Datasets:
run stringclasses 1
value | model stringclasses 1
value | step int64 0 500 | task stringclasses 64
values | attempt int64 0 3 | reward float64 0 1 | judge_scores listlengths 2 2 | judge_assessments listlengths 2 2 | judge_models listlengths 2 2 | regrade_120b_score float64 0 7 ⌀ | regrade_120b_assessment stringlengths 548 1.5k ⌀ | proof stringlengths 240 106k | proof_chars int64 240 106k | completion_tokens int64 595 32.8k | finish_reason stringclasses 2
values | truncated bool 2
classes | trial stringlengths 27 32 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-0420a942ac | 0 | 0 | [
0,
0
] | [
"<score>0</score>\n<assessment>\nThe submitted proof does not satisfy any of the required checkpoints in the marking scheme. It contains numerous arithmetic errors and an inconsistent treatment of the equality case, leading to a contradictory conclusion that the inequality fails for the symmetric point. The proof... | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 0 | ebraic expansion of the numerator to the required form Σa²+3Σab. The proof also fails to apply the standard inequality Σab ≤ Σa², to simplify the numerator, or to use the constraint a²+b²+c²=¼ to conclude the final inequality. Instead, the argument attempts to verify the inequality only for the symmetric case a=b=c and... | # Proof of the Inequality
We are given three positive real numbers $a, b, c$ satisfying the constraint:
$$ a^2 + b^2 + c^2 = \frac{1}{4}. $$
We need to prove the inequality:
$$ \sum_{\text{cyc}} \frac{1}{\sqrt{b^2 + c^2}} \le \frac{\sqrt{2}}{(a + b)(b + c)(c + a)}. $$
Let us define the sum on the left-hand side (LHS)... | 38,866 | 16,131 | stop | false | oe-aops-0420a942ac-64a8c4da |
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-0420a942ac | 1 | 0 | [
0,
0
] | [
"<score>0</score>\n<assessment>The submitted proof does not meet the requirements of the rubric. It does not establish the key QM‑AM bound $\\frac{1}{\\sqrt{b^2+c^2}}\\le\\frac{\\sqrt2}{b+c}$, nor does it sum the resulting terms or combine $\\sum_{\\text{cyc}}\\frac1{b+c}$ into a single fraction. The proof also doe... | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 0 | \frac14$ in the final step. Therefore, 0 pts are earned for this checkpoint.
No zero‑credit items are present, and there are no minor arithmetic or sign errors that trigger the listed deductions. The only shortcomings are the complete lack of the required intermediate steps and the reliance on an informal symmetry ar... | # Proof
Let $a, b, c$ be positive real numbers satisfying $a^2 + b^2 + c^2 = \frac{1}{4}$.
We aim to prove the inequality:
$$ S = \sum_{\text{cyc}} \frac{1}{\sqrt{b^2 + c^2}} \le \frac{\sqrt{2}}{(a+b)(b+c)(c+a)}. $$
### Step 1: Simplify the Left-Hand Side (LHS) using the constraint
Recall the algebraic identity $(a+... | 26,950 | 11,161 | stop | false | oe-aops-0420a942ac-3b94375a |
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-0420a942ac | 2 | 0 | [
0,
0
] | [
"<score>0</score>\n<assessment>The submitted proof does not establish the statement of the problem. The original inequality is \n\\[\n\\frac{1}{\\sqrt{b^2+c^2}}+\\frac{1}{\\sqrt{c^2+a^2}}+\\frac{1}{\\sqrt{a^2+b^2}}\n\\le \n\\frac{\\sqrt{2}}{(a+b)(b+c)(c+a)}.\n\\]\nHowever, the proof replaces the denominator \\((a+... | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 0 | nsion checkpoint (2 pts) is not achieved.
- The optimization step never applies the inequality \(\sum ab\le\sum a^2\) to the correct numerator, nor does it use the constraint \(\sum a^2=\frac14\) to obtain the final bound; the 3‑point optimization checkpoint is also missed.
In addition, the proof contains an arithmeti... | Here is the complete, rigorous proof for the Olympiad problem.
