lean_workbook_plus_79407

⚠️ 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 the value of $f(1)$ from the equation $x^2(f(x)+f(y))=(x+y)f(y(f(x)))$ assuming $f \colon \mathbb{R}^* \to \mathbb{R}^*$ and $f(f(1)) = f(1)$.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_79407 (f : ℝ → ℝ) (hf : ∀ x y, (x ^ 2) * (f x + f y) = (x + y) * f (y * f x)) : f 1 = 1   :=  by sorry

The function may be zero and the source nonzero codomain/fixed-value condition is missing.

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 f:ℝ∖{0}→ℝ∖{0}, quantify the functional equation on its actual domain, include f(f(1))=f(1), and determine f(1); verify all composed arguments lie in the domain.

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.