Split (1)

train · 92 rows

id stringlengths 21 45	source stringclasses 4 values	problem_nl stringlengths 39 973	lean_code stringlengths 67 446	expected_verdict stringclasses 2 values	error_primary stringclasses 6 values	error_secondary listlengths 0 2	error_tags listlengths 0 3	meta_tags listlengths 0 2	gold_explanation stringlengths 70 283
combibench_apmo_1991_p2	CombiBench	Suppose there are 997 points given in a plane. If every two points are joined by a line segment with its midpoint coloured in red, show that there are at least 1991 red points in the plane.	theorem apmo_1991_p2 (points : Fin 997 → ℝ × ℝ) : (red_points points).card ≥ 1991 := by sorry	YES	specification_error	[]	[ "distinctness_missing" ]	[]	If points are not distinct, the bound fails (e.g., all points equal), but the NL assumes distinct points.
combibench_apmo_2023_p1	CombiBench	Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.\nShow that it is possible to arrange these squares in a way ...	theorem apmo_2023_p1 (n : ℕ) (h_n: n ≥ 5) : ∃ position: Fin n → ℝ × ℝ, (∀ n1 n2 : Fin n, n1 ≠ n2 → ¬ touches_interior_or_edge ⟨position n1, n1 + 1⟩ ⟨position n2, n2 + 1⟩) ∧ ∀m : Fin n, ∃ l k, m ≠ l ∧ m ≠ k ∧ l ≠ k ∧ touches ⟨position m, m + 1⟩ ⟨position l, l + 1⟩ ∧ touches ⟨position m, m + 1⟩ ⟨p...	YES	definition_mismatch	[ "formalization_error" ]	[ "encoding_issue", "goal_mismatch" ]	[]	The notion of 'touch / touch only at vertices' and the way 'exactly two touches' is counted appears under-specified/possibly mis-modeled; hard to verify without the definitions of touches*.
combibench_balticway_2015_p7	CombiBench	There are 100 members in a ladies' club. Each lady has had tea (in private) with exactly 56 of the other members of the club. The Board, consisting of the 50 most distinguished ladies, have all had tea with one another. Prove that the entire club may be split into two groups in such a way that, within each group, any l...	theorem balticway_2015_p7 (had_tea: SimpleGraph (Ladies)) [DecidableRel had_tea.Adj] (h_had_tea_with_56: ∀ l : Ladies, had_tea.degree l = 56) (h_board: ∃ board : Finset Ladies, board.card = 50 ∧ (∀ l1 l2: board, had_tea.Adj l1 l2)) : ∃ group1 group2: Finset Ladies, group1 ∪ group2 = Finset.univ ...	YES	definition_mismatch	[]	[ "library_usage_error" ]	[ "tooling_error" ]	∀ l1 l2 : board is ill-typed and does not express a clique; IsClique matches NL.
combibench_brualdi_ch10_34	CombiBench	Let $t$ be a positive integer. Prove that, if there exists a Steiner triple system of index 1 having $v$ varieties, then there exists a Steiner triple system having $v^{t}$ varieties.	theorem brualdi_ch10_34 (t v : ℕ) (ht : t > 1): Nonempty (SteinerTripleSystemOfIndOne 2 3 v) → Nonempty (SteinerTripleSystemOfIndOne 2 3 (v ^ t)) := by sorry	YES	specification_error	[ "definition_mismatch" ]	[ "incorrect_spec", "encoding_issue" ]	[]	NL says 'STS of index 1 ⇒ STS (some index) on v^t'; original conclusion stays in 'index 1' and assumption t>1 vs 'positive integer' is unclear without STS definitions.
combibench_brualdi_ch10_60	CombiBench	Prove that a symmetric, idempotent Latin square has odd order.	theorem brualdi_ch10_60 {n : ℕ} (L : LatinSquare n) : IsIdempotentElem L.1 ∧ L.1.IsSymm → Odd n := by sorry	YES	specification_error	[ "definition_mismatch" ]	[ "base_case_missing", "library_usage_error" ]	[]	NL assumes positive order and idempotence on the diagonal; original predicate name is unclear and no positivity guard was present.
combibench_brualdi_ch11_5	CombiBench	Use the pigeonhole principle to prove that a graph of order n ≥ 2 always has two vertices of the same degree.	theorem brualdi_ch11_5 {V : Type*} (n : ℕ) (h_n: n ≥ 2) (G : SimpleGraph (Fin n)) [DecidableRel G.Adj] : ∃ v1 v2, v1 ≠ v2 ∧ G.degree v1 = G.degree v2 := by sorry	NO	null	[]	[]	[ "unused_parameter", "code_style_issue" ]	Removal of an unused type parameter is refactor-only; original statement was still correct.
combibench_brualdi_ch14_26	CombiBench	How many different necklaces are there that contain four red and three blue beads?	structure PreNecklaces where c : Fin 7 → Fin 2 deriving Fintype	YES	specification_error	[]	[ "missing_hypothesis" ]	[]	NL requires exactly four red and three blue beads, but the original structure had no count constraint. Note: the proposed fix with ∃ S, S.card = 4 only enforces at-least-4-red, not exactly-4-red.
combibench_brualdi_ch1_5	CombiBench	Find the number of different perfect covers of a 3-by-4 chessboard by dominoes.	theorem brualdi_ch1_5 : Fintype.card PerfectCover = brualdi_ch1_5_solution := by sorry	YES	domain_mismatch	[ "definition_mismatch" ]	[ "type_mismatch", "encoding_issue" ]	[]	Refactor to explicit domino modeling and specialization; original already fixed board size, so correctness is unclear.
combibench_brualdi_ch2_36	CombiBench	Determine the total number of combinations (of any size) of a multiset of objects of $k$ different types with finite repetition numbers $n_1, n_2, \ldots, n_k$, respectively.	theorem brualdi_ch2_36 {k : ℕ} (n : Fin k → ℕ) (sols : Finset (Fin k → ℕ)) (h_sols : ∀ f, f ∈ sols ↔ (∀ i, f i ≤ n i)) : sols.card = ((fun n => (∑ i : Fin k, (n i + 1))) : (Fin k → ℕ) → ℕ) n := by sorry	YES	definition_mismatch	[]	[ "wrong_operator" ]	[]	The number of combinations of a multiset with repetition bounds n_i is the product ∏(n_i+1), not the sum ∑(n_i+1). Sum gives wrong answer for any k>1.
combibench_brualdi_ch2_6	CombiBench	How many integers greater than 5400 have both of the following properties? (a) The digits are distinct. (b) The digits 2 and 7 do not occur.	theorem brualdi_ch2_6 (s : Finset ℕ) (hs0 : ∀ n ∈ s, n > 5400) (hs1 : ∀ n ∈ s, (Nat.digits 10 n).Nodup) (hs2 : ∀ n ∈ s, 2 ∉ (Nat.digits 10 n) ∧ 7 ∉ (Nat.digits 10 n)) : s.card = brualdi_ch2_6_solution := by sorry	YES	specification_error	[]	[ "incomplete_spec" ]	[]	Before only required s subset of valid integers, so s.card is not determined; need iff.
combibench_brualdi_ch3_18	CombiBench	Prove that of any five points chosen within a square of side length 2 , there are two whose distance apart is at most $\sqrt{2}$.	theorem brualdi_ch3_18 (points : Fin 5 → Set.Icc (0 : ℝ) 2 × Set.Icc (0 : ℝ) 2) : ∃ i j, i ≠ j ∧ Dist.dist (points i) (points j) ≤ √2 := by sorry	YES	domain_mismatch	[ "definition_mismatch" ]	[ "type_mismatch", "paper_vs_lean_semantics" ]	[ "tooling_error" ]	Original used Set.Icc × Set.Icc which is ill-typed. Also ℝ×ℝ would have wrong metric (sup not Euclidean). EuclideanSpace is correct. Expert confirmed (Joseph Myers).
combibench_brualdi_ch4_35	CombiBench	The complement $\bar{A}$ of an $r$-subset $A$ of $\{1,2, \ldots, n\}$ is the $(n-r)$-subset of $\{1,2, \ldots, n\}$, consisting of all those elements that do not belong to $A$. Let $M=\binom{n}{r}$, the number of $r$-subsets and, at the same time, the number of $(n-r)$ subsets of $\{1,2, \ldots, n\}$. Prove that, if $A...	theorem brualdi_ch4_35 (r n M : ℕ) (hM : M = ((@Finset.univ (Fin n)).powersetCount r).card) (A : Fin M → (Finset.powersetCount r (@Finset.univ (Fin M) _))) : ∀ i j, (List.Lex (fun x1 x2 : Fin M => x1 ≤ x2) (Finset.sort (· ≤ ·) (A i)) (Finset.sort (· ≤ ·) (A j))) → (List.Lex (fun x1 x2 : Fin M => x1 ≤ x2...	YES	domain_mismatch	[]	[ "type_mismatch" ]	[]	Statement is about r-subsets of {1..n}, but before used Fin M elements and lex order on M, so the domain is wrong.
