Datasets:
id stringlengths 19 59 | proof_index int64 0 2 | theorem_group stringlengths 64 64 | source stringclasses 5
values | training_source stringclasses 5
values | category stringclasses 19
values | lean_statement stringlengths 38 1.36k | problem stringlengths 44 1.18k | steps listlengths 1 95 | overall stringclasses 2
values | first_error_step int64 1 49 ⌀ | error_class stringclasses 6
values | known_issue stringclasses 2
values |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
numina_4d4910b6-f920-57d5-bfd2-43c8c69d6e2b | 0 | 2b79461e29e125330b01a54bd133076b48096bbb31583d245d4e5c9d1c57cb53 | NuminaMath-LEAN | numina | Number Theory | theorem negated_number_theory_63098 : {(a, b) : ℕ × ℕ | a > 0 ∧ b > 0 ∧ a ^ 2 + b ^ 2 ∣ a ^ 3 + 1 ∧ a ^ 2 + b ^ 2 ∣ b ^ 3 + 1} = {(1, 1)} := by | We prove that the set of pairs of positive integers (a,b) with a²+b² dividing both a³+1 and b³+1 is exactly the single pair (1,1). | [
{
"index": 1,
"text": "It suffices to show, for an arbitrary pair (a,b), that (a,b) satisfies the divisibility conditions if and only if (a,b) equals (1,1).",
"lean_code": " ext ⟨a, b⟩",
"label": "correct"
},
{
"index": 2,
"text": "Membership in the left-hand set means a>0, b>0, a²+b² ∣... | correct | null | null | |
lean_workbook_hf_29388 | 0 | 0a0afa690051debb4c9d1d782fd676b91717ddd05447af7828f9dbaf4b350a1f | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_29388 (a b c : ℕ) (h₁ : c ∣ b) (h₂ : a ∣ b / c) : a ∣ b | We must show that if c divides b and a divides the quotient b/c (integer division), then a divides b. | [
{
"index": 1,
"text": "Since c divides b exactly, dividing b by c and then multiplying back by c recovers b: that is, (b/c) · c = b.",
"lean_code": " have h3 : b / c * c = b := Nat.div_mul_cancel h₁",
"label": "correct"
},
{
"index": 2,
"text": "Because a divides b/c, multiplying the qu... | correct | null | null | |
lean_workbook_hf_29910 | 0 | 805c1031309dc532f83cf6f0688cab3ae4bece8af97640649b4f343946e35901 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_29910 (A B C : ℂ) : 1 + A / (z - 3) + B / (z - √3 * Complex.I) + C / (z + √3 * Complex.I) = 1 + A / (z - 3) + B / (z - √3 * Complex.I) + C / (z + √3 * Complex.I) | We must show that the expression 1 + A/(z − 3) + B/(z − √3·i) + C/(z + √3·i) equals itself, since both sides of the equation are written identically. | [
{
"index": 1,
"text": "Since the left-hand side and the right-hand side are literally the same expression term for term, the equality holds simply by the reflexive property of equality: any quantity equals itself.",
"lean_code": " rfl",
"label": "correct"
}
] | correct | null | null | |
lean_workbook_hf_34779 | 0 | cdd63f36e918fd04a1049e5b48bc1a4eb408b1ce1a9182f313852f669bfe3f63 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_34779 (hx : x ≥ 1 / Real.sqrt 3) : 3 * x ^ 2 ≥ x * Real.sqrt 3 | We must show that for a real number x satisfying x ≥ 1/√3, the inequality 3x^2 ≥ x√3 holds. | [
{
"index": 1,
"text": "Since 3 is a positive number, its square root √3 is also strictly positive.",
"lean_code": " have h3 : (0:ℝ) < Real.sqrt 3 := Real.sqrt_pos.mpr (by norm_num)",
"label": "correct"
},
{
"index": 2,
"text": "Rewrite the hypothesis x ≥ 1/√3 as 1/√3 ≤ x, and since √3 i... | correct | null | null | |
lean_workbook_hf_54412 | 0 | e46d54dba47841a8106f1bfb41d1af07b3037aeb6198b6c368868c43be654320 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_54412 (x y z : ℤ) (h₁ : x ^ 2 + y ^ 2 = z ^ 2) (h₂ : Odd x ∧ Odd y) : False | We must show that no integers x, y, z with x and y both odd can satisfy x^2 + y^2 = z^2; we derive a contradiction using residues modulo 4. | [
{
"index": 1,
"text": "Split the hypothesis into the two separate facts that x is odd and y is odd.",
"lean_code": " obtain ⟨hx, hy⟩ := h₂",
"label": "correct"
},
{
"index": 2,
"text": "Since x is odd, write x = 2a + 1 for some integer a.",
"lean_code": " obtain ⟨a, ha⟩ := hx",
... | correct | null | null | |
lean_workbook_hf_27740 | 0 | 1746b681818f7ea0e36ecc76b572eb2ceb66cfe857b00983df7467d3925b7851 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_27740 (x : ℝ) (hx : x = 15 / 2 * 1 / 2 * 9 / 2 * 5 / 2) : Real.sqrt x = 15 / 4 * Real.sqrt 3 | We must show that the square root of the number x = 15/2 · 1/2 · 9/2 · 5/2 equals 15/4 times the square root of 3. | [
{
"index": 1,
"text": "Replace x throughout the goal by the explicit product 15/2 · 1/2 · 9/2 · 5/2, so the goal becomes an equation about this specific numeral.",
"lean_code": " subst hx",
"label": "correct"
},
{
"index": 2,
"text": "Compute the numeral directly: 15/2 · 1/2 · 9/2 · 5/2... | correct | null | null | |
lean_workbook_hf_23853 | 0 | ad7cb182f142dba52f57ca5c5ad0da679adde14ce05a741039b1a6af2ba53725 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_23853 (c : ℝ) : ∀ ε : ℝ, ε > 0 → ∃ N : ℕ, ∀ x : ℝ, x > N → |c - c| < ε | We must show that for every positive real number ε, there is a natural number N such that for every real x greater than N, the absolute value of c minus c is less than ε. | [
{
"index": 1,
