PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-06

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Hierarchy theorems do not separate P from NP

Statement

The deterministic and nondeterministic time hierarchy conclusions do not themselves compare $P$ with $N P$ , and therefore do not constitute a proof of either $P = N P$ or $P \neq N P$ .

Facts & Assumptions

Given: the quantified conclusions of the two hierarchy theorems.

[L1]

A binary language is in verifier-defined NP exactly when it belongs to $N T I M E (n^{k})$ for some $k \geq 1$ (The verifier and nondeterministic definitions of NP agree).

Proof

technique · direct

1.1

Each hierarchy compares two deterministic classes or two nondeterministic classes at separated resource bounds, such as $D T I M E (f)$ versus $D T I M E (g)$ .

given

2.1

By [L1] and the definition of $P$ , the $P$ versus $N P$ question compares the unions of deterministic and nondeterministic polynomial-time classes. A separation needs one language in $N P$ outside every deterministic polynomial-time class; equality needs polynomial deterministic simulations for all $N P$ languages. The witnesses in step 1.1 only separate two bounds in the same model, and their bounds and witnesses may vary. These stated conclusions do not themselves provide either of the required cross-model arguments. This is a comparison of what the theorem statements assert, not a formal independence claim about their consequences in an axiomatic theory.

L1step 1.1algebra∎

Depends on

Used by

Why a hierarchy separation is not a P versus NP separation Example

Dependency tree · two levels

11 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 and Barak, Computational Complexity, Chapter 3 (standard reference, not scraped)