Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-02
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.

Polynomial-time many-one reductions transfer P-, NP-, and coNP-membership

Statement

Let A,B{0,1} and suppose that ApB.

  1. If BP, then AP.
  2. If BNP, then ANP.
  3. If BcoNP, then AcoNP.

Facts & Assumptions

Given: Binary languages A,B{0,1} and a fixed total polynomial-time map f witnessing ApB.

[L1]

A polynomial-time many-one reduction is a total polynomial-time membership-preserving map, by Polynomial-time many-one reductions.

[L2]

Composites of polynomial-time many-one reductions are again polynomial-time many-one reductions, by Polynomial-time many-one reductions compose.

[L3]

P is the class of polynomial-time decidable languages, by The class P.

[L4]

NP is the class of languages with polynomial-time verifiers and polynomially bounded certificates, by The class NP via polynomial-time verifiers and Polynomial-time verifiers with polynomially bounded certificates.

[L5]

coNP consists of complements of NP-languages, by The class coNP.

Proof

technique · direct
1.1

Assume BP, and let DB be a polynomial-time decider for B from [L3]. On input x, compute the fixed map f(x) and then run DB on that output. By [L1], the answer is correct because xA iff f(x)B, and the whole procedure is polynomial-time because both stages are. Thus AP.

L1L3givenconstruct
1.2

Assume BNP. By [L4], choose a verifier relation RB and polynomial certificate bound p for B. Using the fixed reduction map f, define a verifier for A by declaring that (x,u) is accepted exactly when (f(x),u)RB. The output length f(x) is polynomial in x, so the bound p(f(x)) is still polynomial in x. Correctness follows from xA    f(x)B    u with up(f(x)) and (f(x),u)RB. Therefore ANP.

L1L4givenconstruct
2.1

Assume BcoNP. Then BcNP by [L5]. The fixed reduction map f also witnesses AcpBc, because xAc    xA    f(x)B    f(x)Bc. Applying step 1.2 to the reduction AcpBc gives AcNP, and then [L5] yields AcoNP. The composition principle [L2] guarantees that chaining such transfers remains inside the same reduction class.

L1L2L5step 1.2given
3.1

Steps 1.1, 1.2, and 2.1 prove the three transfer clauses.

step 1.1step 1.2step 2.1

Depends on

Used by

Dependency tree · two levels

8 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