combibench_brualdi_ch4_59	CombiBench	Let $n \geq 2$ be an integer. Prove that the total number of inversions of all $n$ ! permutations of $1,2, \ldots, n$ equals $\frac{1}{2} n!\binom{n}{2}=n!\frac{n(n-1)}{4}$ (Hint: Pair up the permutations so that the number of inversions in each pair is $\frac{n(n-1)}{2}$.)	def invNum {n : ℕ} (σ : Equiv.Perm (Fin n)) : ℕ := ∑ x ∈ Equiv.Perm.finPairsLT n, if σ x.fst ≤ σ x.snd then 0 else 1	YES	definition_mismatch	[]	[ "wrong_operator" ]	[]	Before counted inversions (pairs where σ(i) > σ(j) for i < j). After counts non-inversions. Both give same sum over all permutations due to symmetry, so theorem remains correct either way.
combibench_brualdi_ch4_9	CombiBench	Show that the largest number of inversions of a permutation of ${1, 2, ... , n}$ equals $\frac{n(n -1)}{2}$.	def invNum {n : ℕ} (σ : Equiv.Perm (Fin n)) : ℕ := ∑ x ∈ Equiv.Perm.finPairsLT n, if σ x.fst ≤ σ x.snd then 0 else 1	YES	definition_mismatch	[]	[ "wrong_operator" ]	[]	Before counted inversions, after counts non-inversions. Both give same maximum n(n-1)/2 (reverse perm has max inversions = identity has max non-inversions), so theorem correct either way.
combibench_brualdi_ch6_9	CombiBench	Determine the number of integral solutions of the equation $x_{1}+x_{2}+x_{3}+x_{4}=20$ that satisfy $1 \leq x_{1} \leq 6,0 \leq x_{2} \leq 7,4 \leq x_{3} \leq 8,2 \leq x_{4} \leq 6$.	theorem brualdi_ch6_9 : {x : Fin 4 → ℕ \| x 0 ∈ Icc 1 6 ∧ x 1 ∈ Icc 0 7 ∧ x 2 ∈ Icc 4 8 ∧ x 3 ∈ Icc 2 6}.ncard = brualdi_ch6_9_solution := by sorry	YES	specification_error	[]	[ "incomplete_spec" ]	[]	Missing constraint x1+x2+x3+x4=20; before counts all tuples in bounds.
combibench_brualdi_ch9_13	CombiBench	Let $A$ be a matrix with $n$ columns, with integer entries taken from the set $S=\{1,2, \ldots, k\}$. Assume that each integer $i$ in $S$ occurs exactly $n r_{i}$ times in $A$, where $r_{i}$ is an integer. Prove that it is possible to permute the entries in each row of $A$ to obtain a matrix $B$ in which each integer $...	theorem brualdi_ch9_13 (n m : ℕ) (k : ℕ+) (A : Matrix (Fin m) (Fin n) ℕ) (hA : ∀ i j, A i j ∈ Finset.Icc 1 k.1) : ∃ (rσ : Fin m → Equiv.Perm (Fin n)), ∀ j : Fin n, ∀ i ∈ Set.Icc 1 k.1, (∑ x : Fin m, if A x ((rσ x).symm j) = x then 1 else 0) * n = (∑ x : Fin m, ∑ y : Fin n, if A x y = i then 1 else 0) :=...	YES	specification_error	[ "quantifier_indexing_mismatch", "formalization_error" ]	[ "missing_hypothesis", "variable_mismatch", "goal_mismatch" ]	[]	Before omits the nr_i counting hypothesis and compares A x ... to x instead of i; it does not encode the NL statement.
combibench_egmo_2022_p5	CombiBench	For all positive integers n, k, let f(n, 2k) be the number of ways an n × 2k board can be fully covered by nk dominoes of size 2 × 1. Find all positive integers n such that for every positive integer k, the number f(n, 2k) is odd.	structure PerfectCover (n k : ℕ) where d_set : Finset (Domino n k) d_card : d_set.card = n * k covers : ∀ i : Fin n × Fin (2 * k), ∃ d ∈ d_set, i ∈ d.carrier	YES	definition_mismatch	[]	[ "encoding_issue" ]	[]	PerfectCover uses Finset of labeled Dominos and distinguishes two squares of each domino. Count is multiplied by (nk)!·2^(nk). Problem asks when count is odd, so this bug matters. Expert confirmed (Joseph Myers).
combibench_hackmath_3	CombiBench	How many four-digit numbers can be formed from the numbers 3 5 8 9 if they are not allowed to be repeated?	theorem hackmath_3 (sol : Finset ℕ) (h_sol : ∀ s ∈ sol, 1000 ≤ s ∧ s ≤ 9999 ∧ (Nat.digits 10 s).toFinset = {3, 5, 8, 9}) : sol.card = hackmath_3_solution := by sorry	YES	specification_error	[]	[ "incomplete_spec" ]	[]	Before only required sol subset of valid numbers, so sol.card is not fixed; need iff.
combibench_hackmath_4	CombiBench	How many people must be in a group for at least two of them to be born in the same month?	theorem hackmath_4 : IsLeast {n \| ∀ f : Fin n → Fin 12, ∃ a b, f a = f b} hackmath_4_solution := by sorry	YES	specification_error	[]	[ "distinctness_missing" ]	[]	Without a ≠ b, the pigeonhole condition is trivially satisfied by taking a=b, so the least n is wrong.
combibench_hackmath_8	CombiBench	A ferry with a capacity of 10 people takes a group of 13 men and 7 women across a river. Find the number of ways in which the group may be taken across the if all women go on the first group.	theorem hackmath_8 (sols : Finset ((Fin 13 → Fin 2) × (Fin 7 → Fin 2))) (h_women : ∀ f ∈ sols, ∀ i, f.2 i = 0) (h_sols : ∀ f, f ∈ sols ↔ ∀ k, ((List.ofFn f.1).count k + (List.ofFn f.2).count k = 10)) : sols.card = hackmath_8_solution := by sorry	YES	specification_error	[]	[ "incomplete_spec" ]	[]	Women constraint was outside the membership iff, so sols could be any subset; NL needs the exact set.
combibench_imo_2000_p4	CombiBench	A magician has one hundred cards numbered $1$ to $100$. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience selects two of the three boxes, chooses one card from each and announces the sum of the numbers on the chosen cards. Given t...	theorem imo_2000_p4 (good_allocations : Finset Allocation) (h : ∀ f ∈ good_allocations, ∃ (t : Trick), trick_works f t) : good_allocations.card = imo_2000_p4_solution := by sorry	YES	specification_error	[ "domain_mismatch" ]	[ "missing_hypothesis", "incomplete_spec", "domain_of_variables_mismatch" ]	[]	Cards should be 1..100 and good_allocations should be exactly surjective allocations with a working trick; before is underconstrained and off by domain.
combibench_imo_2005_p6	CombiBench	In a mathematical competition, in which 6 problems were posed to the participants, every two of these problems were solved by more than 2/5 of the contestants. Moreover, no contestant solved all the 6 problems. Show that there are at least 2 contestants who solved exactly 5 problems each.	theorem imo_2005_p6 {participants : Type} [Fintype participants] [DecidableEq participants] (solved : Fin 6 → Finset participants) (h : ∀ i j, i ≠ j → (solved i ∩ solved j).card > (2 * Fintype.card participants : ℝ) / 5) (h' : ∀ i, (solved i).card < Fintype.card participants) : ∃ s : Finset participants...	YES	quantifier_indexing_mismatch	[]	[ "quantifier_mismatch" ]	[]	NL says each participant missed at least one problem; original only said each problem had someone who missed it.
combibench_imo_2010_p5	CombiBench	Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed: Type 1) Choose a non-empty box $B_j$, $1\leq j \leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$; Type 2) Choose a non-empty box $B_k$, $1\leq k \leq 4$, remove one coin fro...	def op22 : Boxes → Boxes \| ⟨b1, b2, b3, b4, b5, b6⟩ => ⟨b1, b2 - 4, b4, b3, b5, b6⟩	YES	definition_mismatch	[]	[ "wrong_constant" ]	[]	Type-1 operation removes one coin, but op22 subtracted 4; constant was wrong.
combibench_imo_2014_p2	CombiBench	Let $n\ge2$ be an integer. Consider an $n\times n$ chessboard consisting of $n^2$ unit squares. A configuration of $n$ rooks on this board is $\textit{peaceful}$ if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that, for each peaceful configuration of $n$ rooks, there...	theorem imo_2014_p2 (n : ℕ) (hn : n ≥ 2) : IsGreatest {(k : ℕ) \| (k > 0) ∧ ∀ r : peaceful_rooks n, ∃ i j : Fin n, i.val + k - 1 < n ∧ i.val + k - 1 < n ∧ ∀ m n, m.val < k ∧ n.val < k ∧ r.carrier (i + m) (j + n) = false} (imo_2014_p2_solution n) := by sorry	YES	formalization_error	[ "quantifier_indexing_mismatch" ]	[ "connective_mismatch", "variable_mismatch" ]	[]	Original required m<k ∧ n<k for all m,n, which is impossible; implication matches NL. Also had duplicated i.val + k - 1 < n where second should be j.val.
combibench_imo_2014_p6	CombiBench	A set of lines in the plane is in general position if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its finite regions. Prove that for all sufficiently large n, in any set of n lines in general ...	def finite_regions (S : Set (ℝ × ℝ)) (L : Finset (ℝ × ℝ × ℝ)) : Prop := S.Nonempty ∧ Bornology.IsBounded S ∧ ∀ l ∈ L, (∀ p ∈ S, on_line p l) ∨ (∀ p ∈ S, same_side_of_line p l)	YES	definition_mismatch	[]	[ "encoding_issue" ]	[]	finite_regions is true for any singleton point, making the theorem trivially false. Should define actual finite regions created by the line arrangement. Expert confirmed (Joseph Myers).
