lean_workbook_plus_28628

⚠️ Retired — specification defect

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

Choose $a_1, ..., a_{p-1}, p\\in N$ arbitrary and $a_0 = 0$. For all $k\\in N$ and $r\\in\\{0, 1, ..., p-1\\}: f(kp + r) = p(a_r + k)$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_28628 (p : ℕ) (hp : p.Prime) (a : ℕ → ℕ) (f : ℕ → ℕ) (hf: f 0 = 0) (h : ∀ k r : ℕ, (r < p ∧ f (k * p + r) = p * (a r + k))): ∃ k : ℕ, ∀ n : ℕ, (n < p * k ∧ f n = p * (k + a (n % p)))   :=  by sorry

A bounded quantifier became ∀k,r, r<p ∧ …; taking r=p gives p<p. The conclusion also demands every natural n lie below a fixed pk.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

Proposed corrected statement

For p≥1 and natural a_r (0≤r<p), a_0=0, define f(kp+r)=p(a_r+k) for every k≥0 and 0≤r<p. Equivalently f(n)=p(a_(n mod p)+⌊n/p⌋). This is a consistent construction, not the posted impossible universal bound. Recover any further intended classification from the full source.

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.