ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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.

Recovering $J_{3} (0), J_{2} (0), J_{1} (0)$ from ranks of powers

Example

Suppose a nilpotent endomorphism on a six-dimensional space has power ranks $ρ_{0} = 6, ρ_{1} = 3, ρ_{2} = 1, ρ_{3} = ρ_{4} = 0.$ Then its nilpotent Jordan blocks have sizes $3, 2, 1$ , each occurring once.

Facts & Assumptions

Given: The displayed rank sequence.

[L1]

The number of blocks of size at least $k$ is $ρ_{k - 1} - ρ_{k}$ , and the number of size exactly $k$ is $ρ_{k - 1} - 2 ρ_{k} + ρ_{k + 1}$ (Power ranks determine every nilpotent Jordan-block multiplicity).

Verification

technique · computation

1.1

The successive differences $ρ_{k - 1} - ρ_{k}$ are $3, 2, 1, 0$ for $k = 1, 2, 3, 4$ , so there are respectively three, two, one, and zero blocks of size at least those values.

L1algebra

2.1

Taking successive differences again gives one block of each exact size $1, 2, 3$ and none larger; their sizes sum to $6$ , as required.

step 1.1L1algebra∎

Depends on

Power ranks determine every nilpotent Jordan-block multiplicity

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 14 results over 7 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Recovering J3(0),J2(0),J1(0)J3​(0),J2​(0),J1​(0) from ranks of powers