TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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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.

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An oracle separates p from np

Statement

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 $N P^{B} ∖ P^{B}$ .

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

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 $n \geq 1$ and a finite extension $c^{'}$ of $c$ such that every total oracle $B$ extending $c^{'}$ makes $M^{B} (1^{n})$ disagree with membership in $L_{B} = {1^{j} : B \cap {0, 1}^{j} \neq \emptyset}$ . (Oracle diagonalization finite extension).

Proof

1.1

Enumerate all deterministic oracle programs paired with clocks $(n + 2)^{a} + a$ , for positive integers $a$ . Every polynomial-time oracle decider agrees with some clocked program for its oracle. Starting with no commitments, apply the finite-extension lemma once to each pair, always preserving all prior commitments. Let $B$ have precisely the words ever assigned YES; all other words receive NO. This is a consistent total extension of every stage.

2.1

At the assigned input the corresponding clocked program fails by the lemma, so no polynomial-time $B$ decider decides $L_{B}$ . A nondeterministic machine rejects nonunary inputs, guesses one $n$ -bit word on $1^{n}$ , and asks whether it belongs to $B$ . It accepts exactly $L_{B}$ within polynomial time; at $n = 0$ it queries the empty word. Thus the separation follows.

F1step 1.1∎

Depends on

Oracle diagonalization finite extension

Used by

Relativization alone cannot resolve p versus np Corollary
Relativized separations prove unrelativized separations False statement

Dependency tree · two levels

2 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 Theorem3.9, pp71–72. (standard reference, not scraped)