### Proof
**Step 1: Simplify the condition and target inequality**
Let $x = a^2$, $y = b^2$, and $z = c^2$. Since $a, b, c$ are positive real numbers, we have $x, y, z > 0$.
The given condition is:
$$ x + y + z = \frac{1}{4} $$
We need to prove the ineq... | 22,425 | 10,533 | stop | false | oe-aops-0420a942ac-2746ce19 |
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-0420a942ac | 3 | 0 | [
0,
0
] | [
"<score>0</score>\n<assessment>In evaluating the supplied proof against the provided marking scheme, we find that it fails to meet any of the required checkpoints. The proof does not invoke the QM–AM inequality in the form \\( \\frac{1}{\\sqrt{b^2+c^2}}\\le\\frac{\\sqrt{2}}{b+c}\\), nor does it sum these inequalit... | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 0 | e the constraint $a^2+b^2+c^2=\frac14$ in a concluding step. Consequently, no points are earned for the QM‑AM bound (0 pts), the algebraic expansion (0 pts), or the optimization & conclusion (0 pts). The proof also includes several zero‑credit items (attempts at irrelevant bounds and vague statements of classical inequ... | # Proof of the Inequality
Let us define the terms involved in the inequality to make the statement more manageable.
Given positive real numbers $a, b, c$ such that $a^2 + b^2 + c^2 = \frac{1}{4}$.
Let the left-hand side (LHS) of the inequality be $L$.
Let the right-hand side (RHS) of the inequality be $R$.
$$ L = \fr... | 35,827 | 16,528 | stop | false | oe-aops-0420a942ac-3fba72d6 |
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-0476f4e70f | 0 | 0 | [
0,
0
] | ["<score>0</score>\n<assessment>\nThe proof does not satisfy any of the rubric checkpoints. \n1. **(...TRUNCATED) | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 0 | "determinants, Hadamard’s inequality, or any eigenvalue analysis, and provides no valid upper‑bo(...TRUNCATED) | "Here is the complete proof for the problem.\n\n```markdown\n# Proof of the Maximum Product of Segme(...TRUNCATED) | 43,260 | 14,614 | stop | false | oe-aops-0476f4e70f-03b44ae6 |
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-0476f4e70f | 1 | 0 | [
0,
0
] | ["<score>0</score>\n<assessment>The provided proof contains no valid argument for the maximum produc(...TRUNCATED) | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 0 | "umn/row norms bounded by $\\sqrt n$ (or matrix entries bounded by 1), so the 1‑point observation(...TRUNCATED) | "# Proof of the Maximum Product of Segment Lengths\n\n## Problem Statement\nLet $n \\geq 2$ be an in(...TRUNCATED) | 85,857 | 32,768 | length | true | oe-aops-0476f4e70f-2152b3fe |
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-0476f4e70f | 2 | 0 | [
0,
0
] | ["<score>0</score>\n<assessment>In order to award points the solution must satisfy the explicit chec(...TRUNCATED) | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 0 | " V|$. No credit. \n- **Checkpoint 3 (1 pt):** No observation about matrix entries or column/row (...TRUNCATED) | "# Proof of the Maximum Product of Segments\n\n**Problem Statement:**\nGiven an integer $n \\geq 2$ (...TRUNCATED) | 34,492 | 13,604 | stop | false | oe-aops-0476f4e70f-a1ba7834 |
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-0476f4e70f | 3 | 0 | [
0,
0
] | ["<score>0</score>\n<assessment>The proof does not address the problem at hand. It attempts to find (...TRUNCATED) | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 0 | "<score>0</score>\n<assessment>The proof does not address the problem as stated. The task is to maxi(...TRUNCATED) | "# Proof\n\n## Problem Statement\nLet $n \\geq 2$ be an integer. Consider a closed unit disc $D$, wh(...TRUNCATED) | 40,470 | 18,378 | stop | false | oe-aops-0476f4e70f-7d66a29e |
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-069b341fbd | 0 | 0 | [
0,
0
