LemmaStatement: AI-adaptedProof: 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.

Linear-time evaluation from a full cube table

Statement

For every field $F$ , $n \geq 0$ , full table $f : {0, 1}^{n} \to F$ , and $r \in F^{n}$ , one can compute $\tilde{f} (r)$ in $O (2^{n})$ field operations and $O (2^{n} + n)$ storage, measured in field elements with indexed array access.

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

The unique extension is the finite weighted sum in the Boolean interpolation formula (Boolean-cube interpolation).

Proof

1.1givenalgebra

Start with the one-entry array $W_{0} (()) = 1$ . At stage $i$ , replace each entry $W_{i - 1} (b)$ by entries $W_{i} (b, 0) = W_{i - 1} (b) (1 - r_{i})$ and $W_{i} (b, 1) = W_{i - 1} (b) r_{i}$ . After stage $i$ each entry is the product of the appropriate first $i$ factors, directly by this recurrence. This remains valid for zero or unit coordinates.

2.1F1step 1.1algebra

After stage $n$ , return $\sum_{b} f (b) W_{n} (b)$ . The weights equal the interpolation basis evaluated at $r$ , so this is $\tilde{f} (r)$ . Zero tables or zero weights need no separate treatment.

3.1step 1.1step 2.1algebra∎

The weight construction costs $O (\sum_{i = 1}^{n} 2^{i}) = O (2^{n})$ operations, and the final weighted sum costs $O (2^{n})$ . Holding the table, two successive weight arrays and $r$ uses $O (2^{n} + n)$ storage. For $n = 0$ , return the sole table entry; for $n = 1$ , the construction returns $(1 - r_{1}) f (0) + r_{1} f (1)$ . The input still has $2^{n}$ values.

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 3.8, pp.31–32 (standard reference, not scraped)