LemmaStatement: Literature-sourcedProof: 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.

Every nonhalting deterministic configuration has a unique successor

Statement

Let $M$ be a deterministic one-tape Turing machine. Every nonhalting configuration of $M$ has exactly one one-step successor.

Facts & Assumptions

Given: A deterministic one-tape Turing machine $M = (Q, Σ, Γ, ⊔, q_{0}, q_{a c c}, q_{r e j}, δ)$ and a nonhalting configuration $C = (q, h, t)$ of $M$ .

[L1]

The transition function of a deterministic one-tape Turing machine is a function $δ : (Q ∖ {q_{a c c}, q_{r e j}}) \times Γ \to Q \times Γ \times {L, R},$ by Deterministic one-tape Turing machines with designated accept and reject states.

[L2]

The relation $C ⊢_{M} C^{'}$ is obtained by applying the unique transition value $δ (q, t (h))$ , rewriting only the scanned tape cell, and updating the head position by the stated left/right rule, by The one-step configuration relation.

Proof

technique · direct

1.1

Since $C$ is nonhalting, $q \in Q ∖ {q_{a c c}, q_{r e j}}$ . Therefore $(q, t (h))$ lies in the domain of the function $δ$ , so there is a unique triple $(p, b, D) = δ (q, t (h))$ .

givenL1

2.1

Define $t^{'}$ by $t^{'} (h) = b$ and $t^{'} (i) = t (i)$ for $i \neq h$ , and define $h^{'}$ from $h$ and $D$ by the rule in [L2]. Then $C^{'} : = (p, h^{'}, t^{'})$ satisfies $C ⊢_{M} C^{'}$ .

L2step 1.1construct

3.1

If also $C ⊢_{M} C^{''}$ , then [L2] forces $C^{''}$ to use the same unique transition value from step 1.1, the same rewritten tape cell, and the same updated head position. Hence $C^{''} = C^{'}$ .

L1L2step 1.1step 2.1

4.1

So $C$ has exactly one one-step successor.

step 2.1step 3.1∎

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 E. Savage, Models of Computation: Exploring the Power of Computing, Chapter 5 (standard reference, not scraped)
Richard Zach, Sets, Logic, Computation: An Open Introduction to Metalogic (standard reference, not scraped)