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

Integral manifolds have the distribution dimension

Statement

Let $D$ be a rank- $k$ smooth distribution on $M$ , and let $i : N \to M$ be an integral manifold of $D$ . Then $\dim N = k$ .

Facts & Assumptions

Given: A rank- $k$ smooth distribution $D$ and an integral manifold $i : N \to M$ of $D$ .

[A1]

For every $q \in N$ , the image of $d i_{q}$ equals $D_{i (q)}$ .

Proof

technique · direct

1.1

Because $i$ is an immersion, each $d i_{q} : T_{q} N \to T_{i (q)} M$ is injective. [given] Since $i$ is integral, its image is $D_{i (q)}$ , which has dimension $k$ . Therefore $\dim T_{q} N = k$ for every $q \in N$ .

given

1.2

The dimension of a manifold is the common dimension of its tangent spaces, [given] so $\dim N = k$ .

given∎

Depends on

Used by

Nothing in the library uses this result yet.

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

Will J. Merry, Differential Geometry (standard reference, not scraped)