combibench_imo_2015_p1	CombiBench	We say that a finite set $\mathcal{S}$ in the plane is balanced if, for any two different points $A$, $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three points $A$, $B$, $C$ in $\mathcal{S}$, there is no point $P$ in $\mathcal{S}$ suc...	theorem imo_2015_p1 : (∀ n ≥ 3, ∃ (S : Finset (ℝ × ℝ)), balanced S ∧ S.card = n) ∧ {n \| n ≥ 3 ∧ (∃ (S : Finset (ℝ × ℝ)), balanced S ∧ centre_free S ∧ S.card = n)} = imo_2015_p1_solution := by sorry	YES	domain_mismatch	[ "definition_mismatch" ]	[ "type_mismatch", "paper_vs_lean_semantics" ]	[]	ℝ × ℝ uses sup/L∞ metric in Mathlib, not Euclidean. For balanced/centre-free sets with distance conditions, this fundamentally changes the problem. Expert confirmed (Joseph Myers).
combibench_imo_2017_p3	CombiBench	A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit's starting point, $A_0$, and the hunter's starting point, $B_0$, are the same. After $n-1$ rounds of the game, the rabbit is at point $A_{n-1}$ and the hunter is at point $B_{n-1}$. In the nth round of the game, three things occur in order....	theorem imo_2017_p3 (start : ℝ × ℝ) : imo_2017_p3_solution = ∀ (A : ℕ → (Fin 2 → ℝ)), A 0 = ![start.1, start.2] → ∀ n, dist (A n) (A (n + 1)) = 1 → (∃ (P : ℕ → (Fin 2 → ℝ)), ∀ n > 0, dist (P n) (A n) ≤ 1) → (∃ (B : ℕ → (Fin 2 → ℝ)), B 0 = ![start.1, start.2] ∧ ∀ n, dist (B n) (B (n + 1)) = 1 ∧ dist (A (...	YES	domain_mismatch	[ "definition_mismatch" ]	[ "type_mismatch", "paper_vs_lean_semantics" ]	[]	ℝ × ℝ uses sup/L∞ metric in Mathlib, not Euclidean. Hunter/rabbit game with 'distance exactly 1' has different meaning with wrong metric. Expert confirmed (Joseph Myers).
combibench_imo_2017_p5	CombiBench	An integer $N \ge 2$ is given. A collection of $N(N + 1)$ soccer players, no two of whom are of the same height, stand in a row. Sir Alex wants to remove $N(N - 1)$ players from this row leaving a new row of $2N$ players in which the following $N$ conditions hold: ($1$) no one stands between the two tallest players, ($...	theorem imo_2017_p5 (N : ℕ) (h_N : N ≥ 2) (height : Perm (Fin (N * (N + 1)))) : ∃ kept : Fin (2 * N) ↪o Fin (N * (N + 1)), ∀ i j, Even #{l \| height (kept l) < height (kept i)} → (∀ k, height (kept i) < height (kept k) ↔ height (kept j) ≤ height (kept k)) → (∀ k, kept i < kept k ↔ kept j ≤ kept k) ∨ ...	YES	definition_mismatch	[ "formalization_error" ]	[ "encoding_issue", "goal_mismatch" ]	[]	Original ordering condition replaced with count-of-shorter/no-between constraint; unclear which matches NL.
combibench_imo_2018_p3	CombiBench	An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from $...	theorem imo_2018_p3 : imo_2018_p3_solution = ∃ l, IsAntiPascal l ∧ List.range' 1 (∑ i ∈ Finset.Icc 1 2018, i) = List.insertionSort (· ≤ ·) (List.flatten l) := by	YES	definition_mismatch	[]	[ "encoding_issue" ]	[]	Triangle representation changed from lists to indexed function; likely modeling fix but not clearly wrong.
combibench_imo_2018_p4	CombiBench	A site is any point (x, y) in the plane such that x and y are both positive integers less than or equal to 20. Amy and Ben take turns placing stones. On her turn, Amy places a red stone such that the distance between any two red stones is not equal to √5. Find the greatest K such that Amy can ensure she places at least...	-- Site defined as subset of ℝ × ℝ -- Problem: ℝ × ℝ uses sup/L∞ metric in Mathlib, not Euclidean -- dist((1,0), (2,1)) = max(1,1) = 1 with sup metric -- dist((1,0), (2,1)) = √(1²+1²) = √2 with Euclidean -- The "distance √5" condition has wrong meaning!	YES	domain_mismatch	[ "definition_mismatch" ]	[ "type_mismatch", "paper_vs_lean_semantics" ]	[]	ℝ×ℝ (Prod) uses sup/L∞ metric in Mathlib. The problem requires Euclidean distance for 'distance not equal to √5' (knight's move). With sup metric, √5 distance means something completely different. Expert confirmed (Joseph Myers).
combibench_imo_2023_p5	CombiBench	A Japanese triangle consists of 1 + 2 + ... + n circles arranged in an equilateral triangular shape such that for each i = 1, 2, ..., n, the ith row contains exactly i circles. Each circle is colored red or blue. A ninja path starts from a top-row circle and in each step goes to one of the two circles immediately below...	def triangleGraph (n : ℕ+) : Digraph (Index n) where Adj p q := p.atBottomLeft q ∨ p.atBottomRight q ∨ q.atBottomLeft p ∨ q.atBottomRight p	YES	definition_mismatch	[]	[ "encoding_issue", "wrong_constant" ]	[]	Digraph edges go both up and down, but ninja paths must go down only. Also Real.log is natural log, not log₂ as needed. Expert confirmed (Joseph Myers).
combibench_imosl_2011_c6	CombiBench	Let $n$ be a positive integer and let $W=\ldots x_{-1} x_{0} x_{1} x_{2} \ldots$ be an infinite periodic word consisting of the letters $a$ and $b$. Suppose that the minimal period $N$ of $W$ is greater than $2^{n}$. A finite nonempty word $U$ is said to appear in $W$ if there exist indices $k \leq \ell$ such that $U=x...	theorem imosl_2011_c6 (W : ℤ → Fin 2) (n : ℕ+) (N : ℕ) (hN : 2 ^ n.1 < N) (hW : Function.Periodic W N) (hW' : ∀ N' < N, ¬ Function.Periodic W N') : ∃ (x : Fin n ↪ (Σ k, Fin k → Fin 2)), (∀ i, (x i).1 ≠ 0) ∧ (∀ i, ubiquitous W (x i)) := by sorry	YES	specification_error	[]	[ "missing_hypothesis" ]	[]	Period 0 always makes a function periodic, so forbidding all N'<N including 0 makes the premise impossible.
combibench_imosl_2015_c6	CombiBench	Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean.	theorem imosl_2015_c6 (S : Set ℕ+) : ∀ (N : ℕ), ∃ (m : ℕ), N < m ∧ ¬ clean S m := by sorry	YES	specification_error	[]	[ "missing_hypothesis" ]	[]	NL assumes S nonempty; original allowed empty S (statement may still hold), so mismatch is unclear.
combibench_usamo_2000_p4	CombiBench	Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.	theorem usamo_2000_p4 : IsLeast colored_card usamo_2000_p4_solution.1 := by sorry	YES	quantifier_indexing_mismatch	[]	[ "quantifier_mismatch" ]	[]	NL requires all size-n colorings force a right triangle; original only required existence of one such coloring.
formalmath_aime_all_2013_II_1	FormalMath	Time is converted to metric system: 10 metric hours per day, 100 metric minutes per hour. A person wants to wake at the equivalent of 6:36 AM. Find 100A+10B+C where A:BC is the metric time.	import Mathlib.Tactic theorem aime_all_2013_II_1 (A B C : ℕ) (h₀ : 0 ≤ A ∧ A < 10) (h₁ : 0 ≤ B ∧ B < 10) (h₂ : 0 ≤ C ∧ C < 10) (h₃ : (6 * 100 + 36) * 10 / (24 * 60) = 100 * A + 10 * B + C) : 100 * A + 10 * B + C = 275 := by	YES	definition_mismatch	[]	[ "misuse_of_concept", "wrong_constant" ]	[]	Uses Nat division so 6360/1440 truncates to 4; also uses 6100+36 instead of 660+36 for 6:36 AM. Hypothesis can force 100A+10B+C=4, not 275.
formalmath_aime_all_2017_I_10	FormalMath	Let z be the UNIQUE complex number with properties that (z₃-z₁)/(z₂-z₁)·(z-z₂)/(z-z₃) is real and Im(z) is greatest possible. Find Re(z).	import Mathlib theorem aime_all_2017_I_10 (z1 z2 z3 z : ℂ) ... (hzz : ∃ r : ℝ, ...) (hzim : ∀ z', ... → z.im ≥ z'.im) : z.re = 56 := by	YES	specification_error	[]	[ "incomplete_spec" ]	[]	NL specifies a unique maximizer z; the Lean statement (as sketched) does not encode unique existence (∃!) / a uniqueness hypothesis, only properties of a given z.
formalmath_olymid-ref-base_3054	FormalMath	Find all positive integers k and n such that k⁴ + n² is divisible by 7^k - 3^n. Prove that the answer is: (k, n) = (2, 4)	import Mathlib theorem olymid_ref_base_3054 (k n : ℕ) : k > 0 ∧ n > 0 ∧ 7^k - 3^n ∣ k^4 + n^2 ↔ (k = 2 ∧ n = 4) := by	YES	domain_mismatch	[]	[ "truncation_issue" ]	[]	In ℕ, 7^k - 3^n truncates to 0 when 3^n > 7^k; at (2,4) denominator becomes 0, so '0 ∣ 32' is false and the claimed ↔ fails.
