LemmaStatement: Literature-sourcedProof: 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.

Perfect completeness of sum-check

Statement

In sum-check over a supplied finite field, with $n \geq 1$ , trusted individual bounds and trusted evaluation, if $H = \sum_{b} g (b)$ , the honest partial-sum polynomials pass every check for every challenge sequence. Thus the protocol has perfect completeness.

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

The protocol checks coefficient lengths and consistency, samples after each message, and compares the final claim with trusted evaluation; q_i and S_i denote its true partial sums (Sum-check with explicit degree bounds).

Proof

1.1

Use $q_{i}, S_{i}, C_{i}$ from the protocol. Substitution and finite summation cannot raise the degree in the unspecialized variable, so $q_{i}$ has degree at most $d_{i}$ and can be sent in the permitted coefficient format, including when $q_{i} = 0$ or $d_{i} = 0$ .

F1algebra

2.1

Splitting the remaining cube by coordinate $i$ gives $q_{i} (0) + q_{i} (1) = S_{i - 1}$ , while substitution gives $q_{i} (r_{i}) = S_{i}$ . Initially $C_{0} = H = S_{0}$ . If $C_{i - 1} = S_{i - 1}$ and $h_{i} = q_{i}$ , the consistency check passes and the update gives $C_{i} = S_{i}$ . Hence this equality propagates through every round.

F1step 1.1givenalgebra

3.1

At the end $S_{n} = g (r_{1}, \dots, r_{n})$ , since the remaining cube contains the empty tuple. Thus the terminal check passes. For $n = 1$ the same argument uses $q_{1} = g$ . It also covers all-zero or constant polynomials, challenges $0$ or $1$ , and characteristic two without any division.

step 2.1algebra∎

Depends on

Sum-check with explicit degree bounds

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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), Proposition 4.1, p.36 (standard reference, not scraped)