Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30
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.

Eliminating useless symbols preserves the generated language

Statement

For every context-free grammar G there exists a context-free grammar G such that L(G)=L(G) and either:

  1. L(G)= and G is the evident empty-language grammar, or
  2. every variable of G is both generating and reachable.

Facts & Assumptions

Given: A context-free grammar G=(V,Σ,P,S).

[L1]

Variables may be nullable, generating, reachable, or useful exactly as in Nullable, generating, and reachable variables.

[L2]

The generated language is L(G)={wΣ:SGw}, by The language generated by a CFG.

Proof

technique · direct
1.1

Let TV be the generating variables of G, and let P1 consist of the productions whose left-hand side lies in T and whose right-hand side contains no variable outside T. If ST, then no terminal word is derivable from S, so [L2] gives L(G)= and we may take G to be any fixed grammar generating the empty language. If ST, set G1:=(T,Σ,P1,S).

L1L2givenconstruct
1.2

In the case ST, let RT be the variables reachable from S in G1, let P2 consist of the productions in P1 whose left-hand side lies in R and whose right-hand side contains no variable outside R, and set G2:=(R,Σ,P2,S). Every derivation beginning at S stays inside R by definition of reachability.

L1givenconstruct
2.1

Assume now that ST. Any derivation of a terminal word can use only generating variables, because every variable appearing in that derivation must eventually derive a terminal subword. Conversely, every rule kept in G1 was already a rule of G. Hence L(G1)=L(G).

L1L2step 1.1
3.1

Therefore deleting the unreachable variables changes no derivation from S to a terminal word, while every variable remaining in G2 is both generating and reachable. So L(G2)=L(G1)=L(G).

step 2.1step 1.2
4.1

Taking G=G2 in the case ST and the empty-language grammar in the case ST proves the theorem.

step 1.1step 3.1

Depends on

Used by

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