Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Oracle diagonalization finite extension

Statement

Let c be a finite partial assignment of membership bits to binary words, and let M be a deterministic oracle machine with a fixed polynomial clock p(n). There exist n≥1 and a finite extension c′ of c such that every total oracle B extending c′ makes MB(1n) disagree with membership in LB={1j:B∩{0,1}j≠∅}.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

Fix a total language A⊆{0,1}∗. An oracle machine writes a query word and receives its membership bit in A in one answer step. Query writing counts toward time and the query tape toward space. A polynomial time clock bounds every branch for every oracle. PA and NPA are deterministic and nondeterministic polynomial-time oracle classes, respectively; the latter equivalently uses a polynomial-length witness and a deterministic polynomial-time A-oracle verifier. Use the conventions of the stated convention and the stated convention. For Σkp,A and Πkp,A, replace the deterministic predicate in the stated convention by a PA predicate; level zero is PA. Define PSPACEA by deterministic polynomial space under the charged-query convention. For a language class D, PD=⋃B∈DPB and NPD=⋃B∈DNPB. Finally Δk+1p=PΣkp. With a fixed base oracle, a machine may query both A and a language B; encode this by the tagged union A⊕B={0x:x∈A}∪{1x:x∈B}. (Relativized complexity class).

Proof

1.1F1

Choose n larger than every committed word length, with 2n>p(n). Such integers exist: for a fixed degree a, the ratio (n+1)a/2n+1 divided by na/2n tends to 1/2, and is eventually bounded below one, so any polynomial is eventually smaller than 2n. Simulate the clocked machine; answer committed queries consistently and commit every previously unassigned queried word to NO. The clock bounds both query length and count.

2.1step 1.1algebra

If it accepts, commit all n-bit words to NO. If it rejects (including clock expiration), choose an unqueried n-bit word and commit it to YES, committing every other n-bit word to NO. There is an unqueried word because fewer than 2n queries occurred and no earlier commitment had length n. Each extension is finite and consistent.

3.1step 2.1algebra∎

Every total extension retains all answers on the simulated path, including queries longer than n, so the deterministic path is unchanged. In the accepting case 1n∉LB; in the rejecting case 1n∈LB. Thus both outcomes defeat the machine for every total extension.

Depends on

Used by

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Sources