[{"type":"thinking","thinking":"I still have two sorries and undefined hpos1/hpos2 in hmax1. Let me carefully finish the last calc step. The step:\n\ngoal: (Ncount1 : ℝ) + (Ncount2 : ℝ) ≤ A₀·(2^{j+1}X+3)·L where L := log(3·2^{j+1}X+5).\nh1 : Ncount1 ≤ A₀·((2^{j+1}X − (2^jX−1)) + 1)·log(3·max |2^jX−1| |2^{j+1}X| + 5)\nh2 : Ncount2 ≤ A₀·((−2^jX − −2^{j+1}X) + 1)·log(3·max |−2^{j+1}X| |−2^jX| + 5)\n\nmax computations:\nhmax1 : max |2^jX − 1| |2^{j+1}X| = 2^{j+1}X: |2^jX−1| = 2^jX−1 (since 2^jX ≥ 200 > 1: abs_of_pos-ish: 2^jX − 1 ≥ 0 ⟸ 2^jX ≥ 200: h2jX : 0 < 2^jX — need 1 ≤ 2^jX: from X ≥ 200: 2^jX ≥ X ≥ 200 ✓ linarith [h2jX, hX]... h2jX is 0 < 2^jX; X ≥ 200 ⟹ 2^jX ≥ 200 via 2^j ≥ 1: positivity gives 0 < only. Add have h2jX1 : (1:ℝ) ≤ 2^jX: nlinarith... (2^j ≥ 1: one_le_pow₀? `one_le_pow₀ (h : 1 ≤ a)?` : 1 ≤ 2^j — `one_le_pow_iff`? or pow ≥ 1: have : (1:ℝ) ≤ (2:ℝ)^j := one_le_pow₀ (by norm_num) j?? `one_le_pow₀ (ha : 1 ≤ a) (n)`? hmm — `one_le_pow_of_le`? Let me use: `one_le_two_pow`? or nlinarith can't do pow. `have h2j1 : (1:ℝ) ≤ (2:ℝ)^j := by exact one_le_pow₀?? ` — there's `one_le_pow₀ (h : 1 ≤ a) (n : ℕ) : 1 ≤ a ^ n`? I believe `one_le_pow₀` exists (or `one_le_pow_of_le'`). Probe-free: use `pow_le_pow_left₀ (by norm_num : (0:ℝ) ≤ 2)?? no — simplest: `one_le_two_pow (n : ℕ) : 1 ≤ 2^n`? hmm — `one_le_two_pow_iff`? There IS `one_le_two_pow (n) : (1:?) ≤ 2^n`? Not sure. Use: `have : (1:ℝ) ≤ (2:ℝ)^j := by exact_mod_cast Nat.one_le_two_pow` — Nat.one_le_two_pow : 1 ≤ 2^n ✓ exists for ℕ ✓ then cast. OK.\nThen 2^jX ≥ 200: (2^j)·X ≥ 1·200: mul_le_mul h2j1 (le of hX) ✓.\n|2^jX − 1| = 2^jX − 1: abs_of_nonneg (by linarith [that]) ✓.\n|2^{j+1}X| = 2^{j+1}X (pos) ✓.\nmax = 2^{j+1}X ⟺ 2^jX − 1 ≤ 2^{j+1}X: 2^{j+1}X = 2·2^jX ≥ 2·200 = 400 > 2^jX − 1·... 2^{j+1}X = 2^jX·2 ≥ 2^jX·... 2^{j+1} = 2·2^j (pow_succ' : a^(n+1) = a·a^n ✓): 2^{j+1}X = 2·(2^jX): ≥ 2·200 = 400 and 2^jX − 1 ≤ 2^jX ≤ 2^{j+1}X ⟺ 2^jX ≥ 0 ✓: so max_eq_right (2^jX − 1 ≤ 2^{j+1}X) — hmm max_eq_right : a ≤ b → max a b = b ✓ with a := |2^jX−1| = 2^jX−1, b := 2^{j+1}X: 2^jX−1 ≤ 2^{j+1}X ⟺ ... linarith with the values ✓.\n\nLet me restructure the final step to avoid the max entirely: instead of Ncount_le_of_interval_gen's max-form, note that max |a| |b| ≤ max |a| |b| (obviously) and log is monotone — I can bound log(3·max|a||b| + 5) ≤ log(3·(2^{j+1}X) + 5) via max ≤ 2^{j+1}X: max |2^jX−1| |2^{j+1}X| ≤ 2^{j+1}X: max_le: |2^jX−1| ≤ 2^{j+1}X (from the abs forms) and |2^{j+1}X| ≤ 2^{j+1}X ✓. So:\n```lean\n        have hlogle1 : Real.log (3 * max |(2:ℝ)^j*X - 1| |(2:ℝ)^(j+1)*X| + 5)\n            ≤ Real.log (3 * (2:ℝ)^(j+1)*X + 5) := by\n          apply Real.log_le_log (by positivity)\n          have h1a : |(2:ℝ)^j*X - 1| ≤ (2:ℝ)^(j+1)*X := by\n            rw [abs_of_nonneg (by ... : 0 ≤ 2^jX − 1)]\n            nlinarith [...]