"text": "Fix an arbitrary positive real number ε, and assume ε > 0.",
"lean_code": " intro ε hε",
"label": "correct"
},
{
"index": 2,
"text": "Choose N = 0 as the natural number witness; we will show it satisfies the required property for any x greater than 0.",
"lean_... | correct | null | null | |
lean_workbook_hf_37608_18939 | 0 | d084979ddbd39542cde791d2de4481908e06b92cb9f8aa433135f77984567467 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_37608_18939 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a + b + c = 1) : a ^ 2 + b ^ 2 + c ^ 2 = 1 → a ^ 3 / (2 * b + 3 * c) + b ^ 3 / (2 * c + 3 * a) + c ^ 3 / (2 * a + 3 * b) ≥ 1 / 5 | We are given positive reals a, b, c with a+b+c = 1, and we must show that if additionally a^2+b^2+c^2=1, then the sum of the three fractions a^3/(2b+3c) + b^3/(2c+3a) + c^3/(2a+3b) is at least 1/5. We will show that the extra hypothesis a^2+b^2+c^2=1, combined with a+b+c=1 and positivity of a,b,c, is actually impossibl... | [
{
"index": 1,
"text": "Assume the hypothesis a^2 + b^2 + c^2 = 1 holds; call it h2. We now aim to derive a contradiction from this together with a+b+c=1 and a,b,c>0.",
"lean_code": " intro h2",
"label": "correct"
},
{
"index": 2,
"text": "Recall the algebraic identity (a+b+c)^2 = a^2+b^... | correct | null | null | |
lean_workbook_hf_82547 | 0 | 26f2cc45ccf1c414e77fb4dee53d8b2c76c895342cd6a87d7c8784fc81dc996b | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_82547 (X : ℕ → ℝ) (μ : ℝ) (p : ℕ → ℝ) (hp : ∀ x, p x = (Real.exp (-μ) * μ ^ x)/x!) : ∃ M : ℝ → ℝ, ∀ t, M t = ∑' x : ℕ, (p x) * (Real.exp (t * x)) | We want to show that there exists a function M of a real variable t such that for every t, M(t) equals the infinite sum over natural numbers x of p(x) times exp(t·x). To do this we simply exhibit the function defined directly by this formula and check that it satisfies the required equality. | [
{
"index": 1,
"text": "Define M to be exactly the function sending t to the sum ∑' x, p(x)·exp(t·x). With this choice, for every t the value M(t) is by definition equal to that same sum, so the required equation M(t) = ∑' x, p(x)·exp(t·x) holds immediately for all t, since both sides are literally the same ... | correct | null | null | |
lean_workbook_hf_25155 | 0 | 34fe1765192f365f0c8f03996111f477495a26349279ef74d8b6aa482f4d44cc | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_25155 (x : ℝ) : x = x | We must show that for any real number x, x equals itself. | [
{
"index": 1,
"text": "This is immediate, since any quantity is equal to itself by the basic principle of reflexivity of equality: x = x holds identically for every real number x.",
"lean_code": " rfl",
"label": "correct"
}
] | correct | null | null | |
lean_workbook_hf_34349 | 0 | 0aabcb8e9cf1a703947dfbf163d58c993cbb7983c691c2ce4f39a05be5909998 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_34349 : c + d = 7 ∧ c * d = 9 → c ^ 3 + d ^ 3 = 154 | We must show that if two numbers c and d satisfy c + d = 7 and c·d = 9, then c^3 + d^3 = 154. | [
{
"index": 1,
"text": "Assume the two given conditions: h1 states c + d = 7, and h2 states c·d = 9.",
"lean_code": " rintro ⟨h1, h2⟩",
"label": "correct"
},
{
"index": 2,
"text": "We use the algebraic identity (c + d)^3 = c^3 + d^3 + 3·c·d·(c + d), which follows from expanding the cube ... | correct | null | null | |
lean_workbook_hf_24237 | 0 | 6bf82db2e2a15c004124ba9ad437d6539f33b3245f33a9934c040dd82f89e897 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_24237 (R : Type*) [Ring R] [Finite R] (x : R) : ∃ n : ℕ, n • x = 0 | We must show that for a finite ring R and any element x of R, there exists a natural number n such that adding x to itself n times yields zero, that is, n copies of x sum to the additive identity. | [
{
"index": 1,
"text": "We choose the witness n = 0 for the existential statement; it remains to check that adding x to itself zero times gives zero.",
"lean_code": " use 0",
"label": "correct"
},
{
"index": 2,
"text": "By the basic property of repeated addition in any additive monoid, t... | correct | null | null | |
lean_workbook_hf_21534 | 0 | 0e92ff062017f1a3192bf611440ee24c49625d6f38f676d083bf5cc6f830e851 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_21534 : 6 ∣ a + b + c ↔ 6 ∣ a ^ 3 + b ^ 3 + c ^ 3 | We must show that for natural numbers a, b, c, the number 6 divides a+b+c exactly when 6 divides a^3+b^3+c^3. | [
{
"index": 1,
"text": "Since 6 is a nonzero natural number, we record this fact, as it is needed to relate divisibility by 6 to vanishing of residues modulo 6.",
"lean_code": " haveI : NeZero (6 : ℕ) := ⟨by decide⟩",
"label": "correct"
},
{
"index": 2,
"text": "We verify that cubing fix... | correct | null | null | |
numina_0cea2807-3780-5baf-9c23-44688a23cfae | 0 | ae63d9e52e31927c369f4d786307d1ec7905570dced7b8997f3869dbc423521e | NuminaMath-LEAN | numina | Algebra | theorem number_theory_604934 (k : ℝ) (hk : k > 0) (h : k ^ 2 - 1998 * k - 1 = 0)
(x : ℕ → ℕ) (hx0 : x 0 = 1) (hxn : ∀ n, x (n + 1) = ⌊k * x n⌋) :
