Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-12
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.

A variable lies in a CYK table cell exactly when it derives that span

Statement

Let G be a Chomsky-normal-form grammar, let w=a1an be a nonempty word, and let T[i,] be the CYK table of G on w. Then for every variable A and every valid cell (i,), AT[i,]if and only ifAGaiai+1.

Facts & Assumptions

Given: A CNF grammar G, a nonempty word w=a1an, and its CYK table.

[L1]

By The CYK parsing table and span variables, the cells T[i,] are defined by terminal productions for =1 and by binary productions ABC together with a split point for 2.

[L2]

By One-step derivation and finite derivation in a context-free grammar, AGu means that the sentential form A derives the word u by finitely many grammar steps.

Proof

technique · direct
1.1

We induct on the span length . For =1, [L1] says that AT[i,1] exactly when Aai is a production, and by [L2] that is exactly when AGai.

L1L2given
2.1

Assume the claim known for all shorter spans and let 2. If AT[i,], then [L1] provides a split k and a production ABC with BT[i,k] and CT[i+k,k]. By the induction hypothesis, BGaiai+k1 and CGai+kai+1, so one use of ABC followed by those derivations gives AGaiai+1.

L1L2step 1.1induction
2.2

Conversely, suppose AGaiai+1 with 2. Because G is in CNF, the first derivation step from A must be a binary production ABC, and the derived word must split at some k into a prefix from B and a suffix from C. The induction hypothesis then gives BT[i,k] and CT[i+k,k], so [L1] places A in T[i,].

L1L2step 1.1induction
3.1

Steps 1.1, 2.1, and 2.2 prove the equivalence for every cell.

step 1.1step 2.1step 2.2

Depends on

Used by

Dependency tree · two levels

6 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