ExampleConstruction: AI-adaptedVerification: 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.

Np and conp are the first levels

Example

Formula satisfiability and tautology exhibit the existential and universal first levels. For a finite simple graph $G$ and integer $k$ , the language $E X A C T - I N D S E T$ defined by $α (G) = k$ belongs to $Σ_{2}^{p} \cap Π_{2}^{p}$ ; no completeness claim is made for it. Malformed inputs and $k < 0$ or $k > ∣ V (G) ∣$ are rejected.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

$Σ_{1}^{p} = N P$ and $Π_{1}^{p} = c o N P$ . (Np and conp are the first levels).

[F2]

For every fixed $k \geq 1$ , truth of prenex Boolean formulas with $k$ alternating blocks beginning existentially is $Σ_{k}^{p}$ -complete under polynomial-time many-one reductions. The universally starting version is $Π_{k}^{p}$ -complete. The matrix is a general Boolean formula, empty blocks are allowed, and malformed encodings are rejected. (Bounded alternation qbf is level complete).

Verification

1.1

A formula is satisfiable exactly when some assignment makes its polynomial-time evaluation true, and is a tautology exactly when all assignments do. These are first-level predicates, with completeness for the corresponding bounded-QBF encodings. A formula with no variables has one assignment.

F1F2

2.1

For valid $0 \leq k \leq n$ , let $I_{j} (S)$ test that an $n$ -bit indicator represents an independent set of cardinality $j$ . Then the exact condition is $\exists S \forall T [I_{k} (S) \land \neg I_{k + 1} (T)]$ . Reversing the two independent blocks gives the same truth value: in the reverse direction fix any $T$ to obtain a witness $S$ , and every $T$ must fail $I_{k + 1}$ . If both conditions hold one witness $S$ works for every $T$ . At $k = 0$ the empty set witnesses $I_{0}$ ; at $k = n$ , $I_{n + 1}$ is always false. This also covers the graph with no vertices. A deterministic validity test in the matrix rejects all malformed and out-of-range inputs for either polarity.

step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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; Examples5.2–5.3 and5.9, pp91–94. (standard reference, not scraped)