formalmath_olymid-ref-base_4847	FormalMath	Let t=2016 and p=ln 2. Evaluate ∑_{k=1}^∞ (1 - ∑_{n=0}^{k-1} e^{-t}t^n/n!) (1-p)^{k-1} p	import Mathlib open Real theorem olymid_ref_base_4847 (t : ℝ) (p : ℝ) (ht : t = 2016) (hp : p = log 2) : ∑' k : ℕ, ((1 - ∑ n in Finset.range k, exp (-t) * t ^ n / Nat.factorial n) * (1 - p) ^ (k - 1) * p) = 1 - (1 / 2) ^ 2016 := by	YES	quantifier_indexing_mismatch	[]	[ "indexing_mismatch" ]	[]	NL sums k=1..∞ but Lean uses ∑' k : ℕ starting at k=0, adding an extra k=0 term (and k-1 is Nat subtraction, so at k=0 it becomes 0, not −1).
formalmath_olymid-ref-base_5274	FormalMath	Find the smallest positive integer n for which 1!2!⋯(n-1)! > n!²	import Mathlib open Nat BigOperators theorem olymid_ref_base_5274 : IsLeast {n : ℕ \| 0 < n ∧ ∏ i in Finset.Icc 1 (n - 1), i ! > (n !) ^ 2} 8 := by	NO	null	[]	[]	[]	This appears correctly encoded: product over Icc 1 (n-1) is 1!⋯(n−1)!, n>0 is enforced, and 8 is indeed the smallest n satisfying the inequality.
formalmath_olymid-ref-base_6282	FormalMath	Let n ≥ 3. Let a_1 < a_2 < ... < a_n be positive reals with a_{n+1} = a_1. Prove ∑_{k=1}^n (a_k/a_{k+1}) > ∑_{k=1}^n (a_{k+1}/a_k).	import Mathlib open Real theorem olymid_ref_base_6282 {n : ℕ} (hn : 3 ≤ n) (a : Fin (n + 1) → ℝ) (ha : StrictMono a) (hpos : ∀ i, 0 < a i) (han : a n = a 0) : ∑ k, a k / a (k + 1) > ∑ k, a (k + 1) / a k := by	YES	formalization_error	[ "quantifier_indexing_mismatch" ]	[ "premise_translation_error", "indexing_mismatch" ]	[]	StrictMono a on Fin(n+1) implies a 0 < … < a n, but also assumes a n = a 0; premises become inconsistent and mis-encode the cyclic condition a_{n+1}=a_1.
formalmath_olymid-ref-base_8963	FormalMath	The Fibonacci numbers F_0, F_1, F_2, ... are defined by F_0=0, F_1=1, F_{n+1}=F_n+F_{n-1}. Given n ≥ 2, find the smallest size of a set S of INTEGERS such that for every k=2,...,n there exist x,y ∈ S with x-y=F_k.	import Mathlib def fib (n : ℕ) : ℕ := match n with \| 0 => 0 \| 1 => 1 \| n + 2 => fib (n + 1) + fib n theorem olymid_ref_base_8963 (n : ℕ) (hn : 2 ≤ n) : IsLeast {k \| ∃ S : Finset ℕ, S.card = k ∧ ∀ k ∈ Finset.Icc 2 n, ∃ x ∈ S, ∃ y ∈ S, x - y = fib k} (Nat.ceil (n / 2) + 1) := by	YES	domain_mismatch	[]	[ "type_mismatch", "truncation_issue" ]	[]	NL needs a set of integers and differences x−y with sign; Lean uses Finset ℕ and Nat subtraction x-y, which can't represent negative differences and truncates when y>x.
formalmath_olymid_ref_base_2137	FormalMath	Let a₁,...,a₂₀₁₉ be positive integers. Prove: ∃ real x such that aᵢ = ⌊ix⌋ for all i ∈ {1,...,2019} iff for all i,j with i+j ≤ 2019, aᵢ + aⱼ ≤ aᵢ₊ⱼ ≤ aᵢ + aⱼ + 1.	import Mathlib theorem olymid_ref_base_2137 (a : Fin 2019 → ℤ) (ha : ∀ i, 0 < a i) : (∃ x : ℝ, ∀ i, a i = ⌊i * x⌋) ↔ (∀ i j, i + j ≤ 2019 → a i + a j ≤ a (i + j) ∧ a (i + j) ≤ a i + a j + 1) := by	YES	domain_mismatch	[ "quantifier_indexing_mismatch" ]	[ "domain_of_variables_mismatch", "indexing_mismatch", "bound_mismatch" ]	[]	Uses Fin 2019 indices (includes 0, wraps on addition) but NL uses 1..2019 with ordinary addition; i=0 forces ⌊0*x⌋=0 contradicting positivity, and (i+j) in Fin wraps mod 2019 rather than staying ≤2019.
formalmath_omni_theorem_131	FormalMath	What is the smallest integer n, greater than one, for which the root-mean-square of the first n positive integers is an integer? The RMS is defined as sqrt((a_1² + ... + a_n²)/n).	import Mathlib open Real noncomputable def sumOfSquares (n : ℕ) : ℕ := ∑ k in Finset.Icc 1 n, k ^ 2 theorem omni_theorem_131 : IsLeast {n \| 1 < n ∧ ∃ k : ℤ, sqrt (sumOfSquares n / n) = k} 337 := by	YES	definition_mismatch	[]	[ "misuse_of_concept" ]	[]	RMS needs real division sqrt((∑k^2)/n), but Lean computes Nat division sumOfSquares n / n first (truncates) then coerces to ℝ for sqrt.
formalmath_omni_theorem_1586	FormalMath	Let {a_n}_{n≥1} be arithmetic and {g_n}_{n≥1} be geometric such that first four terms of {a_n+g_n} are 0,0,1,0. Find the 10th term.	import Mathlib theorem omni_theorem_1586 (a g : ℕ → ℝ) (ha : ∃ d, ∀ n, a n = a 1 + d * (n - 1)) (hg : ∃ r, ∀ n, g n = g 1 * r ^ (n - 1)) (h : a 1 + g 1 = 0) (h' : a 2 + g 2 = 0) (h'' : a 3 + g 3 = 1) (h''' : a 4 + g 4 = 0) : a 10 + g 10 = -54 := by	YES	domain_mismatch	[]	[ "domain_of_variables_mismatch", "truncation_issue" ]	[]	Sequences are intended for n≥1, but ha/hg quantify ∀ n:ℕ with (n-1) using Nat subtraction; at n=0 this forces a0=a1 and g0=g1, adding unintended constraints.
formalmath_omni_theorem_2830	FormalMath	Find quadratic polynomials f(x) = ax² + bx + c where a,b,c are positive integers, p<q are primes, f(p)=f(q)=17, f(p+q)=47. Find sum of f(pq) mod 100.	import Mathlib theorem omni_theorem_2830 : (∑ᶠ x ∈ {x \| ∃ a b c p q : ℕ, a > 0 ∧ p.Prime ∧ q.Prime ∧ p < q ∧ x = a * (p * q) ^ 2 + b * (p * q) + c ∧ a * p ^ 2 + b * p + c = 17 ∧ a * q ^ 2 + b * q + c = 17 ∧ a * (p + q) ^ 2 + b * (p + q) + c = 47}, x) % 100 = 71 := by	YES	problem_statement_error	[]	[ "unprovable_problem" ]	[]	As written ('a,b,c are positive integers'), the constraints force b = −a(p+q) < 0, so no solutions exist; likely the NL intended b to be an integer (not necessarily positive).
formalmath_omni_theorem_2926	FormalMath	What is the smallest integer that can be placed in the box so that 1/2 < □/9? Prove that the answer is: 5	import Mathlib theorem omni_theorem_2926 : IsLeast {n \| 1 / 2 < n / 9} 5 := by	YES	definition_mismatch	[]	[ "misuse_of_concept" ]	[]	Inferred type is ℕ, so 1/2=0 and n/9 is Nat division; predicate becomes 0 < n/9, whose least solution is 9, not 5.
formalmath_omni_theorem_4006	FormalMath	Let S(n) be the sum of the digits of the positive integer n. Find all n such that S(n)(S(n)-1) = n-1. Prove that the answer is: n ∈ {1, 13, 43, 91, 157}	import Mathlib def S : ℕ → ℕ \| n => (Nat.digits 10 n).sum theorem omni_theorem_4006 {n : ℕ} : (S n) * (S n - 1) = n - 1 ↔ n = 1 ∨ n = 13 ∨ n = 43 ∨ n = 91 ∨ n = 157 := by	YES	specification_error	[]	[ "base_case_missing" ]	[]	NL says positive n, but Lean allows n=0; then (S 0)*(S 0 - 1)=0=(0-1) in ℕ, so LHS true while RHS 'n∈{1,13,43,91,157}' is false.
