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.

Streaming evaluation from a full cube table

Statement

Let $F$ be a field, let $n \geq 1$ and $N = 2^{n}$ , and let a stream supply each indexed pair $(b, f (b))$ for a table $f : {0, 1}^{n} \to F$ exactly once, in any order. Given $r \in F^{n}$ , the multilinear extension can be evaluated at $r$ in one pass with $O (n N)$ field operations and $O (n)$ field-element working storage, plus an $O (n)$ -bit current index. The input is the full table, not a succinct description.

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

The unique extension is the sum of table values times the product basis weights (Boolean-cube interpolation).

Proof

1.1

For each record compute $w_{b} = \prod_{i} (b_{i} r_{i} + (1 - b_{i}) (1 - r_{i}))$ and add $f (b) w_{b}$ to an accumulator initially zero. The interpolation formula says that after all records the accumulator equals $\tilde{f} (r)$ . This uses no divisions, even if a weight or table entry is zero.

F1algebra

2.1

Computing each weight costs $O (n)$ operations; multiplying by its table value and accumulating costs $O (1)$ . There are $N$ records. Store $r$ , the accumulator, the current value and one running product, using $n + O (1) = O (n)$ field elements since $n \geq 1$ . Endpoints $r_{i} = 0, 1$ and $n = 1$ require the same procedure. The work bound is exponential in $n$ and gives no polynomial-time evaluator from a succinct table specification.

step 1.1givenalgebra∎

Depends on

Boolean-cube interpolation

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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), Lemma 3.7, pp.30–31 (standard reference, not scraped)