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.

Ph adaptive oracle transcript normal form

Statement

For fixed k1 and BΣkp, every nondeterministic polynomial-time B-oracle computation has a Σk+1p definition. More generally, for a fixed total base oracle A and BΣkp,A, polynomial nondeterministic access to both A and B has a Σk+1p,A definition.

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=BDPB and NPD=BDNPB. 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 AB={0x:xA}{1x:xB}. (Relativized complexity class).

[F2]

For each fixed k0, both Σkp and Πkp are closed under polynomial-time many-one preimages and polynomially many uniformly indexed conjunctions and disjunctions. Precisely, if the language of pairs (x,i) lies in the class and 0i<q(x) for a polynomial-time computable polynomial bound q, then both “every such i” and “some such i” define languages in that class. Complements interchange Σkp and Πkp, and each is contained in both classes at level k+1. These assertions also hold with one fixed base oracle. (Ph quantifier block closure).

Proof

1.1

Use the oracle-independent polynomial clock to bound the branch string, number and lengths of queries, and the entire transcript. Guess in a first existential block a branch and all answers to B queries. Replay the machine using these answers, making any A calls directly; reject if the replay does not accept. Thus the replay test is in PA.

F1
2.1

Write B(q)=u1u2QkukR(q,u1,,uk) with RPA. For each guessed YES answer, guess its u1 in the first block too; its remaining condition starts universally and has k1 blocks. For each NO answer, use the complementary k-block universal-first formula. Pad YES conditions with an unused last block, and combine the conditions in parallel into k alternating universal-first blocks by tuple conjunction closure. Unused transcript slots impose the constant true condition. At k=1, a YES condition has no remaining quantified variable.

F2step 1.1
3.1

An actually accepting branch supplies its correct transcript and YES witnesses, so the constructed formula is true. Conversely a true formula certifies every YES and every NO answer, and the replay is therefore an actual accepting branch. Tuple lengths are polynomial because both query count and query length are polynomially bounded. Prefixing the guessed first block gives exactly k+1 alternating blocks over PA. For the unrelativized assertion take A=.

step 2.1algebra

Depends on

Used by

Dependency tree · two levels

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Sources