x 1998 % 1998 = 1000 := by | We are given a positive real number k satisfying k² − 1998k − 1 = 0, and a sequence of natural numbers with x₀ = 1 and xₙ₊₁ = ⌊k·xₙ⌋; we must show that x₁₉₉₈ leaves remainder 1000 upon division by 1998. | [
{
"index": 1,
"text": "Rearranging the defining equation k² − 1998k − 1 = 0 gives the convenient form k² = 1998k + 1.",
"lean_code": " have hk2 : k ^ 2 = 1998 * k + 1 := by linarith [h]",
"label": "correct"
},
{
"index": 2,
"text": "Since k(k − 1998) = k² − 1998k = (1998k + 1) − 1998k =... | correct | null | null | |
lean_workbook_hf_22089 | 0 | 8b169fda417fc134f5e06bd00ea4742d6347033b357652668246bb63ef7248d4 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_22089 : ∀ a b c : ℝ, (a ≤ 0 ∧ 0 ≤ b ∧ 0 ≤ c ∨ 0 ≤ a ∧ b ≤ 0 ∧ 0 ≤ c ∨ 0 ≤ a ∧ 0 ≤ b ∧ c ≤ 0) → 2 * (a ^ 2 + b ^ 2 + c ^ 2 - a * b - b * c - c * a) ^ 2 ≥ a ^ 2 * (b - c) ^ 2 + b ^ 2 * (c - a) ^ 2 + c ^ 2 * (a - b) ^ 2 | We must show that for all real numbers a, b, c satisfying one of the three sign patterns (exactly one of a, b, c is nonpositive while the other two are nonnegative), the quantity 2 times the square of a²+b²+c²−ab−bc−ca is at least a²(b−c)²+b²(c−a)²+c²(a−b)². | [
{
"index": 1,
"text": "Fix arbitrary real numbers a, b, c and assume the sign hypothesis h that one of the three disjuncts holds.",
"lean_code": " intro a b c h",
"label": "correct"
},
{
"index": 2,
"text": "We split into the three cases: (i) a ≤ 0 with b, c ≥ 0; (ii) b ≤ 0 with a, c ≥ ... | correct | null | null | |
lean_workbook_hf_6574 | 0 | 10cb4803d2cf70ad632d43c70237bbb96220a80870e29ea97d90b0cb43638a1b | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_6574 (p q : ℕ) (hp : p.Prime) (hq : q.Prime) (hpq : p < q) : ∃ a b, a ≤ (p-1)/2 ∧ b ≤ (p-1)/2 ∧ 1 + a * (p-1) ≡ 1 + b * (p-1) [ZMOD q] | We must show that for any primes p and q with p < q, there exist natural numbers a and b, each at most (p-1)/2, such that 1 + a(p-1) and 1 + b(p-1) are congruent modulo q. | [
{
"index": 1,
"text": "We choose the simplest possible candidates, taking a = 0 and b = 0, and this reduces the problem to establishing three separate facts: that 0 ≤ (p-1)/2, that 0 ≤ (p-1)/2 again, and that 1 + 0·(p-1) is congruent to 1 + 0·(p-1) modulo q.",
"lean_code": " refine ⟨0, 0, ?_, ?_, ?_⟩",... | correct | null | null | |
lean_workbook_hf_17722_14430 | 0 | 2142655371544848de4681bf94be50532275bccdf11b883affcb98ebc6447d24 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_17722_14430 (a b c : ℝ) (ha : a > 0) (hb : b > 0) (hc : c > 0) (hab : a + b + c = 3) : 3 / 2 * (a + b + c) + 1 / a + 1 / b + 1 / c ≥ 15 / 2 | We must show that for positive reals a, b, c summing to 3, the quantity (3/2)(a+b+c) + 1/a + 1/b + 1/c is at least 15/2. | [
{
"index": 1,
"text": "We first establish the tangent-line bound at the point a = 1: for a positive, 1/a is at least 2 − a.",
"lean_code": " have hA : 1/a ≥ 2 - a := by",
"label": "correct"
},
{
"index": 2,
"text": "Since a square of a real number is never negative, (a − 1)^2 ≥ 0.",
... | correct | null | null | |
lean_workbook_hf_30307 | 0 | 03e2943941435d422a8a93f2a43e0284e359f0de959110c32229f26bc09c1b5d | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_30307 : ∀ a b : ℝ, a > 0 ∧ b > 0 → a * b * (a ^ 2 + b ^ 2 - 2) ≥ (a + b) * (a * b - 1) | We must show that for all positive real numbers a and b, the quantity ab(a^2+b^2-2) is at least (a+b)(ab-1). | [
{
"index": 1,
"text": "Fix arbitrary real numbers a and b, and assume the hypothesis that a > 0 and b > 0 holds.",
"lean_code": " intro a b hab",
"label": "correct"
},
{
"index": 2,
"text": "Split the hypothesis into its two components: ha states a > 0, and hb states b > 0.",
"lean_... | correct | null | null | |
lean_workbook_hf_25940 | 0 | df0c86b230d7bf348ea2f53594696026d3e6d6039a053a185e5706f8c137e607 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_25940 : ∀ n : ℤ, Even n → n ^ 2 % 4 = 0 | We must show that for every integer n, if n is even then n squared leaves remainder 0 when divided by 4. | [
{
"index": 1,
"text": "Fix an arbitrary integer n and assume that n is even; we must show n^2 % 4 = 0.",
"lean_code": " intro n hn",
"label": "correct"
},
{
"index": 2,
"text": "Since n is even, there exists an integer k such that n = k + k.",
"lean_code": " obtain ⟨k, hk⟩ := hn",
... | correct | null | null | |
lean_workbook_hf_903_29534 | 0 | af2966e7f6ec45dc129b42a2cb5591dbd919d6588417b1d21251cf82095a4aa9 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_903_29534 (n : ℕ) (h : n % 2 = 0) : ∃ x y, x = 9 * 7 ^ n * 41 ^ (n / 2 - 1) ∧ y = 40 * 7 ^ n * 41 ^ (n / 2 - 1) | We must show that for a natural number n divisible by 2, there exist natural numbers x and y satisfying x = 9·7^n·41^(n/2 − 1) and y = 40·7^n·41^(n/2 − 1). | [
{
"index": 1,
"text": "We choose x to be exactly the quantity 9·7^n·41^(n/2 − 1) and y to be exactly the quantity 40·7^n·41^(n/2 − 1), which are precisely the values required by the existential claim, leaving the two equalities to be verified.",
"lean_code": " refine ⟨9 * 7 ^ n * 41 ^ (n / 2 - 1), 40 *... | correct | null | null | |