formalmath_quantitative_reasoning_zh_blue_41	FormalMath	Given M = {x \| x = a² + 1, a ∈ ℕ}, N = {x \| x = b² - 4b + 5, b ∈ ℕ}, prove M ⊊ N. The proof shows 1 ∈ N (at b=2: 4-8+5=1) but 1 ∉ M.	import Mathlib theorem quantitative_reasoning_zh_blue_41 : {x : ℕ \| ∃ a : ℕ, 0 < a ∧ x = a^2 + 1} ⊂ {x : ℕ \| ∃ b : ℕ, 0 < b ∧ x = b^2 - 4 * b + 5} := by	YES	domain_mismatch	[ "definition_mismatch" ]	[ "truncation_issue", "encoding_issue" ]	[]	b^2 - 4*b + 5 is computed in ℕ with truncating subtraction; the NL witness b=2 gives 1, but Lean gives 5, so N is not the intended set.
formalmath_quantitative_reasoning_zh_blue_891	FormalMath	Given 0 ≤ x_0 < 1, the sequence {x_n} satisfies x_{n+1} = 2x_n - 1 if 1/2 ≤ x_n < 1, else x_{n+1} = 2x_n if 0 ≤ x_n < 1/2. If x_5 = x_0, how many such sequences exist? Answer: 31	import Mathlib open Real open scoped BigOperators theorem quantitative_reasoning_zh_blue_891 : Set.ncard {x : ℕ → ℝ \| ∃ x0 : ℝ, 0 ≤ x0 ∧ x0 < 1 ∧ (∀ n : ℕ, x (n + 1) = if 1 / 2 ≤ x n ∧ x n < 1 then 2 * x n - 1 else 2 * x n) ∧ x 5 = x0} = 31 := by	YES	specification_error	[]	[ "incomplete_spec" ]	[]	Does not enforce x 0 = x0 and does not maintain the invariant x n ∈ [0,1); the piecewise recurrence is only correct under those missing constraints.
formalmath_theorem_proving_zh_blue_548	FormalMath	Find the maximum cardinality of a set S of positive integers ≤100 such that for any two distinct elements a,b ∈ S: (1) there exists c ∈ S with gcd(a,c)=1 and gcd(b,c)=1, and (2) there exists d ∈ S with d≠a, d≠b, gcd(a,d)>1, gcd(b,d)>1. Solution proves \|S\|_max = 72.	import Mathlib theorem algebra_587982 : IsGreatest {n \| ∃ S : Finset ℕ, S.card = n ∧ (∀ x ∈ S, x ≤ 100) ∧ (∀ a ∈ S, ∀ b ∈ S, a ≠ b → ∃ c ∈ S, Nat.gcd a c = 1 ∧ Nat.gcd b c = 1) ∧ (∀ a ∈ S, ∀ b ∈ S, a ≠ b → ∃ d ∈ S, d ≠ a ∧ d ≠ b ∧ Nat.gcd a d > 1 ∧ Nat.gcd b d > 1)} 54 := by	YES	definition_mismatch	[]	[ "wrong_constant" ]	[ "answer_handling_mismatch" ]	Lean claims the maximum is 54, but the provided solution text claims \|S\|_max = 72; the constant in IsGreatest is wrong.
formalmath_theorem_proving_zh_blue_614	FormalMath	If the sequence {a_n} satisfies that for any n ∈ ℕ*, ∑_{d\|n} a_d = 2^n, prove that n \| a_n.	import Mathlib theorem algebra_56552 {a : ℕ → ℕ} (ha : ∀ n, ∑ d in n.divisors, a d = 2 ^ n) : ∀ n, n ∣ a n := by	YES	specification_error	[]	[ "base_case_missing" ]	[]	NL quantifies n∈ℕ* but Lean has ∀ n including 0; Nat.divisors 0 is empty in Mathlib, so ha 0 would require 0=2^0=1 (inconsistent), making the theorem vacuous.
formalmath_theorem_proving_zh_blue_671	FormalMath	Let F be a finite set of integers satisfying: (1) for any x ∈ F, there exist y,z ∈ F with x = y + z; (2) there exists n ∈ ℕ* such that for any 1 ≤ k ≤ n and any x_1,...,x_k ∈ F, we have ∑x_i ≠ 0. Prove \|F\| ≥ 2n + 2.	import Mathlib theorem theorem_proving_zh_blue_671 {F : Finset ℤ} (hF : ∀ x ∈ F, ∃ y z, x = y + z) (hn : ∃ n : ℕ, 0 < n ∧ ∀ k ∈ Finset.Icc 1 n, ∀ x : Fin k → ℤ, (∀ i, x i ∈ F) → ∑ i, x i ≠ 0) : F.card ≥ 2 * n + 2 := by	YES	quantifier_indexing_mismatch	[ "specification_error" ]	[ "variable_mismatch", "incomplete_spec" ]	[]	Conclusion uses free n not bound by the ∃n in hn; also hF forgets to require y,z ∈ F (it only states ∃ y z with x=y+z).
formalmath_theorem_proving_zh_blue_681	FormalMath	Let f(m,n) satisfy f(1,n) = f(m,1) = 1 (m,n ∈ ℕ*), and when m,n ≥ 2, f(m,n) ≤ f(m,n-1) + f(m-1,n). Prove f(m,n) ≤ C(m+n-2, m-1).	import Mathlib theorem algebra_539172 {f : ℕ → ℕ → ℕ} (h₁ : ∀ n, f 1 n = 1) (h₂ : ∀ m, f m 1 = 1) (h₃ : ∀ m n, 2 ≤ m → 2 ≤ n → f m n ≤ f m (n - 1) + f (m - 1) n) : ∀ m n, f m n ≤ (m + n - 2).choose (m - 1) := by	YES	specification_error	[]	[ "missing_hypothesis" ]	[]	NL is for m,n∈ℕ* but Lean concludes ∀ m n without positivity constraints; this introduces m=0 or n=0 cases where (m+n-2).choose(m-1) and n-1 style terms are not the intended spec.
formalmath_u-math_684	FormalMath	An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P, who contracted the flu t days after it broke out is given by the model P = t² - 12t + 180, where 1 ≤ t ≤ 6. Find the day that 160 students had the flu.	import Mathlib theorem u_math_684 (t : ℕ) (ht : 1 ≤ t ∧ t ≤ 6) : let P := t ^ 2 - 12 * t + 180 P = 160 ↔ t = 2 := by	YES	domain_mismatch	[]	[ "truncation_issue" ]	[]	Uses ℕ subtraction in t^2 - 12*t; at t=2, (4-24) truncates to 0 so P=180, not 160, so the ↔ fails.
formalmath_u-math_921	FormalMath	Compute the integral: ∫ -1/(3·sin(x/3)⁶) dx. Answer involves +C.	import Mathlib theorem u_math_921 : ∫ x, -1 / (3 * sin (x / 3) ^ 6) = cos (x / 3) / (5 * sin (x / 3) ^ 5) - 4 / 15 * (-cos (x / 3) / sin (x / 3) ^ 3 - 2 / tan (x / 3)) + C := by	YES	definition_mismatch	[ "quantifier_indexing_mismatch" ]	[ "misuse_of_concept", "variable_mismatch" ]	[]	Lean's '∫ x, …' is a definite integral (not an antiderivative), and the RHS contains free x and free C (autoImplicit/scope bug), so it does not represent '+C' indefinite integration.
formalmath_u-math_945	FormalMath	Compute the integral: ∫ 10/sin(4x)^6 dx	import Mathlib open Real Set open scoped BigOperators theorem u_math_945 : ∫ x, 10 / sin (4 * x) ^ 6 = - (cot (4 * x)) ^ 5 / 2 - 5 / 2 * cot (4 * x) - 5 / 3 * (cot (4 * x)) ^ 3 + C := by	YES	definition_mismatch	[ "quantifier_indexing_mismatch" ]	[ "misuse_of_concept", "variable_mismatch" ]	[]	Same calculus issue: '∫ x, …' is a definite integral, and the RHS contains free x and free C, so it does not formalize an indefinite integral '+C'.
proofnet_artin_exercise_11_4_6b	ProofNet	Prove that $x^2+1$ is irreducible in $\mathbb{F}_7$	theorem exercise_11_4_6b {F : Type*} [Field F] [Fintype F] (hF : card F = 31) : Irreducible (X ^ 3 - 9 : Polynomial F) := sorry	YES	specification_error	[ "formalization_error" ]	[ "incorrect_spec", "goal_mismatch" ]	[]	Original assumes card F = 31 and proves Irreducible (X^3-9), but NL is about F_7 and x^2+1.
proofnet_axler_exercise_3_8	ProofNet	Suppose that $V$ is finite dimensional and that $T \in \mathcal{L}(V, W)$. Prove that there exists a subspace $U$ of $V$ such that $U \cap \operatorname{null} T=\{0\}$ and range $T=\{T u: u \in U\}$.	theorem exercise_3_8 {F V W : Type*} [AddCommGroup V] [AddCommGroup W] [Field F] [Module F V] [Module F W] (L : V →ₗ[F] W) : ∃ U : Submodule F V, U ⊓ (ker L) = ⊥ ∧ (range L = range (domRestrict L U)) := sorry	YES	specification_error	[]	[ "missing_hypothesis" ]	[]	NL assumes V is finite-dimensional, but original Lean omits [FiniteDimensional F V].
proofnet_axler_exercise_6_16	ProofNet	Suppose $U$ is a subspace of $V$. Prove that $U^{\perp}=\{0\}$ if and only if $U=V$	theorem exercise_6_16 {K V : Type*} [RCLike K] [NormedAddCommGroup V] [InnerProductSpace K V] {U : Submodule K V} : U.orthogonal = ⊥ ↔ U = ⊤ := sorry	NO	null	[]	[]	[]	No semantic issue: U.orthogonal = bot iff U = top matches 'U^perp = {0} iff U=V'.
