Alphabeta Math
False statementConstruction: AI-adaptedVerification: 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.

FALSE: nondeterministic acceptance requires every branch to halt

Statement

False claim: a nondeterministic Turing machine accepts an input only when every computation branch on that input halts.

Facts & Assumptions

Given: A nondeterministic machine with states q0,qacc,qrej,qL, input alphabet {1}, tape alphabet {1,⊔}, and allowed transitions Δ(q0,1)={(qacc,1,R),(qL,1,L)},Δ(qL,1)={(qL,1,L)}. Take input word w=1.

[A1]

The statement refuted is: nondeterministic acceptance requires every branch on the input to halt.

[L1]

A nondeterministic machine accepts an input when there exists an accepting computation on that input, by Accepting computations of a nondeterministic machine.

Refutation

technique · direct
1.1given

One allowed first move from the initial configuration is (qacc,1,R). So there is a one-step computation branch that ends immediately in the accept state.

1.2given

Another allowed first move is (qL,1,L). From then on the only allowed move is again (qL,1,L), and the left-boundary rule keeps the head at cell 0, so this branch never leaves the nonhalting state qL and therefore does not halt.

2.1A1L1step 1.1step 1.2∎

By [L1] and step 1.1, the machine accepts w because an accepting computation exists. But step 1.2 gives a branch that does not halt. Therefore [A1] is false.

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