lean_workbook_hf_39808 | 0 | 6ee7675c861f5870132943ae2d13c8ddc241babcef2b93814a1ef305fca1e981 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_39808 (a b c : ℝ) (h₁ : a < b ∧ b < c) (h₂ : a + b + c = 0) : (a ^ 2 + b ^ 2 + c ^ 2) / (c - a)^2 < 2 / 3 | We must show that whenever a < b < c are real numbers with a + b + c = 0, the ratio (a² + b² + c²)/(c − a)² is strictly less than 2/3. | [
{
"index": 1,
"text": "Split the ordering hypothesis into its two parts: a < b and b < c.",
"lean_code": " obtain ⟨hab, hbc⟩ := h₁",
"label": "correct"
},
{
"index": 2,
"text": "Since a < b and b < c, transitivity of the order gives a < c.",
"lean_code": " have hca : a < c := lt_tr... | correct | null | null | |
lean_workbook_hf_78654 | 0 | 3ee2d0df6a826be6bd1b5c462e564ee712b66b7a5452e6541d70e053e8fc8f64 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_78654 (x y : ℝ) (h₁ : x = 200) (h₂ : y = 5/12 * x - 5*110/12) : y = 37.5 | We are given that x equals 200 and that y is defined by the formula y = (5/12)·x − (5·110)/12, and we must show that under these conditions y equals 37.5. | [
{
"index": 1,
"text": "Since x is exactly 200, we replace every occurrence of x by 200 throughout the expression for y, so that y = (5/12)·200 − (5·110)/12.",
"lean_code": " subst h₁",
"label": "correct"
},
{
"index": 2,
"text": "We substitute the defining formula for y into the goal, s... | correct | null | null | |
lean_workbook_hf_51350 | 0 | 69fba148f1f52d8eeaea1ac31a020e7fa7faa15e32b42abdb1468c0ead37cb53 | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_51350 {a b c : ℤ} (h : a + b + c = 0) : (a ^ 2 + b ^ 2 + c ^ 2) ∣ (a ^ 4 + b ^ 4 + c ^ 4) | We must prove that whenever three integers a, b, c satisfy a + b + c = 0, the quantity a² + b² + c² divides a⁴ + b⁴ + c⁴. | [
{
"index": 1,
"text": "Since a + b + c = 0, solving for c gives c = -a - b.",
"lean_code": " have hc : c = -a - b := by linarith",
"label": "correct"
},
{
"index": 2,
"text": "We now replace c everywhere by -a - b, so the problem reduces to a statement purely in terms of a and b: we mus... | correct | null | null | |
lean_workbook_hf_50409 | 0 | feafb24e7f63282449e3dc10bb1ec1c7ba9e64ca84f046674a74b77c442e9385 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_50409 (a b c : ℝ) : (a + b + c) ^ 2 ≥ 3 * (a * b + a * c + b * c) | We must show that for any real numbers a, b, c, the square of their sum is at least three times the sum of their pairwise products. | [
{
"index": 1,
"text": "The square of (a − b) is nonnegative, since any real square is nonnegative.",
"lean_code": " have h1 : (a - b) ^ 2 ≥ 0 := sq_nonneg (a - b)",
"label": "correct"
},
{
"index": 2,
"text": "Similarly, the square of (b − c) is nonnegative.",
"lean_code": " have h... | correct | null | null | |
lean_workbook_hf_23005 | 0 | 7599770b57c3fe0f046d8b4b1b054b92287872b54d1856673726e8e1ae7196be | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_23005 : ∀ a b c : ℝ, a > 0 ∧ b > 0 ∧ c > 0 → a ^ 2 + b ^ 2 + c ^ 2 = 3 → a * b + b * c + c * a ≤ 3 | We must show that for all positive real numbers a, b, c satisfying a² + b² + c² = 3, the sum ab + bc + ca is at most 3. | [
{
"index": 1,
"text": "Fix arbitrary real numbers a, b, c.",
"lean_code": " intro a b c",
"label": "correct"
},
{
"index": 2,
"text": "Assume a, b, c are each positive, giving us ha : a > 0, hb : b > 0, hc : c > 0.",
"lean_code": " rintro ⟨ha, hb, hc⟩",
"label": "correct"
},
... | correct | null | null | |
numina_e81ec851-b2cd-5699-8bb4-47843ae2fad8 | 0 | 5f64b772f3ff1319ce7a4634d9aa67fa8870b05994b076d074fcc4e390c3188a | NuminaMath-LEAN | numina | Algebra | theorem algebra_296307 {n : ℕ} (hn : n ≥ 2)
(f : Fin n → ℝ → ℝ) (k : ℝ) (hk : 0 < k) :
∃ x : Fin n → ℝ,
|∏ i : Fin n, x i - ∑ i : Fin n, f i (x i)| ≥ k := by | We must show that for any family of functions f_0,…,f_{n-1} (with n ≥ 2) and any k > 0 there is a point x with |∏_i x_i − ∑_i f_i(x_i)| ≥ k. | [
{
"index": 1,
"text": "Assume the contrary, namely that no point achieves absolute value at least k.",
"lean_code": " by_contra hcon",
"label": "correct"
},
{
"index": 2,
"text": "Then for every point x we have the strict bound |∏_i x_i − ∑_i f_i(x_i)| < k.",
"lean_code": " push_ne... | correct | null | null | |
lean_workbook_hf_20565 | 0 | 2b066a49a1f7ed6a00f0d7d132e0fb3f5d0a4a82d1b0c2b6d6e875afb668ec40 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_20565 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a + b + c = 1) : 7 * (a * b + b * c + c * a) ≤ 9 * a * b * c + 2 | We must show that for positive reals a, b, c with a+b+c=1, the quantity 7(ab+bc+ca) is at most 9abc+2. | [
{
"index": 1,
"text": "First we bound ab+bc+ca: since the sum of squares (a-b)^2+(b-c)^2+(c-a)^2 is nonnegative, expanding it gives 2(a^2+b^2+c^2) - 2(ab+bc+ca) ≥ 0, i.e. a^2+b^2+c^2 ≥ ab+bc+ca. Since (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = 1, we get 1 = a^2+b^2+c^2+2(ab+bc+ca) ≥ 3(ab+bc+ca), so 3(ab+bc+ca) ≤... | correct | null | null | |
numina_603e0386-259f-50a2-b6cd-1f005eedb498 | 0 | 1e25bb7dfba8ea2674926dfdecb61b0b861e20eb473bdcbeee94eab41b0f2115 | NuminaMath-LEAN | numina | Inequalities | theorem inequalities_63681 (x y z a b c : ℝ) (hx : 0 < x) (hy : 0 < y) (hz : 0 < z)
(ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (h : a + b + c = 1) :
(x ^ 2 + y ^ 2 + z ^ 2) * (a ^ 3 / (x ^ 2 + 2 * y ^ 2) + b ^ 3 / (y ^ 2 + 2 * z ^ 2) +
c ^ 3 / (z ^ 2 + 2 * x ^ 2)) ≥ 1 / 9 := by | We must prove that for positive reals x, y, z, a, b, c with a + b + c = 1, the product of x² + y² + z² with the cyclic sum a³/(x²+2y²) + b³/(y²+2z²) + c³/(z²+2x²) is at least 1/9. | [
{
"index": 1,
"text": "Introduce the abbreviation S = x² + y² + z² for the common factor outside the parentheses.",
"lean_code": " set S := x ^ 2 + y ^ 2 + z ^ 2 with hSdef",
"label": "correct"
},
{
"index": 2,
"text": "Write d₁ = x² + 2y² for the first denominator.",
"lean_code": "... | correct | null | null | |
numina_5c9fa94c-6f3d-52f8-a4ed-54f56568caf0 | 0 | 84cd4467bf18fa923fc85da025fac01cf782916b5ed47019b2659b5c8a8002ad | NuminaMath-LEAN | numina | Inequalities | theorem inequalities_598920 (θ : ℝ) (hθ : θ ∈ Set.Ioo 0 1) (a b c : ℝ)
(ha : a ∈ Set.Ioo 0 1) (hb : b ∈ Set.Ioo 0 1) (hc : c ∈ Set.Ioo 0 1)
(h : θ < a ∧ b - c < 1) :
Real.sqrt (a * b * c) + Real.sqrt ((1 - a) * (1 - b) * (1 - c)) < 1 := by | We must show that for a, b, c all strictly between 0 and 1, the sum of the square root of abc and the square root of (1-a)(1-b)(1-c) is strictly less than 1. The hypotheses about θ turn out not to be needed for this fact. | [
{
"index": 1,
"text": "Write out that a satisfies 0 < a and a < 1.",
"lean_code": " obtain ⟨ha0, ha1⟩ := ha",
"label": "correct"
},
{
"index": 2,
"text": "Write out that b satisfies 0 < b and b < 1.",
"lean_code": " obtain ⟨hb0, hb1⟩ := hb",
"label": "correct"
},
{
"ind... | correct | null | null | |
lean_workbook_hf_12002 | 0 | 1bc5ef2c08aa211fbfe54ce2e7cc432653d9ce8d9a0b0b08a5769367a67258ea | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_12002 (x y z : ℝ) : x ^ 3 + y ^ 3 + z ^ 3 = 3 * x * y * z + (x + y + z) * (x ^ 2 + y ^ 2 + z ^ 2 - (x*y + y*z + z*x)) | We must show that for all real numbers x, y, z, the sum of cubes x^3 + y^3 + z^3 equals 3xyz plus the product (x + y + z) times (x^2 + y^2 + z^2 − xy − yz − zx). This is the classical factorization identity for the sum of cubes minus three times the product, and we verify it by direct algebraic expansion. | [
{
"index": 1,
"text": "Expanding the product (x + y + z)(x^2 + y^2 + z^2 − xy − yz − zx) term by term: multiplying x, y, z respectively by x^2 + y^2 + z^2 − xy − yz − zx and collecting all resulting monomials, every mixed term of the form x^2 y, x y^2, and similar permutations cancels in pairs, leaving exac... | correct | null | null | |
lean_workbook_hf_46902 | 0 | 3e891f4ba168db044a8dadeec8a3cd465cf973d793ed6959d2d6a8efa6ffb161 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_46902 (a b : ℝ) : (a * (a + 1) + b * (b + 1))^2 ≥ (8:ℝ) / 3 * (a * b * (a * b + 1) + (a + b) * (a ^ 2 + b ^ 2)) | We must show that for all real numbers a and b, the square of a(a+1)+b(b+1) is at least 8/3 times ab(ab+1)+(a+b)(a²+b²). The strategy is to show that three times the difference of these two sides can be written as an explicit sum of manifestly nonnegative terms, using the substitution s = a+b−2 and d = a−b. | [
{
"index": 1,
"text": "We verify the key algebraic identity: three times the left-hand side minus eight times ab(ab+1)+(a+b)(a²+b²) equals ((1/2)(a+b−2)(a+b))² plus ((5/2)(a+b−2)²+9(a+b−2)+10)(a−b)² plus ((1/2)(a−b)²)². This can be checked by direct expansion: writing s = a+b−2 and d = a−b, both sides expan... | correct | null | null | |
lean_workbook_hf_36246 | 0 | 3582d9e97c03174797b0612b3c3ea6d3078b64d9a9176c24125df126f5f104f9 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_36246 : ¬ (∃ a b : ℤ, a ^ 4 + b ^ 4 + a * b ^ 3 + 1 = 4 * a ^ 2 * b ^ 2 + a ^ 3 * b) | We must show that no pair of integers a, b satisfies a^4 + b^4 + a b^3 + 1 = 4a^2b^2 + a^3 b. | [
{
"index": 1,
"text": "Suppose, for contradiction, that integers a and b exist satisfying this equation; call the equation hab.",
"lean_code": " rintro ⟨a, b, hab⟩",
"label": "correct"
},
{
"index": 2,
"text": "We first establish the algebraic identity (a^2 - 2ab - b^2)(a^2 + ab - b^2) ... | correct | null | null | |
lean_workbook_hf_34745 | 0 | d0c73ecd839226570442d145d1e602697967f9d7e64db3fccba96702c82e253e | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_34745 (x y z : ℝ) (hx : 0 < x ∧ x < 1) (hy : 0 < y ∧ y < 1) (hz : 0 < z ∧ z < 1) (h : x * y + y * z + z * x = 1) : 1 ≤ x + y + z + x * y * z ∧ x + y + z + x * y * z ≤ 2 | We must show that if x, y, z are each strictly between 0 and 1 and satisfy xy + yz + zx = 1, then the sum x + y + z + xyz lies between 1 and 2 inclusive. | [
{
"index": 1,
"text": "Split the hypothesis on x into its two parts: 0 < x and x < 1.",
"lean_code": " obtain ⟨hx0, hx1⟩ := hx",
"label": "correct"
},
{
"index": 2,
"text": "Split the hypothesis on y into its two parts: 0 < y and y < 1.",
"lean_code": " obtain ⟨hy0, hy1⟩ := hy",
... | correct | null | null | |
