lean_workbook_plus_60009

⚠️ 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.

There is no such $f\colon \mathbb{R} \to \mathbb{R}$ . By the formula given, we need $f(x)\neq 0$ for all $x$ . But $\dfrac {1}{f(f(x))} = f^{-1}(f(x)) = x$ , and for $x=0$ we get a contradiction.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_60009 : ¬∃ f : ℝ → ℝ, ∀ x, f x = 1 / f (f x)   :=  by sorry

The source distinguishes compositional inverse and reciprocal and requires nonzero values; the target states a different equation admitting f≡0.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Formalize f as a bijectionℝ→ℝ and assert its compositional inverse equals x↦1/f(x), with f nowhere zero. Then derive impossibility, e.g. evaluating at f(0). Recover the exact original inverse relation.

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.