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.

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

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

Fix a total language $A \subseteq {0, 1}^{*}$ . 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. $P^{A}$ and $N P^{A}$ 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 $Σ_{k}^{p, A}$ and $Π_{k}^{p, A}$ , replace the deterministic predicate in the stated convention by a $P^{A}$ predicate; level zero is $P^{A}$ . Define ${P S P A C E}^{A}$ by deterministic polynomial space under the charged-query convention. For a language class $D$ , $P^{D} = ⋃_{B \in D} P^{B}$ and $N P^{D} = ⋃_{B \in D} N P^{B}$ . Finally $Δ_{k + 1}^{p} = P^{Σ_{k}^{p}}$ . With a fixed base oracle, a machine may query both $A$ and a language $B$ ; encode this by the tagged union $A \oplus B = {0 x : x \in A} \cup {1 x : x \in B}$ . (Relativized complexity class).

Proof

1.1

Choose $n$ larger than every committed word length, with $2^{n} > p (n)$ . Such integers exist: for a fixed degree $a$ , the ratio $(n + 1)^{a} / 2^{n + 1}$ divided by $n^{a} / 2^{n}$ tends to $1 / 2$ , and is eventually bounded below one, so any polynomial is eventually smaller than $2^{n}$ . 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.

2.1

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 $2^{n}$ queries occurred and no earlier commitment had length $n$ . Each extension is finite and consistent.

step 1.1algebra

3.1

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 $1^{n} \notin L_{B}$ ; in the rejecting case $1^{n} \in L_{B}$ . Thus both outcomes defeat the machine for every total extension.

step 2.1algebra∎

Depends on

Relativized complexity class

Used by

An oracle separates p from np Theorem

Dependency tree · two levels

4 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 proof of Theorem3.9, p72. (standard reference, not scraped)