lean_workbook_plus_30666

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Proved status carries no information about that problem. Do not import this node or use it as a dependency.

Now $(1,y)$ gives

and $(-1,y)$ gives

, if we let $k=y+1$ and subtract the 2 eqs we get

$(-1,-1)$ gives $f(1)=1$ so $\boxed{f(x)=-f(-x)}$ for all $x$ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_30666  (f : ℤ → ℤ)
  (h₀ : ∀ x, ∀ y, f (x + 1) = f 1 + (f 1) * f y)
  (h₁ : ∀ x, ∀ y, -f (-x - 1) = 1 + (f 1) * f y) :
  ∀ x, f x = -f (-x)   :=  by sorry

Independent x,y quantifiers turn the two equations into an inconsistent system.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

Proposed corrected statement

Tie each left argument to y: f(y+1)=f(1)+f(1)f(y) and−f(−y−1)=1+f(1)f(y), and include the source’s separately derived f(1)=1.

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.