefully about the piece-3 design and write the Lean plan.\n\n**Goal (piece 3):** bound the far difference series\n```\n‖∑' ρ, if ρ ∈ nearZerosSub t then 0 else seriesDiffTerm s s₀ ρ‖ ≤ 12.333·C0hat(t+1) + 6.014\n```\nwhere s₀ = 2 + tI, t ≥ 10, s.re ∈ [−5/2, 3], s.im = t, and ζ(s) ≠ 0 (s not a zero — needed for 1/(s−ρ) to make sense; actually for the norm bound we need s ≠ ρ, i.e. s not a nontrivial zero. Since ζ(s) ≠ 0, s is not a zero at all. Good — the capstone has hζ : riemannZeta s ≠ 0).\n\n**Per-term bound.** For ρ ∉ nearZerosSub t, x := |ρ.im − t| ≥ 1:\n- ‖s₀ − s‖ = |2 − s.re| ≤ 4.5 (have this from one_div_diff_le's hsdiffle).\n- ‖s − ρ‖ ≥ |s.im − ρ.im| = |t − ρ.im| = x.\n- ‖s₀ − ρ‖ ≥ √((2 − ρ.re)² + (t − γ)²) ≥ √(1 + x²) since 2 − ρ.re > 1.\n\nSo ‖seriesDiffTerm s s₀ ρ‖ = m·‖s₀−s‖/(‖s−ρ‖·‖s₀−ρ‖) ≤ 4.5·m/(x·√(1+x²)).\n\nFor k ≤ x < k+1 (k = ⌊x⌋ ≥ 1): ≤ 4.5·m/(k√(1+k²)) (since x√(1+x²) is increasing in x for x > 0).\n\n**Decomposition strategy.** This is the hard part in Lean: we have a tsum over the subtype, and we want to split by intervals. The approach used in LiCircleBound for similar purposes: define finite finsets covering everything up to some bound, and a tail.\n\nActually — here's a cleaner idea avoiding tsum gymnastics. Note that for the far series, we can split into:\n\n1. The finite part: zeros with x = |t − γ| ≤ t + 14 wait, no. Hmm.\n\nLet me think about the ranges:\n- γ ∈ (0, ∞): NoLowZeros gives γ ≥ 14.\n  - x = |t − γ|. For γ ∈ [14, ...]: near window excluded (x ≥ 1).\n- γ ∈ (−∞, 0): |γ| ≥ 14, x = t + |γ| ≥ t + 14 ≥ 24.\n\nSplit at x ≤ 2t (say): \n- **Finite part:** zeros with 1 ≤ x ≤ 2t+1 (or some bound). These have |γ| ≤ t + x ≤ 3t+1, a bounded set — finite. Split by intervals k ≤ x < k+1 for k = 1, ..., K where K = ⌈2t⌉. Per-interval counts:\n  - Right interval (γ ∈ [t+k, t+k+1)): ZeroCountProp at t+k needs t+k ≥ 2 ✓, gives count ≤ C0hat(t+k+1) ≤ C0hat(2t+2) (for k ≤ t+1) — and we need C0hat(2t+2) ≤ C0hat(t+1) + 0.55.\n  - Left interval (γ ∈ (t−k−1, t−k]): for t−k−1 ≥ 2 i.e. k ≤ t−3: ZeroCountProp at t−k−1 gives ≤ C0hat(t−k) ≤ C0hat(t+1).\n    For k > t−3 (i.e. t−k < 5): the interval [t−k−1, t−k] ⊂ (−∞, 5). By NoLowZeros + conj, no zeros with |γ| ≤ 14, so if t−k ≤ 14 the interval is EMPTY. Since k > t − 3 means t − k < 3 < 14 ✓ empty. \n\n    Wait: left interval γ ∈ (t−k−1, t−k]. If t−k ≤ 14, then γ ≤ 14 and γ > t−k−1. Hmm, γ could still be < −14 if t−k−1 < −14. For k ≤ 2t, t−k−1 ≥ t−2t−1 = −t−1 ≥ −(t+1). If t−k ≤ 14 and t−k−1 ≥ −14 (i.e. k ≤ t+13), then γ ∈ (−14, 14) — empty by NoLowZeros (since |γ| < 14 ⟹ no zeros). For k ∈ (t−3, t+13]: the left interval is empty.