Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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.

A space-bounded machine has exponentially many configurations

Statement

Fix a finite k-tape machine M, a single input x of length n, and an integer visited-cell bound S. Configurations reached from this input while using at most S visited cells have injective fixed-length binary encodings of length OM(S+log(n+2)), and there are at most 2OM(S+log(n+2)) of them. The decoder may use the fixed x. If Sn the bounds are OM(S) and 2OM(S) for nonempty feasible configurations.

Facts & Assumptions

Given: The fixed machine in the deterministic k-tape model of Multitape and nondeterministic Turing machine variants, the standard initial configuration with every head at zero, x on tape one and every other tape blank, and the fixed space bound; no bound across different inputs is asserted.

[F1]

Space is the number of distinct scanned tape/cell pairs, includes every initial head cell, and does not count initially occupied but unvisited input cells (Polynomial space and nondeterministic polynomial space).

[F2]

A one-tape configuration consists of state, head position and whole finite-support tape content (Configurations of a deterministic one-tape Turing machine).

[F3]

Initially the input occupies the first tape starting at zero, all other cells are blank, and halting predicates are machine-relative (Initial tapes and machine-relative halting configurations).

Proof

1.1

Every head begins at zero and moves by one cell per transition, with a left move at zero unable to leave the right-infinite tape. Hence the cells visited on each tape form an initial segment. A transition writes only at the currently scanned cell, so every unvisited cell retains its initial content. Suppose Sk. For each tape let ri1 be the visited prefix length; iriS. Encode the state by a fixed-width word, each ri and head position by a width-log2(S+1) field, and the concatenation of the visited tape prefixes by at most Slog2(Γ+1) bits. Pad the last field to this fixed length with zeros. Reconstruct the unvisited suffix on tape one from the fixed x, and all other unvisited cells as blank. These data recover the full configuration, extending F2's triple to the fixed number of tapes.

givenF1F2F3
2.1

To make the map canonical on configurations rather than histories, use on tape i the least prefix containing its head and every cell whose current content differs from its initial content. This prefix is nonempty and no longer than the visited prefix, because writes require a visit. Apply the encoding just described to these canonical prefixes. Their lengths sum to at most S, and decoding recovers the exact configuration, so equal codes imply equal configurations. Different histories reaching the same configuration now give the same code.

step 1.1F1F3
3.1

The code length is at most cM+2klog2(S+1)+Slog2(Γ+1)=OM(S) for Sk1. There are at most 2m binary words of a fixed length m, proved by doubling the count for each appended bit, starting with one empty word. Hence the number of configurations is at most 2OM(S), in particular the stated weaker bound with the input-length logarithm. If S<k, even the initial head cells exceed the bound and the reachable set is empty; there is then a vacuous encoding. Empty input and blank-only tape data need no special extra field. No uniform code independent of the fixed input is claimed for an arbitrary unvisited input suffix.

step 2.1F1

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Sources