CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

One point evaluation for an exponentially large cube

Statement refuted

The assertion that a sum-check verifier must evaluate $g$ separately at every Boolean vertex is false. For $n \geq 1$ , take $g (X_{1}, \dots, X_{n}) = \prod_{i} X_{i}$ over a supplied finite field, with individual bounds $d_{i} = 1$ . Its cube sum is one although the cube has $2^{n}$ points. The verifier uses $O (n)$ field work, including one evaluation of $g$ with at most $n$ multiplications. False initial claims have acceptance probability at most $\min (1, n / ∣ F ∣)$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

The verifier checks coefficient messages round by round and evaluates g at the terminal challenge point (Sum-check with explicit degree bounds).

[F2]

Verifier field work is O(D) plus a single terminal point evaluation with D=sum_i(d_i+1) (Sum-check verifier resources).

[F3]

The false-claim acceptance probability is at most min(1,sum_i d_i/|F|) (Adaptive-prover soundness of sum-check).

Counterexample

1.1givenalgebra

At the all-one vertex the product equals one, and at every other Boolean vertex a zero coordinate makes it zero. Hence the cube sum is one. There are two independent choices per coordinate, giving $2^{n}$ vertices, also when $n = 1$ .

1.2F1F2algebra

The specified protocol uses one terminal evaluation. Implement it by multiplying the $n$ supplied coordinates starting from one, at a cost of at most $n$ multiplications. Here $D = \sum_{i} (1 + 1) = 2 n$ , so the resource lemma bounds all verifier field work, including this evaluation, by $O (n)$ . Zeros among the challenge coordinates simply make the product zero, without any division.

2.1F3step 1.1step 1.2algebra∎

The soundness theorem gives the stated false-claim bound from $\sum_{i} d_{i} = n$ , including the cap at one when the field is small. Thus the verifier does not need $2^{n}$ separate evaluations of $g$ . The claim concerns verifier work with a supplied evaluator and imposes no general efficiency assertion on the prover.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Justin Thaler, Proofs, Arguments, and Zero-Knowledge (2023), §4.1 protocol and resource analysis pp.33–38 (standard reference, not scraped)