proofnet_axler_exercise_6_3	ProofNet	Prove that $\left(\sum_{j=1}^{n} a_{j} b_{j}\right)^{2} \leq\left(\sum_{j=1}^{n} j a_{j}{ }^{2}\right)\left(\sum_{j=1}^{n} \frac{b_{j}{ }^{2}}{j}\right)$ for all real numbers $a_{1}, \ldots, a_{n}$ and $b_{1}, \ldots, b_{n}$.	theorem exercise_6_3 {n : ℕ} (a b : Fin n → ℝ) : (∑ i, a i * b i) ^ 2 ≤ (∑ i : Fin n, i * a i ^ 2) * (∑ i, b i ^ 2 / i) := sorry	YES	quantifier_indexing_mismatch	[]	[ "indexing_mismatch" ]	[]	NL sums j=1..n; original uses i:Fin n (0..n-1), so weights use i not (i+1) and introduces division by i=0.
proofnet_dummit_foote_exercise_4_5_13	ProofNet	Prove that a group of order 56 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.	theorem exercise_4_5_13 {G : Type*} [Group G] [Fintype G] (hG : card G = 56) : ∃ (p : ℕ) (P : Sylow p G), P.Normal := sorry	YES	specification_error	[]	[ "incomplete_spec" ]	[]	Same as 4_5_14: missing p.Prime and p \| card G; otherwise trivial p=1-style witnesses are allowed.
proofnet_dummit_foote_exercise_4_5_14	ProofNet	Prove that a group of order 312 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.	theorem exercise_4_5_14 {G : Type*} [Group G] [Fintype G] (hG : card G = 312) : ∃ (p : ℕ) (P : Sylow p G), P.Normal := sorry	YES	specification_error	[]	[ "incomplete_spec" ]	[]	Original misses the constraints 'p is prime' and 'p \| \|G\|'; otherwise p=1 gives a trivial normal Sylow 1-subgroup-style witness.
proofnet_herstein_exercise_2_2_5	ProofNet	Let $G$ be a group in which $(a b)^{3}=a^{3} b^{3}$ and $(a b)^{5}=a^{5} b^{5}$ for all $a, b \in G$. Show that $G$ is abelian.	def exercise_2_2_5 {G : Type} [Group G] (h : ∀ (a b : G), (a b) ^ 3 = a ^ 3 * b ^ 3 ∧ (a * b) ^ 5 = a ^ 5 * b ^ 5) : CommGroup G := sorry	YES	definition_mismatch	[]	[ "paper_vs_lean_semantics" ]	[]	Conclusion 'CommGroup G' can mean existence of some commutative group structure on the type G, not commutativity of the given [Group G] multiplication; NL needs forall a b, ab=ba.
proofnet_herstein_exercise_2_8_15	ProofNet	Prove that if $p > q$ are two primes such that $q \mid p - 1$, then any two nonabelian groups of order $pq$ are isomorphic.	def exercise_2_8_15 {G H: Type} [Fintype G] [Group G] [Fintype H] [Group H] {p q : ℕ} (hp : Nat.Prime p) (hq : Nat.Prime q) (h : p > q) (h1 : q ∣ p - 1) (hG : card G = pq) (hH : card G = pq) : G ≃ H := sorry	YES	quantifier_indexing_mismatch	[ "specification_error" ]	[ "variable_mismatch", "missing_hypothesis" ]	[]	Original typo uses hH : card G = p*q instead of card H; also missing 'G nonabelian' and 'H nonabelian' hypotheses required by NL.
proofnet_ireland_rosen_exercise_2_4	ProofNet	If $n$ is a positive integer such that $n \equiv 7 \pmod{8}$, show that $n$ is not a sum of three squares.	theorem exercise_2_4 {n : ℕ} (hn : n % 8 = 7) : ¬∃ (a b c : ℤ), n = a^2 + b^2 + c^2 := sorry	YES	domain_mismatch	[]	[ "type_mismatch" ]	[ "tooling_error" ]	Original equates n:N with a^2+b^2+c^2:Z; needs explicit cast (n:Z). Likely does not typecheck as written.
proofnet_ireland_rosen_exercise_4_11	ProofNet	Prove that $1^{k}+2^{k}+\cdots+(p-1)^{k} \equiv 0(p)$ if $p-1 \nmid k$ and $-1(p)$ if $p-1 \mid k$.	theorem exercise_4_11 {p : ℕ} (hp : p.Prime) (k s: ℕ) (s := ∑ n : Fin p, (n : ℕ) ^ k) : ((¬ p - 1 ∣ k) → s ≡ 0 [MOD p]) ∧ (p - 1 ∣ k → s ≡ 0 [MOD p]) := sorry	YES	quantifier_indexing_mismatch	[ "definition_mismatch" ]	[ "indexing_mismatch", "wrong_constant" ]	[]	Original sums over Fin p (includes 0, excludes 1-indexing) and states the divisible case is =0, but NL says it should be = -1 (mod p).
proofnet_munkres_exercise_38_6	ProofNet	Let $X$ be completely regular. Show that $X$ is connected if and only if the Stone-Čech compactification of $X$ is connected.	theorem exercise_38_6 {X : Type} (X : Type) [TopologicalSpace X] [RegularSpace X] (h : ∀ x A, IsClosed A ∧ ¬ x ∈ A → ∃ (f : X → I), Continuous f ∧ f x = (1 : I) ∧ f '' A = {0}) : IsConnected (univ : Set X) ↔ IsConnected (univ : Set (StoneCech X)) := sorry	YES	definition_mismatch	[]	[ "encoding_issue" ]	[ "code_style_issue" ]	Original 'completely regular' hypothesis uses f '' A = {0}, which fails for A=empty (image is empty), so the encoding is wrong/vacuous; also has redundant/shadowed X binder.
proofnet_rudin_exercise_2_28	ProofNet	Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable.	theorem exercise_2_28 (X : Type*) [MetricSpace X] [SeparableSpace X] (A : Set X) (hA : IsClosed A) : ∃ P₁ P₂ : Set X, A = P₁ ∪ P₂ ∧ IsPerfect P₁ ∧ Set.Countable P₂ := sorry	NO	null	[]	[]	[]	No semantic issue: statement matches NL (perfect set + countable set decomposition of closed sets in separable metric spaces).
proofnet_rudin_exercise_3_21	ProofNet	If $\left\{E_{n}\right\}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X$, if $E_{n} \supset E_{n+1}$, and if $\lim _{n \rightarrow \infty} \operatorname{diam} E_{n}=0,$ then $\bigcap_{1}^{\infty} E_{n}$ consists of exactly one point.	theorem exercise_3_21 {X : Type*} [MetricSpace X] [CompleteSpace X] (E : ℕ → Set X) (hEcl : ∀ n, IsClosed (E n)) (hEne : ∀ n, (E n).Nonempty) (hEbdd : ∀ n, Bornology.IsBounded (E n)) (hE : ∀ n, E n ⊇ E (n + 1)) (hE' : Tendsto (λ n => Metric.diam (E n)) atTop (𝓝 0)) : ∃ a, Set.iInter E = {a} := sorry	NO	null	[]	[]	[]	No semantic issue: standard Cantor intersection theorem variant is correctly expressed.
proverbench_aime_2024ii_p13	ProverBench	Let ω≠1 be a 13th root of unity. Find the remainder when ∏_{k=0}^{12}(2-2ω^k+ω^{2k}) is divided by 1000. Show that it is 321.	import Mathlib import Aesop set_option maxHeartbeats 0 open BigOperators Real Nat Topology Rat theorem aime_2024ii_p13 (ω : ℂ) (h₀ : ω = Complex.exp (2 * ↑Real.pi * Complex.I / 13)) : (∏ k in Finset.range 13, (2 - 2 * ω ^ k + ω ^ (2 * k))) % 1000 = 321 := by	YES	domain_mismatch	[]	[ "type_mismatch" ]	[ "tooling_error" ]	Applies % 1000 to a complex-valued product; the intended workflow is: prove the product is an integer N, then take N % 1000. As written it's ill-typed or meaningless.
proverbench_calculus__p1	ProverBench	The integral of \[(x^3 - 3cosx) dx\] is \[\frac{x^4}{4} - 3sinx + C\].	import Mathlib /- The integral of \[(x^3 - 3cosx) dx\] is \[\frac{x^4}{4} - 3sinx + C\]. -/ -- Define the function f(x) noncomputable def f (x : ℝ) : ℝ := x^3 - 3 * Real.cos x -- Define the integral of f(x) noncomputable def integralf (x : ℝ) : ℝ := (x^4 / 4) - 3 * Real.sin x -- Theorem statement theorem integral_...	YES	definition_mismatch	[ "quantifier_indexing_mismatch" ]	[ "misuse_of_concept", "quantifier_mismatch" ]	[]	Lean uses ∫ x, f x (a definite integral over ℝ with measure), not an indefinite integral/antiderivative; also C is treated as arbitrary, but the constant of integration should be existential/handled via derivatives.
