lean_workbook_plus_10818

⚠️ 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.

Let $f$ be a function defined in the set of positive integers such that: \n\n

for all $x$ and $y$ positive integers. If $f (2004) = 1002,$ find $f (5555).$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_10818 (f : ℕ → ℕ) (hf : ∀ x y : ℕ, f (x + y) = f x + f y + 1002) : f 2004 = 1002 → f 5555 = 15220   :=  by sorry

All-natural quantification introduces x=y=0 and makes f(0)=2f(0)+1002 impossible; the natural-valued codomain also cannot model the source values, and 15220 is a wrong inserted answer.

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

Let f:ℕ_{>0}→ℝ satisfy f(x+y)=f(x)+f(y)+1002 for positive x,y and f(2004)=1002. Determine f(5555). Equivalently prove f(n)=n−1002 for every positive n, yielding f(5555)=4553.

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.