lean_workbook_plus_50484

⚠️ 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.

$f(0)^2 = f(0)$ , so $f(0)\in \{0,1\}$ . If $f(0)= 0$ , then $P(x,0)$ gives $f\equiv 0$ , which does not work. So $f(0) = 1$ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_50484  (f : ℝ → ℝ)
  (h₀ : f 0^2 = f 0)
  (h₁ : ∀ x y, f (x^2 - y^2) = f x^2 - f y^2) :
  f 0 = 1   :=  by sorry

The provided source fragment does not state why the zero function fails, so the necessary condition cannot be reconstructed.

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

Proposed corrected statement

Restore the actual original functional equation and the nonzero/nondegeneracy hypothesis used to reject f≡0.

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.