lean_workbook_plus_69498

⚠️ 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.

$f(x)\ge x\quad\forall x>0$ . If $f(x)=x+d$ for some $d>0$ and $x\ge 2$ we get that contracdition and $f(x)=x\quad\forall x\ge 2$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_69498  (d : ℝ)
  (f : ℝ → ℝ)
  (h₀ : 0 < d)
  (h₁ : ∀ x, 0 < x → f x ≥ x)
  (h₂ : ∀ x, 2 ≤ x → f x = x + d) :
  ∀ x, 0 < x → f x = x   :=  by sorry

The source refers to an omitted contradiction from earlier conditions; the formal hypotheses admit f(x)=x+d.

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

Proposed corrected statement

Recover the original functional equation and prove identity on the intended range x≥2; do not conclude identity on all positive x from these two conditions.

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.