False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 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.

FALSE: converting to Chomsky normal form preserves epsilon without a special start-symbol exception

Statement

False claim: converting a context-free grammar to Chomsky normal form preserves $ε$ without any special exception.

Facts & Assumptions

Given: The grammar with one variable and one production, $S \to ε .$

[A1]

The statement refuted is: CNF conversion preserves $ε$ without a special start-symbol exception.

[L1]

Chomsky normal form allows an $ε$ -production only in the special case $S \to ε$ with $S$ absent from every right-hand side, by Chomsky normal form.

[L2]

Every context-free grammar is equivalent to one in Chomsky normal form, by Every context-free language has an equivalent grammar in Chomsky normal form.

Refutation

technique · direct

1.1

The given grammar generates exactly the language ${ε}$ . So any equivalent CNF grammar must still generate $ε$ .

givenL2

1.2

Without the exception described in [L1], every CNF production would have the form $A \to B C$ or $A \to a$ . Such rules can produce only nonempty terminal words, because every terminal step contributes one symbol and no rule deletes all remaining symbols. Therefore a grammar with only those two rule types cannot generate $ε$ .

2.1

Step 1.2 contradicts the requirement from step 1.1, so the special start-symbol exception is genuinely necessary. The claim [A1] is false.

A1step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

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

John Watrous, Introduction to the Theory of Computing, Lecture 8 (standard reference, not scraped)