lean_workbook_plus_22787

⚠️ 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.

Let $Y_1 = Y_{X+1}, Y_2 = Y_{X^2+X+1}, Y_3 = Y_{X^2+1}$ , and let $Y_4 = Y_1+Y_3$ . Note that if $S$ modulo $4$ has $a,b,c,d$ of $0,1,2,3$ , respectively, then $Y_1$ occurs if and only if $a+c = b+d$ , and $Y_3$ occurs if and only if $a = c, b=d$ . These two never coincide, so $Y_4$ is the probability that $P$ is divisible by one of $X+1, X^2+1$ . Furthermore, $Y_1 = \frac{\binom 63}{2^6} = \frac{5}{16}$ , while $Y_3 = \frac{5}{16} \cdot \frac{5}{16}$ , so $Y_4 = \frac{105}{256}$ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_22787  (x : ℕ)
  (y : ℕ → ℕ)
  (h₀ : y 1 = (Nat.choose 6 3) / 2^6)
  (h₁ : y 3 = (Nat.choose 6 3) / 2^6 * (Nat.choose 6 3) / 2^6)
  (h₂ : y 4 = y 1 + y 3)
  : y 4 = 105 / 256   :=  by sorry

Natural floor divisions make all displayed probabilities and the answer zero; the event model is absent.

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

Use real probabilities y₁=5/16,y₃=25/256,y₄=y₁+y₃=105/256. Separately define the actual events and prove disjointness before equating their union probability with a sum.

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.