| Model | d | e | pass@1 | pass@8 |
| mathlib-valid | ||||
| PACT | 512 | 16 | 48.4% | |
| 0 | 512 | 16 | 48.5% | 57.6% |
| 0 | 512 | 8 | 46.7% | 57.5% |
| 01 | 512 | 8 | 56.3% | 66.3% |
| 0 | 512 | 8 | 55.6% | 65.9% |
| Model | d | e | pass@1 | pass@8 |
| miniF2F-valid | ||||
| miniF2F | 128 | 16 | 23.9% | 29.3% |
| 128 | 16 | 27.6% | 31.8% | |
| 0 | 512 | 8 | 28.4% | 33.6% |
| 01 | 512 | 8 | 28.5% | 35.5% |
| 0 | 512 | 8 | 28.3% | 34.7% |
| Model | pass@1 | pass@8 | pass@64 | pass@1 | pass@8 | pass@64 |
| mathlib-valid | mathlib-test | |||||
| PACT (Han et al.,2022) | 48.4% | = | ||||
| 01 | 56.3% | 66.3% | 72.0% | 56.5% | 66.9% | 73.7% |
| 62.6% | 70.7% | 75.8% | 63.0% | 71.5% | 77.1% | |
| G | 61.7% | 69.8% | 75.3% | 62.9% | 71.6% | 76.3% |
| miniF2F-valid | miniF2F-test | |||||
| PACT (Zheng et al., 2022) | 23.9% | 29.3% | 24.6% | 29.2% | ||
| 01 | 28.5% | 35.5% | 41.2% | 25.9% | 31.1% | 33.6% |
| ggahibh | 31.3% | 38.3% | 44.1% | 27.2% | 33.0% | 35.2% |
| 33.6% | 41.2% | 47.3% | 29.6% | 34.5% | 36.6% |
| Dataset | Size | Mix |
| Github Python | 179 GB | 25% |
| arXiv Math | 10 GB | 25% |
| Math StackExchange | 2GB | 25% |
| PACT mix2 | 28GB | 17% |
| Math Overflow | 200 M | 5% |
| ProofWiki | 30M | 2% |
| PlanetMath | 25M | 1% |
| Compositions | AddLeAdd Bernoulli 99 c AddLeAdd SelfDivConst（(a）/（f)）6 LeMulOfRatio SelfDivConst c 70 DivLeDiv Cauchy（(a）/(f)）dc(log（((59:R）+ f))) Young((a)/(f)）a((3:R)/(2:R)）((3:R)/(1:R)) theorem synthetic_ineq_nb_seed_var_4_depth_4_p_13 |
| Statement | (abcdef:R) (h0 :0<a) (h1 :0<b) (h2 :0<c) (h3:0<d) (h4 :0<e) (h5:0<f): (1:R)+(99:R)*c+(a/f/(6:R)+a*(a/f)/ （(d^2+a^2/f^2)* (real.log（(59:R）+f)^2+c^2))）≤ （(a/f）^((3:R)/(2:R)）/（(3:R)/(2:R)）+ a^3/(3:R)）/ (real.log（(59:R）+f）*d+a/ f*c)^2 * (c/(c/(70:R))）+a/f+(c+(1:R)）^99 := sorry |
| Compositions | Sqnonneg a ((a)+((-68:R))) |
| Statement | theorem synthetic_ineq_nb_seed_var_4_depth_0_p_4 (ab:R) (h0 :0<a) (h1:0<b): (2:R)*(a*（a+-(68:R))）≤ (a +-(68:R))^2 +a^2 := sorry |
| Compositions | AmGm a b (67:R)((1:R)/(10:R)）((1:R)/(10:R)) ((8:R)/(10:R)) |
| Statement | theorem synthetic_ineq_nb_seed_var_0_depth_0_p_1 (ab:R) (h0:0<a) (h1:0<b): (67:R)^（(8:R)/(10:R)）*b^(10:R)-1 * a^(10:R)-1≤ (8:R)/(10:R)* (67:R) + (10:R)-1 * a+b* (10:R)-1 := sorry |
| Statement | lemma comap_eq_of_inverse {f：filter α} {g：filter β}{Φ:α→β} (φ:β→α)(eq:γ○Φ=id)(hΦ : tendsto f g)(hψ :tendsto 𝜑 g f):comap Φ g = f := |
| Ground-truth | beginrefine((comap_mono $map_le_iff_le_comap.1 h).trans_).antisymm(map_le_iff_le_comap.1 hΦ),rw [comap_comap，eq，comap_id],exact le_rflend |
| Model proof | beginrefine le_antisymm _ (filter.map_le_iff_le_comap.1 h),refine 入s hs,-,rw mem_comap,use [ -1' s,h hs],rw[← preimage_comp，eq，preimage_id]end |
| Statement | lemma sum_range_sub_sum_range {α :Type*}[add_comm_group α]{f：N→α}{nm：N}(hnm:n≤m）:∑k in range m，fk-∑kin range n，fk =∑k in (range m).filter (λk，n≤k)，f k := |
