lean_workbook_plus_38713

⚠️ 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.

Find all sets $A \subset \mathbb{N}^*$ finitely and satisfied $|A| \ge 2$ , $\forall a,b \in A, a \ne b$ then $\dfrac{a+b}{\gcd(a,b)} \in A$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_38713 (A : Finset ℕ) (hA: A.card >= 2) (hA2: ∀ a ∈ A, ∀ b ∈ A, a ≠ b → (a + b) / Nat.gcd a b ∈ A) : (A = {1,2} ∨ A = {1,3} ∨ A = {1,4} ∨ A = {1,5} ∨ A = {1,6} ∨ A = {1,7} ∨ A = {1,8} ∨ A = {1,9} ∨ A = {1,10} ∨ A = {2,3} ∨ A = {2,5} ∨ A = {3,4} ∨ A = {3,7} ∨ A = {3,8} ∨ A = {4,7} ∨ A = {5,7} ∨ A = {5,8} ∨ A = {7,8})   :=  by sorry

The finite list of eighteen sets omits an infinite family, including the positive set {11,110}; positivity alone does not repair the answer.

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

Proposed corrected statement

Restore the original mathematical task without the inserted answer: Find all sets $A \subset \mathbb{N}^*$ finitely and satisfied $|A| \ge 2$ , $\forall a,b \in A, a \ne b$ then $\dfrac{a+b}{\gcd(a,b)} \in A$ Require both necessity and sufficiency for a find-all/classification task; no replacement fixed answer is asserted in this audit.

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.