lean_workbook_plus_60321

⚠️ 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.

We claim the answer is all $y$ such that $y \neq 0$ and $y \neq 2023+1/2022$ . Let $g(x)=x-f(x)$ . Then, the condition rewrites as $g(g(x))=1/(1-x)$ for all $x \neq 1$ . Equivalently, we need to prove that for all $y \neq 2023$ and $y \neq 1/(1-2023)$ , there is a function $g$ satisfying the above condition such that $g(2023)=y$ . Note that $g(2023)$ cannot equal $2023$ , as otherwise $1/(1-2023)=g(g(2023))=2023$ , a contradiction. Moreover, it cannot equal $1/(1-2023)$ either, as otherwise $g(-1/2022)=g(g(2023))=-1/2022$ , and so $g(u)=u$ for some $u$ , implying that $1/(1-u)=g(g(u))=u,$ hence $u-u^2=1,$ a contradiction. Now, it remains to prove that all other values are attainable.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_60321 :
  ∀ y, y ≠ 0 ∧ y ≠ 2023 + 1 / 2022 → ∃ f : ℝ → ℝ, ∀ x ≠ 1, f (f x) = x / (x - 1) ∧ f 2023 = y   :=  by sorry

The source transformation g=x−f is lost, the composition equation changes from 1/(1−x) to x/(x−1), and the allowed values are mixed between f and g.

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

Proposed corrected statement

Choose one function (original f or transformed g), state its exact equation and corresponding excluded values consistently, and then prove every allowed value at 2023 is attainable.

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.