| Ground-truth | beginrw[← sum_sdiff(@filter_subset _（λ k，n ≤k）_(range m))，sub_eq_iff_eq_add,← eq_sub_iff_add_eq，add_sub_cancel'],refine finset.sum_congr(finset.ext $入 a，< h，by simp at *；finish,入 h，have ham ：a<m :=lt_of_lt_of_le (mem_range.1 h）hnm,by simp * at *>)(λ__，rfl)end |
| Model proof | beginrw[← sum_Ico_eq_sub_ hnm],congr,apply finset.ext,simp [Ico.mem，*],tautoend |
| Statement | lemma prod_inv_distrib :(II× in s，(f x)-1)=①I×in s,f x)-1 := |
| Ground-truth | beginclassical,by_casesh:×∈s，fx=0,{simpa [prod_eq_zero_iff.mpr h，prod_eq_zero_iff]using h },{ push_neg at h,have h' := prod_ne_zero_iff.mpr h,have hf :∀x ∈ s，(f x)-1 * fx=1 :=λx hx,inv_mul_cancel (h × hx)，apply mul_right_cancel'h',simp[h，h'，← finset.prod_mul_distrib,prod_congr rfl hf]}end |
| Model proof | beginclassical；induction s usingfinset.induction_on with a s has ih,{simp，},simp only[has，prod_insert has，mul_inv_rev'],finishend |
| Natural language | Solve the system of equations:x+y+z=αx²+y²+2²=b²xy= x²where α and b are constants.Give the conditions that α and b must satisfy sothat x,y,z (the solutions of the system) are distinct positive numbers.Note: theformalized statement in miniF2F is a weaker problem as it focuses on the secondpart of the question, providing the actual conditions,and asking for a proof that therequirement entails them. |
| Model proof | theorem imo_1961_p1(xyzab:R)(ho:0<x>0<y>0<z)(h1:×≠y)(h2 :y≠z)(h3:z≠x)(h4 :x+y+z= a)(h5 :x^2 +y^2 + z^2 = b^2)(h6 :x *y= z^2) :0<a∧ b^2<a^2 > a^2<3*b^2 :=beginrevert_all,intros,rw mul_comm,split,{nlinarith [sq_nonneg (x - y)，sq_nonneg(y- z)]，},split,{nlinarith [sq_nonneg(z - 1)]，},revert h3 h4,field_simp [mul_comm a b],rw [mul_comm，← h5],contrapose!,rw mul_comm at h6,rw mul_comm,intro h,nlinarith [sq_nonneg (x - y)，sq_nonneg (y - z)]end |
| Comments | The model is able to close this problem by spliting into cases,contraposing for thelast case and using nlinarith.It must be noted that the arguments for the first twonlinarith uses are not necessary,however the [sq_nonneg (x - y)，sq_nonneg(y- z)] argument provided on the last line is crucial to close the goal and arecompletely exogenous (present in no form in the tactic state before). |
| Natural language | Suppose a,b,c are the sides of a triangle.Prove thata²(b+c-a)+b²(c+a-b)+c²(a+b-c)≤3abc |
| Model proof | theorem imo_1964_p2(abc:R)(ho:0<a>0<b>0<c)(h1:c<a+b)(h2:b<a+c)(h3:a<b+c):a^2*(b+c-a)+b^2*(c+a-b)+c^2*(a+b-c)≤3*a*b*c:=beginnlinarith [sq_nonneg (b - a)，sq_nonneg (c - b),sq_nonneg (a - c)，sq_nonneg (c - a)]end |
| Comments | The model is able to close an IMO problem in one-line.It correctly providesexogenous arguments to nlinarith,which are necessary to close the goal. Notethat either one of the last two arguments in the sequence [sq_nonneg (b - a),sq_nonneg(c -b)，sq_nonneg(a - c)，sq_nonneg (c- a)]can be omitted. |