\n          have h1b : |(2:ℝ)^(j+1)*X| ≤ (2:ℝ)^(j+1)*X := le_refl _  -- |pos| = itself: abs_of_nonneg\n          ...\n```\nUgh — the abs forms again. You know what — simplify by NOT using the _gen form here; instead note the intervals have b ≥ a ≥ −... for the FIRST interval a = 2^jX−1 ≥ 199 > 0 — use the NON-general `Ncount_le_of_interval` (needs 0 ≤ a): a = 2^jX − 1 ≥ 0 ✓: gives log(b+4) with b = 2^{j+1}X ✓ NO max! For the second (negative) interval: a = −2^{j+1}X, b = −2^jX: Ncount_le_of_interval needs 0 ≤ a — fails. Hmm. But — the negative side: zeros with im ∈ (−2^{j+1}X, −2^jX] — use the CONJUGATE symmetry to map to the positive interval!! The conj map: zeros with im ∈ (a, b] negative ↔ conj zeros with im ∈ (−b, −a] positive — and zeroMult (conj ρ) = zeroMult ρ?? — the conj-mult invariance: NOT built (session 030 noted this). Ugh.\n\nOK simplest: just do the max computations properly. Let me rewrite the whole final calc step cleanly with all the aux haves. And double check `one_le_pow₀` name. Also `Finset.sum_image` signature (journal-011: takes Set.InjOn, map = g, summand = f): `Finset.sum_image (h : Set.InjOn g s) : ∑ x ∈ s.image g, f x = ∑ x ∈ s, f (g x)` — my usage: `rw [Finset.sum_image (fun x _ y _ h => Subtype.val_injective h)]` — the injOn proof : Set.InjOn Subtype.val FS — `fun x _ y _ h => Subtype.val_injective h` ✓ — the rewrite direction: sum_image : ∑ x ∈ s.image g, f x = ∑ x ∈ s, f (g x): my hsum_img : ∑_{FS} m∘val = ∑_{image} m: so rw [Finset.sum_image ...] at... I wrote `rw [Finset.sum_image (...)]` in the proof of hsum_img : LHS(∑ FS) = RHS(∑ image): the rw turns ∑_{image} into ∑_{FS}∘val — the goal's RHS: rw [Finset.sum_image h] rewrites the RHS pattern ✓ then goal ∑ FS (m∘val) = ∑ FS (m∘val)... rfl? The (zeroMult ↑ρ : ℝ) vs (zeroMult (Subtype.val ρ)) — same ✓ rfl-closes ✓.\n\nLet me now write the final step fully and also fix `one_le_pow₀` check. Actually — `one_le_pow₀`: let me just probe quickly along with a couple of other names, then write the final version.","signature":""},{"type":"text","text":"The final calc step needs careful handling of the max/abs forms. Let me probe the pow-one name then write the complete step."},{"type":"tool_use","id":"Bash_381","name":"Bash","input":{"command":"cd D:/Python/Quantum/research/lean/NguyenRH1 && cat > Probe.lean << 'EOF'\nimport Mathlib\nopen Real\nexample (j : ℕ) : (1 : ℝ) ≤ (2 : ℝ) ^ j := by exact?\nEOF\nexport PATH=\"$HOME/.elan/bin:$PATH\" && cd D:/Python/Quantum/research/lean/NguyenRH1 && lake env lean Probe.lean 2>&1 | head -6","description":"Probe 1 ≤ 2^j","timeout":300000},"extra_content":null}]