CorollaryStatement: 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.

Relativization alone cannot resolve p versus np

Statement

Neither an argument for $P = N P$ nor an argument for $P \neq N P$ can be valid unchanged relative to every total oracle.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

An argument about complexity classes relativizes if its asserted reasoning remains valid when every machine in it receives the same arbitrary total membership oracle, using the resource conventions of the stated convention. This is a property of an argument, not a claimed algorithm for classifying all proofs. (Relativizing proof technique).

[F2]

There is a total oracle $B$ for which $P^{B} \neq N P^{B}$ . In fact $L_{B} = {1^{n} : B \cap {0, 1}^{n} \neq \emptyset}$ lies in $N P^{B} ∖ P^{B}$ . (An oracle separates p from np).

[F3]

For $A = T Q B F$ , $P^{A} = N P^{A} = P S P A C E$ . (An oracle collapses p and np).

Proof

1.1

By the definition of relativization, an argument for equality with this property would prove $P^{B} = N P^{B}$ for every total $B$ . The separating oracle supplies a counterinstance.

F1F2

2.1

An argument for inequality with this property would prove $P^{A} \neq N P^{A}$ for every total $A$ . The TQBF oracle instead gives equality. These two obstructions concern the property of the argument, and imply neither an unrelativized equality nor an unrelativized separation.

F1F3∎

Depends on

Used by

Relativization alone cannot resolve p versus np Example

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

Arora–Barak, Computational Complexity, 2007 draft; §3.5 conclusion, p72. (standard reference, not scraped)