TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-02

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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.

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The Turing-machine acceptance problem is undecidable

Statement

The language $A_{T M}$ of The Turing-machine acceptance problem $A_{T M}$ is not decidable.

Facts & Assumptions

Given: The acceptance language $A_{T M}$ .

[L1]

A language is decidable exactly when some deterministic Turing machine halts on every input and answers membership correctly, by Decidable and recognizable languages.

[L2]

If a deterministic Turing machine $H$ decides $A_{T M}$ , then there exists a deterministic one-tape machine $D_{H}$ with input alphabet $Σ_{2}$ such that, for every coded machine $M$ whose input alphabet contains $Σ_{2}$ , $D_{H} accepts \overline{⌜ M ⌝} ⟺ M does not accept \overline{⌜ M ⌝},$ by A hypothetical decider for $A_{T M}$ yields a diagonal self-application machine.

[L3]

Every coded deterministic one-tape machine has one definite chosen code $⌜ M ⌝$ , by A fixed effective binary encoding of deterministic one-tape Turing machines.

Proof

technique · contradiction

1.1assume-contraL1L2construct

Assume for contradiction that $A_{T M}$ is decidable. Then [L1] gives a deterministic decider $H$ for $A_{T M}$ . Apply [L2] to that $H$ and obtain the corresponding machine $D_{H}$ . Because $D_{H}$ itself has input alphabet $Σ_{2}$ , the diagonal equivalence from [L2] applies to $M = D_{H}$ .

2.1L2L3step 1.1contradiction

Let $d = ⌜ D_{H} ⌝$ , whose existence is guaranteed by [L3]. Substituting $M = D_{H}$ into [L2] yields $D_{H} accepts \overline{d} ⟺ D_{H} does not accept \overline{d}$ . That is impossible.

3.1step 2.1discharge-contradiction: the assumed decider for $A_{TM}$ cannot exist∎

The contradiction in step 2.1 discharges the assumption from step 1.1. Therefore $A_{T M}$ is undecidable.

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

A. M. Turing, On Computable Numbers, with an Application to the Entscheidungsproblem (standard reference, not scraped)
EECS 376 Course Notes, Part 6: Computability (standard reference, not scraped)