Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Ph complete language forces collapse

Statement

If a language C∈PH is polynomial-time many-one hard for all of PH, then PH collapses to a finite level. In particular, PH=PSPACE implies such a collapse.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

For each fixed k≥0, 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 0≤i<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).

[F2]

For fixed k≥1, Σkp=Πkp implies PH=Σkp=Πkp. Also P=NP implies PH=P. (Sigma k equals pi k implies ph collapse).

[F3]

TQBF lies in PSPACE, and every language in PSPACE polynomial-time many-one reduces to TQBF. Therefore TQBF is PSPACE-complete. (TQBF is PSPACE-complete).

Proof

1.1F1F2

Choose a fixed k≥1 with C∈Σkp, padding upward if necessary. Every L∈Πkp belongs to PH by padding and reduces to C, hence lies in Σkp by preimage closure. Complementing gives the reverse inclusion, so the finite-level collapse theorem applies.

2.1F3step 1.1∎

Under PH=PSPACE, TQBF belongs to PH and every PH language reduces to it, by its exact PSPACE-completeness statement. It therefore serves as C in the first step. No strict containment between PH and PSPACE is assumed.

Depends on

Used by

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Dependency tree · two levels

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Sources