PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 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.

$P \subseteq N P \cap c o N P$

Statement

Every language in $P$ belongs to both $N P$ and $c o N P$ .

Facts & Assumptions

Given: A language $L \subseteq {0, 1}^{*}$ in $P$ .

[L1]

$L \in P$ means that some deterministic Turing machine decides $L$ in polynomial time, by The class P.

[L2]

$L \in N P$ means that $L$ has a polynomial-time verifier with polynomially bounded certificates, by The class NP via polynomial-time verifiers and Polynomial-time verifiers with polynomially bounded certificates.

[L3]

$L \in c o N P$ means that the complement of $L$ belongs to $N P$ , by The class coNP.

Proof

technique · direct

1.1L1L2givenconstruct

By [L1], choose a polynomial-time decider $D$ for $L$ . Use the empty certificate $u = ε$ and define the verifier relation $R_{L}$ by $(x, ε) \in R_{L}$ exactly when $D$ accepts $x$ , with no other certificates allowed. Then the paired language for $R_{L}$ is decidable in polynomial time by running $D$ on the first component and checking that the second component is empty. Thus [L2] gives $L \in N P$ .

2.1L1L2L3step 1.1construct

Swap the accepting and rejecting halting states of $D$ to obtain a polynomial-time decider for the complement ${0, 1}^{*} ∖ L$ . Repeating step 1.1 with the same empty-certificate idea shows that the complement belongs to $N P$ . Therefore [L3] gives $L \in c o N P$ .

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 show that every language in $P$ lies in $N P \cap c o N P$ .

Depends on

Used by

A regular parity language refutes 'NP means not polynomial' Counterexample
An even-length language sits in P, hence also in NP and coNP Example
FALSE: NP means not polynomial-time solvable False statement
An NP-complete language in P forces P=NP Proposition
Max-3SAT has no PTAS unless P=NP Theorem
Maximum independent set has no PTAS unless P=NP 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

Sanjeev Arora and Boaz Barak, Computational Complexity: A Modern Approach (standard reference, not scraped)
Michael Sipser, MIT 18.404J Theory of Computation, Lecture 14: P and NP, SAT, Poly-time Reducibility (standard reference, not scraped)

P⊆NP∩coNPP⊆NP∩coNP