] | ["<score>0</score>\n<assessment>\nThe proof fails to meet any of the rubric checkpoints. It does not(...TRUNCATED) | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 0 | "proof misdefines the quantities $S$ and $T$, confuses $S^2-T\\ge0$ with the original inequality, an(...TRUNCATED) | "Here is the complete, rigorous proof for the given olympiad problem.\n\n### Proof\n\nLet $S$ denote(...TRUNCATED) | 35,041 | 15,290 | stop | false | oe-aops-069b341fbd-f450786a |
fp-rl2-2b | Qwen/Qwen3.5-2B | 0 | aops-069b341fbd | 1 | 0 | [
0,
0
] | ["<score>0</score>\n<assessment> \nThe submitted proof contains several fundamental errors that inv(...TRUNCATED) | [
"openai/gpt-oss-20b:groq",
"openai/gpt-oss-20b:groq"
] | 5 | "extbook target), earning 1 pt.\n\n**Polynomial analysis (3 pts out of 4)** \n- Definition of(...TRUNCATED) | "# Proof\n\n**Problem Statement:**\nGiven real numbers $a_i \\ge 0$ ($i=1, 2, \\ldots, n$) and real (...TRUNCATED) | 27,254 | 13,835 | stop | false | oe-aops-069b341fbd-4394680c |
FineProof RL runs
Everything the FineProof RL run produced, in three tables.
- The run:
fp-rl2-2b, Qwen3.5-2B trained with RL from the base model (no SFT) on FineProofEnvs for 500 updates. - The model: published as FineEnvs/Qwen3.5-2B-FineProof-RL.
- Held-out mean reward: 0.217 → 0.407 with the gpt-oss-20b judge, and 0.259 → 0.464 with a gpt-oss-120b regrade.
The same run as charts and traces: the Trackio dashboard. The recipe and the write-up: RESULTS.md.
eval_attempts (default)
One row per proof written in a held-out eval: 64 validated FineProofs-RL-test problems × 4 attempts, for each of the 14 evaluated checkpoints. Step 0 is the base model.
| Column | |
|---|---|
run, model, step |
which run and checkpoint |
task, attempt |
the problem (a FineProofEnvs test task) and the attempt, 0–3 |
proof |
the model's whole reply, as the judge read it |
reward |
the eval reward: mean of judge_scores / 7 |
judge_scores, judge_assessments, judge_models |
each gpt-oss-20b grade (0–7), its written assessment and the provider that gave it |
regrade_120b_score, regrade_120b_assessment |
gpt-oss-120b (high effort) regrade, at steps 0, 40, 100, 250, 450 and 500. The assessment keeps its last 1,500 characters |
completion_tokens, finish_reason, truncated |
the generation; a proof cut off at 32,768 tokens has finish_reason = length |
proof_chars, trial |
the proof's length, and the Harbor trial it came from |
from datasets import load_dataset
attempts = load_dataset("FineEnvs/fineproof-rl-runs", "eval_attempts", split="train")
final = attempts.filter(lambda r: r["run"] == "fp-rl2-2b" and r["step"] == 500)
train_updates
One row per training update: the reward and its spread, KL, entropy, gradient norm and loss from TRL. Also the rollout and judge statistics over the rollouts that finished since the last update: graded share, truncation, tokens, judge disagreement and cost. Plus timings and throughput.
train_rollouts
One row per training rollout (128 per update):
reward;judge_reward, the judge's grade of a proof that was cut off and so scored 0;- the judge's scores, the completion's length and finish reason, and wall time.
The training proofs themselves are not in this table. Every 10th update's are in the Trackio dashboard's traces.
Provenance
These tables are converted from the run's own logs (rollouts.jsonl, train.log, each eval's Harbor trials), with
no filtering. A repeat of the step-100 eval, launched by mistake when the run resumed, is left out; the original
step-100 eval is kept. The judge is an LLM grading against QED-Nano's marking schemes. A hand audit found it lenient
on proofs that keep contradictions around a correct argument, so read reward as the judge's view, not a
verified proof check.
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