| Natural language | Find ax+ by if the real numbers a,b,x,and y satisfy the equations ax+by =3, ax²+by²=7, ax²+by³=16, ax²4 +by4 = 42. Note: the formalized statement in miniF2F provides the answer and asks for a proof of it. theorem aime_1990_p15 |
| Model proof | (abxy:R) (ho:a*x+b*y=3) (h1 :a* x^2+b*y^2= 7) (h2 :a* x^3 +b*y^3= 16) (h3 :a* x^4 +b*y^4= 42) : a * x^5+b*y^5=20 := begin revert_all, intros ab ×y hg hi h2 h4, ring_nf at hi h2, rw ← sub_eq_zero at h1, nlinarith [sq_nonneg (× - y)，sq_nonneg (a + b - 2), sq_nonneg (x + y - (2:R))，sq_nonneg (a -b - 2)] |
| Comments | end The model is able to close a challenging AIME problem by providing crucial ex- ogenous arguments sq_nonneg (x - y) and sq_nonneg (x + y - (2 :R)) to nlinarith,which are required to close the goal (while the other two can be removed). |
| Natural language | Let f(x)= Ax+B and g(x)=Bx +A,where A≠B. If f(g(x)) - g(f(x))= B-A,what is A+B? Note: the formalized statement in our curriculum provides the answer and asks for a proof of it. theorem mathd_train_algebra_217 |
| Model proof | (ab:R) (fg:R→R) (ho:∀x，fx=a*x+b) (h1:∀×，fx=b*x+a) (h2 :a≠b) (h3 :∀x，f (gx)-g(f x)=b-a): a+b=0 := begin revert_all, intros a b, intros f g, contrapose!, rintro <ho，<hi，h2>>, use (0 :R), simp only[sub_eq_iff_eq_add，ho，mul_zero]at *, simp only[*，zero_add], |
| Comments | norm_num at ho end The model is able to close the goal by contraposing,supplying a witness by the use of use (O :R）and finally leveraging the simp and norm_num. This example demonstrates the model's ability to chain multiple non-trivial steps of reasoning including the generation of witnesses. |
| Natural language (A) a multiple of 4 (D) a perfect square | For all integers n ≥ 9, the value of (n+2)!-(n+1)! | |
| n! is always which of the following? (B) a multiple of 10 (C)a prime number | ||
| Model proof | theorem amc12b_2020_p6 (n :N) (ho:9≤n): x :N，(x:R)^2 = (nat.factorial (n + 2) - nat.factorial (n + 1)）/ nat.factorial n := begin revert_all, intros n hn, use n + 1, | field_simp [nat.factorial_ne_zero，pow_succ'], |
| Comments | ring_exp end The model exhibits here its ability to implicitly interpret factorial expressions by guessing the correct answer with the use of use n + 1 and straightforwardly proving |
| Natural language | The expression 24x² - 19x - 35 can be written as (Ax - 5)(2Bx + C),where A, B,and C are positive numbers.Find AB- 3C.Note: the formalized statement in miniF2F provides the answer and asks for a proof of it. |
| Model proof | theorem mathd_algebra_140 (abc:R) (ho:0<a>0<b>0<c) (h1:∀×，24* ×^2-19*×-35= (((a*x)-5)*((2*(b*x))+c))）: a*b-3*c=-9:= begin revert_all, rintro abc h1, rw mul_comm, rw ←sub_eq_zero, field_simp, rw sub_eq_add_neg, rw←eq_neg_iff_add_eq_zero, rw [mul_comm，← sub_eq_add_neg，sub_eq_add_neg, ← sub_eq_add_neg], rw ←sub_eq_zero, simp only [mul_add，neg_mul_eq_neg_mul_symm，mul_comm, add_left_comm], norm_num, simp only [pow_two], intro ho, rw← sub_eq_zero, linarith [hi.2，hi.1，ho 3，ho 1，ho 2] |
