lean_workbook_plus_51482

⚠️ 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.

Find all function $f:\mathbb {R}^{+} \rightarrow \mathbb {R}^{+}$ such that: $yf(y)f(f(x+y))=yf(x)f(y)+1$ $,\forall x,y \in \mathbb{R}^{+}$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_51482 (f : ℝ → ℝ) (hf: ∀ x y : ℝ, (0 < x ∧ 0 < y) → y * f y * f (f (x + y)) = y * f x * f y + 1) : ∃ x :ℝ, ∀ y : ℝ, (0 < x ∧ 0 < y) → f y = 1/y   :=  by sorry

The accepted witness x=0 makes the concluding implication false-premise for every y; it never uses the functional equation. The advertised f(y)=1/y should not be assumed to solve the written source equation.

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

Use f:ℝ_{>0}→ℝ_{>0} and state the complete solution classification under the exact functional equation. Remove the existential x guarding the conclusion; any claimed formula must be checked against the equation.

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.