proverbench_calculus__p10	ProverBench	The integral $\displaystyle \int \frac{1}{\sqrt{81-x^2}}\,dx$ is equivalent to $\arcsin\left(\frac{x}{9}\right)+C$.	import Mathlib /- The integral $\displaystyle \int \frac{1}{\sqrt{81-x^2}}\,dx$ is equivalent to $\arcsin\left(\frac{x}{9}\right)+C$. -/ open Real theorem integral_of_one_over_sqrt_eighty_one_minus_x_squared (x C : ℝ): ∫ x, 1 / sqrt (81 - x^2) = arcsin (x / 9) + C :=	YES	definition_mismatch	[ "quantifier_indexing_mismatch" ]	[ "misuse_of_concept", "quantifier_mismatch" ]	[]	Mis-models an indefinite integral as a definite integral; C is treated as an arbitrary parameter rather than part of an antiderivative characterization.
proverbench_calculus__p12	ProverBench	The integral of $\displaystyle \frac{1}{x^2-9}\,dx$ is equal to $\frac{1}{6}\ln\left\|\frac{x-3}{x+3}\right\|+C$.	import Mathlib /- The integral of $\displaystyle \frac{1}{x^2-9}\,dx$ is equal to $\frac{1}{6}\ln\left\|\frac{x-3}{x+3}\right\|+C$. -/ open Real theorem integral_of_rational_function_with_log : ∀ (x : ℝ), x ≠ 3 ∧ x ≠ -3 → (∫ (dx : ℝ), 1 / (x^2 - 9)) = (1 / 6) * log (abs ((x - 3) / (x + 3))) + C :=	YES	definition_mismatch	[ "quantifier_indexing_mismatch" ]	[ "misuse_of_concept", "variable_mismatch" ]	[ "autoimplicit_bug" ]	C is unbound; also the integral is over dx but the integrand doesn't depend on dx (it depends on x), so it's not an antiderivative statement and is semantically wrong.
proverbench_calculus__p14	ProverBench	The value of the integral ∭_D e^{x²+y²} dV where D is bounded by planes y=0, z=0, y=x, and paraboloid z=4-x²-y² is (e^4-5)π/8.	import Mathlib open MeasureTheory variable {D : Type} [MeasurableSpace D] [MeasureSpace D] variable (D : Set (Fin 3 → ℝ)) theorem integral_value_over_region (D : Set (Fin 3 → ℝ)) (hD : D = {p : Fin 3 → ℝ \| p 2 ≥ 0 ∧ p 3 ≥ 0 ∧ p 2 ≤ p 1 ∧ p 3 ≤ 4 - p 1^2 - p 2^2}) : let integrand := fun p : Fin 3 → ℝ => Real.exp ...	YES	quantifier_indexing_mismatch	[]	[ "indexing_mismatch" ]	[]	Uses p 3 with p : Fin 3 → ℝ; in Lean (3 : Fin 3) = 0, so 'p 3' is actually p 0, breaking the intended coordinate constraints for the 3D region.
proverbench_calculus__p15	ProverBench	Prove that $\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac12$.	import Mathlib /- Prove that $\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac12$. -/ open Filter Topology /-- Version 1: Using explicit domain restriction -/ theorem limit_of_cosine_difference_quotient : ¬ (Tendsto (fun x => if x ≠ 0 then (1 - Real.cos x) / x^2 else 1/2) (𝓝 0) (𝓝 (1/2)) ) :=	YES	formalization_error	[]	[ "connective_mismatch" ]	[]	The Lean statement negates the limit (¬ Tendsto ...), while the NL statement asserts the limit equals 1/2.
proverbench_calculus__p17	ProverBench	The integral $\int \frac{1}{\cos^2 x} \, dx$ equals to $\tan x + C$.	import Mathlib /- The integral $\int \frac{1}{\cos^2 x} \, dx$ equals to $\tan x + C$. -/ -- Define the constant C for the integration constant variable (C : ℝ) -- Theorem: The indefinite integral of 1/cos^2(x) with respect to x is tan(x) + C. theorem integral_of_one_over_cos_sq (x : ℝ) : ∫ (x : ℝ), 1 / (Real.c...	YES	definition_mismatch	[ "quantifier_indexing_mismatch" ]	[ "misuse_of_concept", "quantifier_mismatch" ]	[]	Indefinite integral is mis-modeled as ∫ x, ... (definite integral), and '+ C' is encoded as an arbitrary parameter rather than antiderivative up to a constant.
proverbench_calculus__p3	ProverBench	The local maximum of the function f(x)=x^4/4 - x^3/3 - 2x^2 + 4x + 5 on the interval [-3,3] is 41/4 and is attained at x=3, and the local minimum is -13/3 and it is attained for x=-2.	import Mathlib noncomputable def f (x : ℝ) : ℝ := x^4 / 4 - x^3 / 3 - 2 * x^2 + 4 * x + 5 theorem extrema_on_interval : ¬ ((∃ x : ℝ, x ∈ Set.Icc (-3) 3 ∧ (∀ y ∈ Set.Icc (-3) 3, f y ≤ f x) ∧ f x = 41/4 ∧ x = 3) ∧ (∃ z : ℝ, z ∈ Set.Icc (-3) 3 ∧ (∀ y ∈ Set.Icc (-3) 3, f z ≤ f y) ∧ f z = - 13 / 3 ∧ z = -2)) :=	YES	formalization_error	[]	[ "connective_mismatch" ]	[]	The Lean statement is negated (¬(...)) but NL asserts the extrema claims are true; it proves the opposite.
proverbench_calculus__p5	ProverBench	The integral of $\int \frac{1}{x^2+1} \, dx$ is $\arctan x + C$.	import Mathlib /- The integral of $\int \frac{1}{x^2+1} \, dx$ is $\arctan x + C$. -/ open Real noncomputable def f (x:ℝ ):ℝ := 1 / (x^2 + 1) theorem integral_of_inv_sq_plus_one (x C: ℝ) : ∫ x , f x = arctan x + C :=	YES	definition_mismatch	[ "quantifier_indexing_mismatch" ]	[ "misuse_of_concept", "quantifier_mismatch" ]	[]	Indefinite integral '= arctan x + C' is not expressed by ∫ x, f x = ...; C is treated as arbitrary rather than handled via derivative/∃ constant.
proverbench_calculus__p8	ProverBench	The surface integral ∬_G (x²+y²+2z) dS, where G is the part of the paraboloid z=7-x²-y² that lies above the xy-plane is equal to (957√29-47)/20 π.	import Mathlib variable {G : Set (ℝ × ℝ × ℝ)} (hG : G = {p : ℝ × ℝ × ℝ \| p.2.2 = 7 - p.1^2 - p.2.1^2 ∧ p.2.2 ≥ 0}) theorem paraboloid_integral_value : ∫ (x : ℝ) in Set.Icc (-Real.sqrt 7) (Real.sqrt 7), ∫ (y : ℝ) in Set.Icc (-Real.sqrt (7 - x^2)) (Real.sqrt (7 - x^2)), (x^2 + y^2 + 2(7 - x^2 - y^2)) = (957 Real.sqr...	YES	definition_mismatch	[]	[ "misuse_of_concept" ]	[]	NL asks for surface integral ∬ f dS but Lean computes double integral ∬ f dA. Missing surface element √(1 + (∂z/∂x)² + (∂z/∂y)²) = √(1 + 4x² + 4y²).
proverbench_calculus__p9	ProverBench	The integral of $\sin^3 x\cos^{6}x\,dx$ is equal to $-\frac{\cos^7 x}{7} + \frac{\cos^9 x}{9} + C$.	import Mathlib /- The integral of $\sin^3 x\cos^{6}x\,dx$ is equal to $-\frac{\cos^7 x}{7} + \frac{\cos^9 x}{9} + C$. -/ theorem integral_sin_cube_cos_six_power (x : ℝ) : ∫ (x : ℝ), Real.sin x ^ 3 * Real.cos x ^ 6 = -(Real.cos x ^ 7 / 7) + Real.cos x ^ 9 / 9 + C :=	YES	definition_mismatch	[ "quantifier_indexing_mismatch" ]	[ "misuse_of_concept", "variable_mismatch" ]	[ "autoimplicit_bug" ]	Uses ∫ x, ... as a definite integral, and C is free/unbound in the statement (scope/autoImplicit issue).
proverbench_elementary_algebra__p1	ProverBench	For a polynomial $P$ of degree $n$ where $P(i)$ equals the remainder of $i$ modulo 2 for each $i=0,1,\dots,n$, $P(n+1)$ is 1 if $n$ is even and 0 if $n$ is odd.	import Mathlib /- For a polynomial $P$ of degree $n$ where $P(i)$ equals the remainder of $i$ modulo 2 for each $i=0,1,\dots,n$, $P(n+1)$ is 1 if $n$ is even and 0 if $n$ is odd. -/ variable {P : Polynomial ℤ} {n : ℕ} (hP : ∀ i ∈ Finset.range (n + 1), P.eval (i : ℤ) = i % 2) theorem polynomial_parit...	YES	problem_statement_error	[ "specification_error" ]	[ "unprovable_problem", "incomplete_spec" ]	[]	Even with degree = n, the claimed parity value at n+1 is not generally true over ℤ (fails for n=0, n=1, n=2); also Lean statement omits the degree constraint entirely.
proverbench_elementary_algebra__p15	ProverBench	Prove that if the difference of two consecutive cubes is n², n ∈ ℕ, then 2n-1 is a square.	import Mathlib variable (n : ℕ) theorem consecutive_cubes_difference_implies_square : ( (n + 1) ^ 3 - n ^ 3 = n ^ 2 ) → ∃ k : ℕ, k ^ 2 = 2 * n - 1 :=	YES	formalization_error	[ "quantifier_indexing_mismatch" ]	[ "premise_translation_error", "variable_mismatch" ]	[]	Uses same variable n for both the cube index and the square. NL says 'difference of two consecutive cubes is n²' meaning (m+1)³ - m³ = n² for some m. Correct formalization needs separate variables. Current premise is never satisfiable for n ∈ ℕ, making theorem vacuously true.