\n    Hmm, but for k ≤ 2t we need emptiness up to k = 2t, and t+13 < 2t when t > 13. Hmm!\n\n    Let me redo: for the left side with k ∈ (t−3, 2t], the interval (t−k−1, t−k] has t−k ≤ t−(t−3) = 3 wait no: k > t−3 ⟺ t−k < 3. And t−k ≥ t−2t = −t. So γ ∈ (t−k−1, t−k] ⊆ (−t−1, 3). Zeros here have |γ|... γ could be very negative (down to −t−1). Not empty by NoLowZeros alone (|γ| could exceed 14). \n\n    So for the left side at k ∈ (t−3, 2t] (i.e. γ ∈ (−t−1, 3)): these are zeros with γ < 14, and by NoLowZeros + the fact they're in (−t−1, 3) — if γ ∈ [−14, 3) they're excluded. So only γ ∈ (−t−1, −14) survive. For those, x = |t−γ| = t−γ = t+|γ| ≥ t+14. Hmm wait, that means x ≥ t+14, i.e. k ≥ t+14 (well, k = ⌊x⌋ ≥ t+13ish). So actually for k < t+13, the left interval (t−k−1, t−k] intersected with {|γ| ≥ 14} is empty unless t−k ≤ −14, i.e. k ≥ t+14. So:\n    - For k ∈ [1, t+13]: left interval nonempty only possibly when t−k > 14 (i.e. k < t−14): then it's the ZeroCountProp window.\n    - Hmm wait, I need to be careful: for k ∈ [t−14, t+13], the left interval (t−k−1, t−k] ⊆ (−14−1+1, 14] hmm let me redo: k ≥ t−14 ⟺ t−k ≤ 14. k ≤ t+13 ⟺ t−k ≥ −13, so t−k−1 ≥ −14. So γ ∈ (t−k−1, t−k] ⊆ (−14, 14] — |γ| ≤ 14, empty by NoLowZeros. ✓\n    - For k ≥ t+14: left interval ⊆ (−∞, −14), zeros possible; these are in the \"tail\" (x ≥ t+14).\n\n- **Tail:** zeros with x > 2t (k > 2t): γ > 3t or γ < −t. Bound via the band lemma + dyadic-ish sum ≤ 0.5828-ish. Hmm.\n\nHmm, this is getting complicated. Let me re-read the journal-035 piece-3 map once more:\n\n\"per-interval C0hat counts from the perimeter (left ≤ C0hat(t+1), right ≤ C0hat(2t+1) ≤ C0hat(t+1) + 0.55); the negative-γ side via isNontrivialZero_conj; the height-< 14 intervals empty by NoLowZeros; the sum S₃ := Σ 1/(k√(1+k²)) ≤ 1.3703 via k ≤ 3 exact ... + the telescope for k ≥ 4 (Σ = 1/3); the k > t tail ≤ 0.5828 (Σ log(3k)/k² and Σ 1/k² via the same telescopes). Total piece 3 ≤ 4.5·[2·S₃·C₀ + 0.55·S₃ + 0.5828] = 12.333·C₀ + 6.014.\"\n\nOK so the intended structure:\n- For each k = 1, 2, ...: TWO intervals (left and right), each with count ≤ C₀ = C0hat(t+1) (right needs +0.55 correction once: right ≤ C0hat(2t+1) ≤ C₀ + 0.55).\n- \"2·S₃·C₀\" = sum over k of [left count + right count]·4.5/(k√(1+k²)) ≤ Σ 4.5·(2C₀ + 0.55)/(k√(1+k²)) = 4.5·(2C₀+0.55)·S₃.\n- The k > t tail: zeros with x > t (γ < 0 — count via conj to positive side? or γ > 2t). Bounded NOT by C0hat counts but by... the band lemma? \"Σ log(3k)/k² and Σ 1/k²\" suggests: count in band at height ~t+k via band lemma ≤ 3·A₀·3·log(3(t+k+1)+5) ≤ 9·A₀·log(9k+8)-ish, then Σ_{k>t≥10} 4.5/(k√(1+k²))·9·A₀·log(9k+8). Hmm, with A₀ ≥ 1 that would be ≥ 40·Σ log(9k)/k² > 40·(log(99)/11 + ...) — way bigger than 0.5828. That can't be right.