lean_workbook_plus_17797

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

And so we can study the sequences $a_n,b_n$ defined as : $a_0=6$ , $b_0=0$ , $a_{k+1}=\frac 6{b_k+1}$ and $b_{k+1}=\frac 6{a_k+1}$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_17797 (a b : ℕ → ℝ) (a0 b0 : ℝ) (a_rec b_rec : ℕ → ℝ) (ha : a = a_rec) (hb : b = b_rec) (hab: a0 = 6 ∧ b0 = 0) (h1 : a_rec 0 = 6 / (b0 + 1)) (h2 : b_rec 0 = 6 / (a0 + 1)) (h3 : ∀ n, a_rec (n + 1) = 6 / (b_rec n + 1)) (h4 : ∀ n, b_rec (n + 1) = 6 / (a_rec n + 1)) : ∃ n, a n = b n   :=  by sorry

The prose defines a sequence, and the formalizer invents an equality-at-a-finite-index goal.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Set a(0)=6,b(0)=0 directly and retain the coupled recurrence; remove the unsupported finite-index equality conclusion. A substantive replacement is convergence of both sequences to2, while b(n)<a(n) for every n.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.