lean_workbook_hf_17227 | 0 | b39afecde8978ac0aa3715e198c811cc19d7adf33f403b50bb1ad08092725795 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_17227 (k : ℝ) : k ^ 2 + 10 * k + 16 = 0 ↔ k = -2 ∨ k = -8 | We must show that a real number k satisfies k² + 10k + 16 = 0 exactly when k equals −2 or k equals −8. | [
{
"index": 1,
"text": "First we note the algebraic factorization k² + 10k + 16 = (k + 2)(k + 8), which can be verified by expanding the right-hand side: (k+2)(k+8) = k² + 8k + 2k + 16 = k² + 10k + 16.",
"lean_code": " have factored : k ^ 2 + 10 * k + 16 = (k + 2) * (k + 8) := by ring",
"label": "co... | correct | null | null | |
numina_9bc11617-8d8f-5885-a85f-6a87cf80d608 | 0 | 4b7b84243806fed9377799797c3cf8d757c6b32bc16260ec68d59f95f83794da | NuminaMath-LEAN | numina | Inequalities | theorem inequalities_44746 {m n : ℕ} (hm : 0 < m) (hn : 0 < n)
(a : ℕ → ℕ → ℝ) (ha : ∀ i ∈ Icc 1 m, ∀ j ∈ Icc 1 n, 0 ≤ a i j)
(hij : ∀ i ∈ Icc 1 m, ∀ j ∈ Icc 1 n, a i j ≥ a i (j + 1)) (hji : ∀ i ∈ Icc 1 m, ∀ j ∈ Icc 1 n, a i j ≥ a (i + 1) j)
(X Y : ℕ → ℕ → ℝ) (hX : ∀ i ∈ Icc 1 m, ∀ j ∈ Icc 1 n, X i j = ∑ k ... | We must show that the product of the partial sums X i j taken over all rows i and columns j is at least as large as the product of the reflected partial sums Y i j, given that the array a is nonnegative and weakly decreasing along both rows and columns. | [
{
"index": 1,
"text": "First we establish that within a fixed column l, the array value decreases as the row index increases: for any indices i1 ≤ i2 lying between 1 and m, a(i2,l) ≤ a(i1,l). This is proved by induction on i2 starting from i1: the base case is trivial equality, and at each inductive step we... | correct | null | null | |
lean_workbook_hf_26556 | 0 | 730a1f89096677baa2ac594405a98dd8f5c4725773c120154e8b1cc2789f6789 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_26556 (m n a b c : ℤ) (h₀ : 0 < m ∧ 0 < n ∧ 0 ≤ a ∧ 0 ≤ b ∧ 0 ≤ c) (h₁ : m ^ 2 + n ^ 2 + a ^ 2 + b ^ 2 + c ^ 2 = 1989) (h₂ : m ^ 2 + n ^ 2 + a + b + c = 19) (h₃ : m ^ 2 + a ^ 2 + b ^ 2 + c ^ 2 = 44) (h₄ : n ^ 2 + a ^ 2 + b ^ 2 + c ^ 2 = 25) : m ^ 2 = 81 ∧ n ^ 2 = 36 | We are given five nonnegative or positive integers m, n, a, b, c satisfying four equations relating their squares (and, in one case, linear terms). We want to show m² = 81 and n² = 36. We will show instead that the hypotheses are jointly contradictory, from which any conclusion, including the desired one, follows. | [
{
"index": 1,
"text": "We break apart the conjunction of sign conditions into its five separate facts: m > 0, n > 0, a ≥ 0, b ≥ 0, c ≥ 0.",
"lean_code": " obtain ⟨hm, hn, ha, hb, hc⟩ := h₀",
"label": "correct"
},
{
"index": 2,
"text": "Subtracting the equation m² + a² + b² + c² = 44 fro... | correct | null | null | |
pythagoras_55fd91d739371385e289 | 0 | e260e16a2f318737d6efdb7ae179f66b0cb724bdecbd18d5b3afef1b5ca1a6eb | Pythagoras | pythagoras | Pythagoras_SFT_N/A | theorem my_favorite_problem :
let total_cards : ℕ := 52
let ace_cards : ℕ := 4
let non_ace_cards : ℕ := 48
let hand_size : ℕ := 5
let prob_exactly_two_aces : ℚ := (Nat.choose ace_cards 2 * Nat.choose non_ace_cards (hand_size - 2)) / Nat.choose total_cards hand_size
let prob_at_least_one_ace : ℚ := 1 - (Nat.choose... | We want to show that, when drawing a 5-card hand from a standard 52-card deck, the conditional probability of drawing exactly two aces given that at least one ace is drawn equals 1081/9236. This requires computing the total number of 5-card hands, the number of hands with exactly two aces, and the number of hands with ... | [
{
"index": 1,
"text": "First, the number of ways to choose 2 aces out of the 4 aces in the deck is C(4,2) = 6.",
"lean_code": " have hc1 : Nat.choose 4 2 = 6 := by decide",
"label": "correct"
},
{
"index": 2,
"text": "Next, the number of ways to choose the remaining 3 cards of the hand ... | correct | null | null | |
lean_workbook_hf_14683 | 0 | 8deedee6016db5b9800962c9bc70d3281612cb8980226a432721cfaa6be89885 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_14683 (x y z : ℝ) : x ^ 4 + y ^ 4 + z ^ 4 ≥ x ^ 3 * y + y ^ 3 * z + z ^ 3 * x | We must show that for any real numbers x, y, z, the sum of fourth powers x^4+y^4+z^4 is at least as large as the cyclic sum x^3 y + y^3 z + z^3 x. | [
{
"index": 1,
"text": "For any two real numbers a and b, we have the algebraic identity 3a^4 + b^4 − 4a^3 b = (a − b)^2 · (2a^2 + (a + b)^2). This can be checked by expanding both sides: (a−b)^2 = a^2−2ab+b^2, and 2a^2+(a+b)^2 = 3a^2+2ab+b^2; multiplying these two quadratics out and collecting terms yields ... | correct | null | null | |
lean_workbook_hf_78082 | 0 | e63cb253f98d788fd16f8d34b8c00f78e0c80c9dd0e01b246965236f816f2120 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_78082 : ∀ a b c : ℕ, a ∈ ({5, 49, 51} : Finset ℕ) ∧ b ∈ ({5, 49, 51} : Finset ℕ) ∧ c ∈ ({5, 49, 51} : Finset ℕ) → a + b = c ∨ a * b = c ∨ a = c / 2 → 3 ∣ c | We must show: for any natural numbers a, b, c each drawn from the set {5, 49, 51}, if a+b=c, or a·b=c, or a equals c divided by 2 (with integer division), then c must be divisible by 3. | [
{
"index": 1,
