LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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.

A diagonal set at each arithmetical level

Statement

For every $n \geq 0$ , there is a $Σ_{n + 1}^{0}$ set which is not $Π_{n + 1}^{0}$ ; its complement is $Π_{n + 1}^{0}$ and not $Σ_{n + 1}^{0}$ .

Facts & Assumptions

Given: a finite level $n \geq 0$ .

Proof

technique · contradiction

1.1givenassume-contra

Enumerate the $0^{(n)}$ -c.e. sets as $(W_{e}^{0^{(n)}})_{e}$ and set $D_{n} = {e : e \notin W_{e}^{0^{(n)}}}$ . If $D_{n}$ were itself $0^{(n)}$ -c.e., say $D_{n} = W_{d}^{0^{(n)}}$ , substituting $e = d$ gives $d \in D_{n} ⟺ d \notin D_{n}$ .

2.1step 1.1discharge-contradiction∎

Hence $D_{n}$ is not $0^{(n)}$ -c.e., while its complement is. Post's theorem identifies the complement as $Σ_{n + 1}^{0}$ and $D_{n}$ as $Π_{n + 1}^{0}$ ; exchanging them gives the claimed pair.

Depends on

Used by

The arithmetical hierarchy is strict Theorem

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

Ludovic Patey, Computability Theory, §§4.5–4.6 (standard reference, not scraped)