lean_workbook_plus_52626

⚠️ 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.

so $f(x)=0$ for all $x\ge0$ , then by plugging $x>0$ into $P(-x,x$ we get $f(x)=0$ for all $x<0$ too.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_52626  (f : ℝ → ℝ)
  (h₀ : ∃ x, 0 ≤ x ∧ f x ≠ 0)
  (h₁ : ∀ x, ∀ y, x ≥ 0 ∧ y ≥ 0 → f (x + y) = f x + f y)
  (h₂ : ∀ x, ∀ y, x ≤ 0 ∧ y ≤ 0 → f (x + y) = f x + f y)
  (h₃ : ∀ x, ∀ y, x ≥ 0 ∧ y ≤ 0 → f (x + y) = f x + f y)
  (h₄ : ∀ x, ∀ y, x ≤ 0 ∧ y ≥ 0 → f (x + y) = f x + f y) :
  ∀ x, f x = 0   :=  by sorry

The source already established f=0 on nonnegative inputs; the formal premise instead asserts a nonzero value there.

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 ∀x≥0,f(x)=0 together with the original functional equation to extend to negative x; remove the contradictory-direction existential.

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.