Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-02
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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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An NP-complete language in P forces P=NP

Statement

If some NP-complete language belongs to P, then P=NP.

Facts & Assumptions

Given: An NP-complete language C with CP.

[L1]

NP-complete means both CNP and every language in NP reduces to C in polynomial time, by NP-hard and NP-complete languages.

[L2]

Polynomial-time many-one reductions transfer P-membership backward: if ApB and BP, then AP, by Polynomial-time many-one reductions transfer P-, NP-, and coNP-membership.

[L3]

Every language in P belongs to NP, by PNPcoNP.

Proof

technique · direct
1.1

Let LNP be arbitrary. By [L1], because C is NP-hard, there is a polynomial-time many-one reduction LpC. Then [L2] and the hypothesis CP give LP.

L1L2given
2.1

Step 1.1 shows NPP. By [L3], every language in P also lies in NP, so PNP. Therefore P=NP.

L3step 1.1

Depends on

Used by

Dependency tree · two levels

9 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