Exponents_of_Primes_in_Prime_Decomposition_are_Less_iff_Divisor
Disproved⚠️ Retired — specification defect
The Lean statement below does not encode the problem shown on this page, so its
Disprovedstatus carries no information about that problem. Do not import this node or use it as a dependency.
a divides b iff for every prime p, the exponent of p in a's factorization is at most that in b's factorization.
Why this node was retired
The posted statement is
theorem Exponents_of_Primes_in_Prime_Decomposition_are_Less_iff_Divisor (a b : ℕ) (ha : 0 < a) (hb : 0 < b) : a ∣ b ↔ ∀ p : ℕ, Nat.Prime p → p.factorization a ≤ p.factorization b := by sorry
Factorization arguments are reversed.
The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.
Proposed corrected statement
Use a.factorization p≤b.factorization p, retaining a,b>0.
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.
import Mathlib.Data.Nat.Factorization.Basic
theorem Exponents_of_Primes_in_Prime_Decomposition_are_Less_iff_Divisor (a b : ℕ) (ha : 0 < a) (hb : 0 < b) : a ∣ b ↔ ∀ p : ℕ, Nat.Prime p → p.factorization a ≤ p.factorization b := by sorry