proverbench_elementary_algebra__p19	ProverBench	Proves the inequality: a²(s-a) + b²(s-b) + c²(s-c) ≤ (3/2)abc where a, b, c are real numbers.	import Mathlib theorem inequality_statement {a b c s : ℝ} : a^2 * (s - a) + b^2 * (s - b) + c^2 * (s - c) ≤ (3/2) * a * b * c :=	YES	problem_statement_error	[ "specification_error" ]	[ "incomplete_statement", "missing_hypothesis" ]	[]	Both NL and Lean are missing the constraint that s = (a+b+c)/2 (semi-perimeter). The NL problem statement is incomplete - this inequality only holds when s is the semi-perimeter. Without this constraint, the statement is FALSE. Counterexample: a=b=c=1, s=100 gives LHS=297 > RHS=1.5.
proverbench_linear_algebra__p13_1	ProverBench	Let T: Z × Z → Z be a linear function satisfying T(1, 0) = 3 and T(0, 1) = -5. Then for all (x, y) ∈ Z × Z, T(x, y) = 3x - 5y.	import Mathlib /- Let T: Z × Z → Z be a linear function satisfying T(1, 0) = 3 and T(0, 1) = -5. Then for all (x, y) ∈ Z × Z, T(x, y) = 3x - 5y. -/ open Int AddMonoidHom variable (T : ℤ × ℤ →+ ℤ) axiom T10 : T (1, 0) = 3 axiom T01 : T (0, 1) = -5 lemma T_expression : ∀ (x y : ℤ), T (x, y) = 3 * x - 5 * y :=	YES	formalization_error	[]	[ "premise_translation_error" ]	[ "unsound_axiom" ]	Constraints T(1,0)=3 and T(0,1)=-5 were introduced as global axioms (effectively for all T), rather than hypotheses about a specific T; this makes the context inconsistent (e.g. T=0).
proverbench_linear_algebra__p13_2	ProverBench	Let T: Z × Z → Z be a linear function satisfying T(1, 0) = 3 and T(0, 1) = -5. Then for all (x, y) ∈ Z × Z, T(x, y) = 3x - 5y.	import Mathlib /- Let T: Z × Z → Z be a linear function satisfying T(1, 0) = 3 and T(0, 1) = -5. Then for all (x, y) ∈ Z × Z, T(x, y) = 3x - 5y. -/ open Int AddMonoidHom variable (T : ℤ × ℤ →+ ℤ) axiom T10 : T (1, 0) = 3 axiom T01 : T (0, 1) = -5 axiom T_expression : ∀ (x y : ℤ), T (x, y) = 3 * x - 5 * y theore...	YES	formalization_error	[]	[ "premise_translation_error" ]	[ "unsound_axiom" ]	Same issue as p13_1: axioms assert properties for all T and even directly assert the conclusion, making the theorem trivial/unsound rather than a conditional statement.
proverbench_number_theory__p16	ProverBench	The smallest possible value for n in a library with n books, where if books are arranged in boxes with 7 books in each box, then 5 books remain, if arranged with 9 books in each box, then 3 books remain, and if arranged with 11 books in each box, then 7 books remain, is 579.	import Mathlib theorem smallest_library_books_value {n : ℕ} (h1 : n % 7 = 5) (h2 : n % 9 = 3) (h3 : n % 11 = 7) : n = 579 :=	YES	specification_error	[]	[ "missing_hypothesis" ]	[]	Missing minimality constraint. The theorem claims ALL n satisfying the CRT conditions equal 579, but n = 1272 (= 579 + lcm(7,9,11)) also satisfies h1, h2, h3 and 1272 ≠ 579. Should add hypothesis that n is the smallest such n.
proverbench_number_theory__p2	ProverBench	For the equation x² + y² - 1 = 4xy its general solution in the integers is given by x + u√3 = (2 + √3)^n, where u is the substitute for y - 2x.	import Mathlib theorem general_solution_quadratic_equation (x y : ℤ) (u : ℤ) (n : ℕ) : x^2 + y^2 - 1 = 4 * x * y → x + u * Real.sqrt 3 = (2 + Real.sqrt 3)^n := sorry	YES	problem_statement_error	[ "specification_error", "domain_mismatch" ]	[ "unprovable_problem", "missing_hypothesis", "type_mismatch" ]	[]	NL/formula is wrong as written (the Pell reduction gives u^2-3x^2=1 so the standard form is u + x√3 = (2+√3)^n); also missing u = y - 2x; Lean mixes ℤ variables with ℝ sqrt without proper spec/casts.
proverbench_number_theory__p26	ProverBench	For any real number $\alpha$, there exist infinitely many pairs of positive integers $(p, q)$ such that $\left\|\alpha - \frac{p}{q}\right\| < \frac{1}{q^2}$	import Mathlib /- For any real number $\alpha$, there exist infinitely many pairs of positive integers $(p, q)$ such that $\left\|\alpha - \frac{p}{q}\right\| < \frac{1}{q^2}$ -/ theorem infinitely_many_positive_integer_pairs_for_real_approximation (α : ℝ) : ∃ (p q : ℕ) (h : q > 0), ∀ (n : ℕ), n > 0 → ∃ (p_n q...	YES	problem_statement_error	[ "quantifier_indexing_mismatch" ]	[ "unprovable_problem", "quantifier_mismatch" ]	[]	As stated with p,q positive, the NL claim fails for negative α (e.g. α = -2); the Lean quantifier structure also doesn't encode 'infinitely many distinct pairs' or unbounded q. Variables p,q are unused (bound then ignored).
proverbench_number_theory__p27	ProverBench	Prove that $4kxy-1$ does not divide the number $x^m+y^n$ for any positive integers $x,y,k,m,n$.	import Mathlib /- Prove that $4kxy-1$ does not divide the number $x^m+y^n$ for any positive integers $x,y,k,m,n$. -/ theorem no_division {x y k m n : PNat} (h : ∃ z : PNat, (x : ℕ)^(m : ℕ) + (y : ℕ)^(n : ℕ) = z) : ¬((4 * k * x * y - 1) ∣ z) :=	YES	quantifier_indexing_mismatch	[]	[ "variable_mismatch" ]	[ "autoimplicit_bug" ]	z in the divisibility conclusion is not bound to the z from the ∃-hypothesis (autoImplicit-style scope bug), making the statement meaningless/false.
proverbench_number_theory__p28	ProverBench	Let $p$ be a prime number and $ N=\prod _{k=1}^{p-1}(k^2+1)$. The remainder of $N$ upon division by $p$ is 4 if $p$ is congruent to 3 modulo 4, and 0 if $p$ is congruent to 1 modulo 4.	import Mathlib /- Let $p$ be a prime number and $ N=\prod _{k=1}^{p-1}(k^2+1)$. The remainder of $N$ upon division by $p$ is 4 if $p$ is congruent to 3 modulo 4, and 0 if $p$ is congruent to 1 modulo 4. -/ open Finset theorem prime_remainder_theorem (p : ℕ) (hp : Prime p) : let N := (range (p - 1)).pr...	YES	quantifier_indexing_mismatch	[ "specification_error" ]	[ "indexing_mismatch", "missing_hypothesis" ]	[]	range (p-1) indexes k=0..p-2, not 1..p-1 as NL specifies; additionally 'remainder = 4' should be encoded as N % p = 4 % p (or require p>4), otherwise p=3 breaks literal remainder equality.
proverbench_number_theory__p6	ProverBench	The solutions for the equation x² - dy² = 1 in the set of rational numbers, for a given integer d, are x = (dt² + 1)/(dt² - 1) and y = 2t/(dt² - 1) where t ∈ ℚ.	import Mathlib variable (d : ℤ) theorem rational_solutions_of_pell_equation (t : ℚ) : ∃ (x y : ℚ), x = (d * t^2 + 1) / (d * t^2 - 1) ∧ y = 2 * t / (d * t^2 - 1) ∧ x^2 - d * y^2 = 1 :=	YES	specification_error	[]	[ "incomplete_spec", "division_by_zero_risk" ]	[]	Only proves 'for each t, the formulas give a solution' (one direction); NL reads like a full characterization of all rational solutions (↔). Also denominator d*t^2-1 can be 0, which needs an explicit guard in a faithful spec.
proverbench_number_theory__p7	ProverBench	Exactly one of the equations x² - py² = ±2 has an integral solution where p is a prime of the form 4k + 3.	import Mathlib variable {p : ℤ} (hp : Prime p) (hpForm : p % 4 = 3) theorem exactly_one_has_integral_solution: (∃ (x y : ℤ), x^2 - p * y^2 = 2) ∨ (∃ (x y : ℤ), x^2 - p * y^2 = -2) ∧ ¬((∃ (x y : ℤ), x^2 - p * y^2 = 2) ∧ (∃ (x y : ℤ), x^2 - p * y^2 = -2)) :=	YES	formalization_error	[]	[ "connective_mismatch" ]	[]	Due to precedence, Lean parses A ∨ (B ∧ ¬(A ∧ B)) instead of (A ∨ B) ∧ ¬(A ∧ B); that weakens 'exactly one'.

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.