TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-05

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.

Vertex cover polynomial-time many-one reduces to set cover

Statement

VERTEX COVER polynomial-time many-one reduces to SET COVER.

Facts & Assumptions

Given: A VERTEX COVER instance $(G, k)$ , where $G = (V, E)$ .

[L1]

A polynomial-time many-one reduction is a total polynomial-time function preserving membership in both directions, by Polynomial-time many-one reductions.

[F1]

A vertex cover is a subset of vertices meeting every edge, by Clique, independent set, and vertex cover decision problems.

[F2]

A set cover chooses at most $k$ members of a family of subsets whose union is the whole universe, by The set cover decision problem.

Proof

technique · direct

1.1

Let the SET COVER universe be the edge set $U : = E$ . For each vertex $v \in V$ , define the subset $S_{v} : = {e \in E : v \in e}$ . Output the SET COVER instance $(U, {S_{v} : v \in V}, k)$ . This is computable in polynomial time by scanning the incidence relation of $G$ .

F1F2givenconstruct

2.1

If $C \subseteq V$ is a vertex cover with $∣ C ∣ \leq k$ , then every edge of $G$ has an endpoint in $C$ . Equivalently, every element of $U = E$ lies in one of the sets $S_{v}$ with $v \in C$ . Hence ${S_{v} : v \in C}$ is a set cover of size at most $k$ .

F1F2step 1.1

2.2

Conversely, if ${S_{v} : v \in C}$ is a set cover of size at most $k$ , then every edge $e \in E$ belongs to some $S_{v}$ with $v \in C$ . By definition of $S_{v}$ , that means $v$ is an endpoint of $e$ . Therefore $C$ is a vertex cover of size at most $k$ .

F1F2step 1.1

3.1

Steps 2.1 and 2.2 prove $(G, k) \in V E R T E X C O V E R ⟺ (U, {S_{v}}_{v \in V}, k) \in S E T C O V E R$ . By [L1], the construction in step 1.1 is a polynomial-time many-one reduction.

L1step 1.1step 2.1step 2.2∎

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

Fabrizio Iozzi and Luca Trevisan, Handout NP3 (standard reference, not scraped)