"text": "Fix arbitrary natural numbers a, b, c for which we will verify the implication.",
"lean_code": " intro a b c",
"label": "correct"
},
{
"index": 2,
"text": "Assume a, b, c each belong to {5, 49, 51} (giving membership facts ha, hb, hc), and assume the disjunctive h... | correct | null | null | |
numina_a7af0b16-6032-5960-a5e7-7dbf2dc48399 | 0 | 15a3e0622ee088b9e246692d9a5b2878dbc3ae90c1db892af3e260193e334416 | NuminaMath-LEAN | numina | Inequalities | theorem inequalities_54864
(a b c : ℝ)
(ha : 0 < a)
(hb : 0 < b)
(hc : 0 < c)
(h : a * b + b * c + c * a = 1 / 3) :
1 / (a + b + c) ≤ a / (a ^ 2 - b * c + 1) + b / (b ^ 2 - c * a + 1) + c / (c ^ 2 - a * b + 1) := by | We are given positive reals a, b, c with ab + bc + ca = 1/3, and we must show that 1/(a+b+c) is at most the sum a/(a²−bc+1) + b/(b²−ca+1) + c/(c²−ab+1). | [
{
"index": 1,
"text": "Since a, b, c are all positive, their sum a + b + c is positive.",
"lean_code": " have hs : 0 < a + b + c := add_pos (add_pos ha hb) hc",
"label": "correct"
},
{
"index": 2,
"text": "The first denominator is positive: using 1 = 3(ab + bc + ca) we get a² − bc + 1 =... | correct | null | null | |
lean_workbook_hf_15032 | 0 | 85c9fa1808f46af6022d5240b4f6807ca9cfbed0cc0b3b72060a768626ed9d4b | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_15032 (x y : ℝ) (h : x ≠ y) (h1 : x ^ 3 - x = y ^ 3 - y) : -2 ≤ x ^ 3 * y + x * y ^ 3 ∧ x ^ 3 * y + x * y ^ 3 ≤ 2/9 | We must prove that for real numbers x and y with x ≠ y satisfying x³ − x = y³ − y, the quantity x³y + xy³ lies between −2 and 2/9. | [
{
"index": 1,
"text": "We use the algebraic identity x³ − y³ = (x³ − x) − (y³ − y) + (x − y). Substituting the hypothesis x³ − x = y³ − y makes the first two bracketed terms cancel, leaving x³ − y³ = x − y.",
"lean_code": " have hcube : x ^ 3 - y ^ 3 = x - y := by\n have hid : x ^ 3 - y ^ 3 = (x ^ 3... | correct | null | null | |
numina_12f17825-21c6-5cae-b844-482753d7d0c9 | 0 | dde582c02220de1feb68e1ad0567a11f4007f352794bf767331ab3544c0c3aca | NuminaMath-LEAN | numina | Algebra | theorem algebra_63440 {n : ℕ} (a : ℕ → ℝ) (x : ℝ) (hx : 0 < x ∧ x < 1)
(hsum : ∑ i ∈ Finset.range (n + 1), a i / (1 - x ^ (i + 1)) = 0) :
∃ y : ℝ, 0 < y ∧ y < 1 ∧ ∑ i ∈ Finset.range (n + 1), a i * y ^ i = 0 := by | We must show that if 0 < x < 1 and the weighted sum ∑ a_i/(1 − x^{i+1}) over i = 0,…,n vanishes, then the polynomial ∑ a_i y^i has a root y strictly between 0 and 1. | [
{
"index": 1,
"text": "From the hypothesis on x we record the two facts that 0 < x and x < 1 separately.",
"lean_code": " obtain ⟨hx0, hx1⟩ := hx",
"label": "correct"
},
{
"index": 2,
"text": "Introduce the polynomial p(y) = ∑_{i=0}^{n} a_i y^i, so that the desired conclusion is exactly... | correct | null | null | |
lean_workbook_hf_14812 | 0 | 5b1e5c5bd6ef3c5aa355ec24e2bc2751e6ae1ed5fd3ad9f43fdfc5e12095c788 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_14812 (f : ℝ → ℝ) (h₁ : Monotone f) (h₂ : ∀ x, f (f x) = (f x)^2) (h₃ : ∀ x, f (-f x) = (f x)^2) : ∀ x ∈ Set.range f, f x = f (-x) | We must show that for every real number x lying in the range of f, we have f(x) = f(-x). | [
{
"index": 1,
"text": "Fix an arbitrary real number x that belongs to the range of f, and let us show f(x) = f(-x) for this x.",
"lean_code": " intro x hx",
"label": "correct"
},
{
"index": 2,
"text": "Since x is in the range of f, there exists some real number y such that f(y) = x.",
... | correct | null | null | |
lean_workbook_hf_30190 | 0 | 3905d9c65b981cbd964b802eacfc574e12a66928a8fca63fb01dd956fb30f0ad | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_30190 : ¬∃ x y : ℤ, 7*x ^ 2 - 13*y ^ 2 = 1 | We must show that there are no integers x and y satisfying 7x² − 13y² = 1. | [
{
"index": 1,
"text": "Suppose, for contradiction, that such integers x and y exist, with the equation h: 7x² − 13y² = 1 holding.",
"lean_code": " rintro ⟨x, y, h⟩",
"label": "correct"
},
{
"index": 2,
"text": "Reduce the equation h modulo 13 by applying the natural map from the integer... | correct | null | null | |
lean_workbook_hf_14211_13550 | 0 | 89dc758b183d3e3c7b70e665ea91abffc552ec9ca8f2ed30c35362058eb13182 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_14211_13550 (x y : ℝ) (f : ℝ → ℝ) (hf: f x + f y = (Real.sqrt (x / y) + Real.sqrt (y / x)) * f (Real.sqrt (x * y))) : ∃ f: ℝ → ℝ, ∀ x y : ℝ, (x > 0 ∧ y > 0) → f x + f y = (Real.sqrt (x / y) + Real.sqrt (y / x)) * f (Real.sqrt (x * y)) | We must show that there exists a function f from the reals to the reals satisfying, for all positive real numbers x and y, the identity f(x) + f(y) = (√(x/y) + √(y/x))·f(√(xy)). It suffices to exhibit one such function and verify the identity holds for it. | [
{
"index": 1,
"text": "Take as our candidate the function that sends every real number to 0, that is f(t) = 0 for all t.",
"lean_code": " use fun _ => 0",
"label": "correct"
},
{
"index": 2,
"text": "Fix arbitrary real numbers x and y, and assume both are positive, so that x > 0 and y >... | correct | null | null | |
