Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-30
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FALSE: complementing a DFA complements its transition graph

Statement

False claim: to complement the language of a DFA, one complements its transition graph.

Facts & Assumptions

Given: The one-state DFA D=({q},{0,1},δ,q,{q}) with δ(q,0)=q and δ(q,1)=q.

[A1]

The statement refuted is: complementing a DFA means complementing the transition graph.

[L1]

The correct complement construction keeps the same transition function and replaces the accepting set F by QF, by Complementing the accepting states complements the recognized language.

[L2]

A DFA must have exactly one outgoing transition for each input letter from each state, by Deterministic finite automata.

Refutation

technique · direct
1.1

By [L1], the complement of L(D)={0,1} is the empty language, obtained by keeping the same looping transitions and changing only the accepting set from {q} to .

L1given
1.2

If one literally complements the transition graph instead, both loop edges are removed because they were the only edges present. The resulting picture has no outgoing transition on either input letter, so by [L2] it is not even a DFA.

givenL2
2.1

Thus graph complementation does not produce the complement DFA from step 1.1 and does not in general preserve the DFA conditions. 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

8 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