| Comments | end The model mostly "struggles" to make progress up to the last line. There, it presents us with the idea to specialize the statement given in h1 (which became ho in the process of the proof) at three consecutive natural numbers 1,2,3 which closes the goal with nlinarith. This proof is interesting as it demonstrates the model's ability to evaluate symbolic expressions implicitly. |
| Natural language | Find the value of a2 +a4 +a6 +ag +...+ ag if a1,a2,a3... is an arithmetic progression with common difference 1,and a1 +a2 + a3 +...+ag8 = 137.Note: the formalized statement in miniF2F provides the answer and asks for a proof of it. |
| Model proof | theorem aime_1984_p1 (u:N→Q) (ho :∀n，u(n+1)=un+1) (hi ：∑k in finset.range 98，u k.succ = 137）: ∑ k in finset.range 49，u (2 * k.succ)= 93 := begin revert_all, simp [finset.sum_range_succ], ring, simp[pow_succ，mul_comm，add_left_comm，add_assoc], simp [two_mul], rintro p, revert p, simp [add_comm], intros p hp, simp [hp]，ring_nf, intros, nlinarith |
| Comments | end The model finds a rather short proof of the statement compared to typ- ical human formalizations (see https://gist.github.com/MantasBaksys/ 3efd5c0d42f440d16c09076db42822f2). |
| Natural language | Consider the sequence = 1 3 ，，，. Prove that this sequence tends to 1 as n→ |
| Model proof | theorem aopsbook_v2_c8_ex1 (u:N→R) (ho:∀n，un=n/(n+1)): filter.tendsto u filter.at_top (N 1) := begin revert_all, simp [← nnreal.coe_one], norm_cast, intros, revert ho, assume h, simp [tendsto_const_nhds，← nnreal.coe_one，h], revert u, assume f, norm_num, rw tendsto_iff_norm_tendsto_zero, assume H, convert tendsto_norm_zero.comp tendsto_one_div_add_at_top_nhds_0_nat, funext n, have ho ：（((n:R）+1）:R)≠0, { norm_cast，exact n.succ_ne_zero，}, rwH, field_simp [ho，norm_neg], ring, rw [← sub_eq_zero], simp [← sub_eq_zero], simp[sub_eq_add_neg], |
| Comments | end An interesting example of a generated have statement,denoted by ho,which is introduced as a cut in order to simplify the expression containing divisions by using field_simp[ho，norm_neg] ata later step. |
| Natural language | What is the sum of the units digits of all the multiples of 3 between O and 5O? Note:the formalized statement in miniF2F provides the answer and asks for a proof of it. |
| Model proof | theorem mathd_numbertheory_447 :∑ k in finset.filter (入 ×，3|x)(finset.erase (finset.range 50) 0)，(k % 10) = 78 :=beginreflend |
| Comments | Because the predicate 入 ×，3|× is registered as decidable over N,we can state theproblem by using finset.filter,which is computable.Hence,refl is able toclose the goal. |