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.

Complementing the accepting states complements the recognized language

Statement

Let $D = (Q, Σ, δ, q_{0}, F)$ be a DFA and let $D^{c} : = (Q, Σ, δ, q_{0}, Q ∖ F) .$ Then $L (D^{c}) = Σ^{*} ∖ L (D) .$

Facts & Assumptions

Given: A DFA $D = (Q, Σ, δ, q_{0}, F)$ over $Σ$ .

[L1]

For languages over a fixed alphabet, the complement of $L$ is $L^{c} = Σ^{*} ∖ L$ , by Boolean operations on languages over a fixed alphabet.

[L2]

A word $w$ is accepted by a DFA exactly when the final state $δ^{*} (q_{0}, w)$ lies in its accepting set, by Acceptance of a word by a DFA and the recognized language.

Proof

technique · direct

1.1given

The machines $D$ and $D^{c}$ have the same state set, the same start state, and the same transition function. Therefore, for every word $w \in Σ^{*}$ , they follow exactly the same run and end in the same state $δ^{*} (q_{0}, w)$ .

2.1L2step 1.1

By [L2], the word $w$ is accepted by $D^{c}$ exactly when this common final state lies in $Q ∖ F$ , which is equivalent to saying that it does not lie in $F$ . So $w$ is accepted by $D^{c}$ exactly when $w$ is not accepted by $D$ .

3.1L1step 2.1∎

Since this equivalence holds for every $w \in Σ^{*}$ , the language of $D^{c}$ is precisely $Σ^{*} ∖ L (D)$ , which is $L (D)^{c}$ by [L1].

Depends on

Used by

Dependency tree · two levels

10 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 4 (standard reference, not scraped)