\n\nHold on — maybe the tail uses a different counting: the RvM Ncount directly: Ncount(a, a+1] ≤ A₀·log(|a|+3). For the tail at height t+k (k > t): ≤ A₀·log(t+k+3). Σ_{k>t} 4.5·A₀·log(t+k+3)/(k√(1+k²)) — still has A₀ ≥ 1, so ≥ 4.5·Σ_{k≥11} log(k)/(k√(1+k²)) ≈ 4.5·(log 11/11 + ∫...) ≈ 4.5·(0.218 + 0.24) ≈ 2. Still > 0.5828.\n\nHmm, so A₀ must be a small explicit constant? What's the actual value of A0count? From Zeta23's RvM.zeta_local_zero_count — [Tit86 Thm 9.2]: N(T+1)−N(T) ≤ A·log T-ish with some explicit A. Let me check what value Zeta23 proves. The journal says \"A0count from Zeta23's RvM local count\". Titchmarsh 9.2: N(T+H)−N(T) ≤ C·log T for H ≥ 1 with explicit C? The RvM paper (Trudgian?) — \"zeta_local_zero_count\" — let me grep Zeta23 for the actual statement.\n\nActually, maybe the tail bound is computed differently: not per-unit-interval counts but a GLOBAL tail count: zeros with |γ| > Y have total mult-weighted count... no, that's infinite.\n\nHmm, alternatively the tail per-interval x-weight: for the tail, use the count for the WINDOW of x's: zeros with x ∈ [k, k+1) on the right have γ ∈ [t+k, t+k+1), count via ZeroCountProp (since t+k ≥ 2 always!): ≤ C0hat(t+k+1). Then C0hat(t+k+1) = 0.433·log(t+k+3) + 0.886·log log(t+k+3) + 2.9 ≤ 0.433·log(2k+3)+... using t ≤ k: t+k+3 ≤ 2k+3 ≤ 3k for k ≥ 3/2... hmm, then Σ_{k>t≥10} 4.5·C0hat(3k-ish)/(k√(1+k²)) ≈ 4.5·Σ (0.433 log(3k) + 0.886 log log(3k) + 2.9)/k² — the 2.9 alone gives 4.5·2.9·Σ_{k≥11}1/k² ≈ 13·0.095 ≈ 1.24. Still > 0.5828. \n\nHmm! So 0.5828 can't come from per-interval C0hat counts either. Let me reconsider.\n\nOH WAIT. I think I misread. Maybe the tail is over k > t but using a SMARTER weight: the per-term bound 4.5/(x√(1+x²)) summed with the GLOBAL count function. I.e., instead of per-interval × per-interval-count, use: Σ_zeros 4.5/(x√(1+x²)) = 4.5·∫ count-density... In finite terms: sort the tail zeros... This is getting too speculative. \n\nLet me look at how D-056 §2 (the original analysis) does the far sum. Let me read D-056.","signature":""},{"type":"text","text":"Let me check D-056 §2 for the original far-sum analysis, and the actual RvM count constant in Zeta23."},{"type":"tool_use","id":"Bash_13","name":"Bash","input":{"command":"ls \"D:/Python/Quantum/research/derivations/\" | grep -i \"D-05[56]\"","description":"Find D-055/D-056 files"},"extra_content":null},{"type":"tool_use","id":"Grep_14","name":"Grep","input":{"pattern":"zeta_local_zero_count","path":"D:\\Python\\Quantum\\research\\lean","output_mode":"content","-n":true,"head_limit":10},"extra_content":null}]