lean_workbook_hf_70031 | 0 | 672b0b6b39a36c72cb7736afd803c9ceb7e88a434574575fb1030c225b350708 | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_70031 : 5555 ^ 5555 ≡ 5 [ZMOD 9] | We want to show that 5555 raised to the power 5555, viewed as an integer, is congruent to 5 modulo 9. | [
{
"index": 1,
"text": "First, note that 5555 = 9·617 + 2, so 5555 leaves remainder 2 when divided by 9; hence 5555 is congruent to 2 modulo 9.",
"lean_code": " have h1 : (5555 : ℤ) ≡ 2 [ZMOD 9] := by decide",
"label": "correct"
},
{
"index": 2,
"text": "Raising both sides of the congrue... | correct | null | null | |
lean_workbook_hf_35597 | 0 | 76af5b7b3a66a1e0ad99b8192c0cabc9b0abf3886d0cfea9cbc2b9b5b498645c | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_35597 (n x y z : ℝ) (hn : n > 0) (hx : x > 0) (hy : y > 0) (hz : z > 0) (h : n * y * z ≥ 2 * x) : n / 2 ≥ x / (y * z) | We must show that, given positive reals n, x, y, z with n·y·z ≥ 2x, the ratio n/2 is at least x/(yz). | [
{
"index": 1,
"text": "Since y and z are both positive, their product yz is positive.",
"lean_code": " have hyz : y * z > 0 := mul_pos hy hz",
"label": "correct"
},
{
"index": 2,
"text": "The number 2 is positive.",
"lean_code": " have h2 : (0:ℝ) < 2 := by norm_num",
"label": "... | correct | null | null | |
lean_workbook_hf_12421 | 0 | 1853079f008e1d962bbc72bf5009cce6fdd996703b599928863643f88a50084e | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_12421 (x : ℝ) (hx: 0 ≤ x ∧ x ≤ 1) : 0 ≤ x - x ^ 2 ∧ x - x ^ 2 ≤ 1/4 | We must show that for a real number x satisfying 0 ≤ x ≤ 1, the quantity x − x² is both nonnegative and at most 1/4. | [
{
"index": 1,
"text": "Split the hypothesis into its two parts: x ≥ 0 and x ≤ 1.",
"lean_code": " obtain ⟨hx0, hx1⟩ := hx",
"label": "correct"
},
{
"index": 2,
"text": "It suffices to prove the two claims separately: first that x − x² ≥ 0, and second that x − x² ≤ 1/4.",
"lean_code"... | correct | null | null | |
lean_workbook_hf_36804_18771 | 0 | 33f90bf92756b701010b5b34872c35f35afd2ce2b88d6bf06de76ef5450982b9 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_36804_18771 (f : ℝ → ℝ) (hf : ∀ x y, x * f y = y * f x) : ∃ k, ∀ x, f x = k * x | We must show that there exists a real number k such that f(x) = k·x for every real x, given that x·f(y) = y·f(x) holds for all real x and y. | [
{
"index": 1,
"text": "We claim that k = f(1) works, so we take this value as our candidate constant.",
"lean_code": " use f 1",
"label": "correct"
},
{
"index": 2,
"text": "Let x be an arbitrary real number; we must show f(x) = f(1)·x.",
"lean_code": " intro x",
"label": "corr... | correct | null | null | |
lean_workbook_hf_46031 | 0 | 5f08a6e58679b14ecf0ffc3db8a66f7b24b2c549cb2cea86fc389888e7270761 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_46031 (b : ℕ) (p : ℕ) (k : ℕ) (h₁ : b = 4095) (h₂ : p ^ k = 13) : (p ^ k)^3 = 2197 | We are given that p^k equals 13, and we must show that (p^k)^3 equals 2197. | [
{
"index": 1,
"text": "Since p^k = 13, we substitute this value into the expression (p^k)^3, reducing the goal to showing that 13^3 equals 2197.",
"lean_code": " rw [h₂]",
"label": "correct"
},
{
"index": 2,
"text": "We compute directly: 13^3 = 13 · 13 · 13 = 169 · 13 = 2197, confirming... | correct | null | null |
End of preview. Expand in Data Studio
LeanPRM dataset and benchmark
Natural-language proofs with aligned Lean tactics and step-level correctness labels. The dataset contains 10,634 proofs and 152,005 proof steps.
| Resource | Proofs | Files | HF split |
|---|---|---|---|
| LeanPRM training data | 10,264 | data/train.parquet | train |
| LeanPRM validation data | 170 | data/validation.parquet | validation |
| LeanPRM-Bench | 200 | benchmark/test.parquet | test |
LeanPRM dataset
The data folder contains the training and validation proofs.
from datasets import load_dataset
train = load_dataset("pavelslab-nyu/LeanPRM", split="train")
validation = load_dataset("pavelslab-nyu/LeanPRM", split="validation")
LeanPRM-Bench
The benchmark folder contains the held-out test set, with 100 correct and 100 incorrect proofs. It is included in this repository as the test split.
benchmark = load_dataset("pavelslab-nyu/LeanPRM", split="test")
Labels and sources
Positive labels come from Lean compilation and model review of the natural-language alignment. Negative labels mark the first mathematical error and all later steps. Labels are not manual annotations.
No theorem group is shared across the final splits. Thirty benchmark examples came from the original validation set and were available during historical checkpoint selection. The provenance folder preserves that validation set and the original weighted training order for reproduction.
Each row includes the source, theorem, aligned proof steps, labels, and error class. first_error_step is one-based and null for correct proofs. Counts and source-file hashes are in manifest.json.
Source statements come from NuminaMath-LEAN, Lean-Workbook, Goedel's Lean-workbook proofs, Pythagoras, Herald, and ProofNet. Source materials retain their upstream terms. Release licensing has not been assigned.
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