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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
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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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