ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-11

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.

The content of a concrete three-dimensional parallelepiped computed from its spanning matrix

Example

Let $v_{1} = (2, 0, 0), v_{2} = (1, 3, 0), v_{3} = (0, 1, - 2) .$ The parallelepiped spanned by these vectors has Jordan content $12$ .

Facts & Assumptions

Given: The three spanning vectors in the statement.

[L1]

The content of a square-matrix parallelepiped is the absolute value of its determinant (The Jordan content of the parallelepiped spanned by the columns of a square real matrix is the absolute value of its determinant).

[L2]

For a commutative ring $R$ , $n \geq 1$ , and $A = (a_{i j}) \in M_{n} (R)$ , the determinant is the finite signed-permutation sum $\det (A) = \sum_{σ \in S_{n}} sgn (σ) \prod_{i < n} a_{σ (i), i}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

Verification

technique · computation

1.1

Put the vectors into the columns of the upper-triangular matrix $A = (\begin{array}{ccc} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & - 2 \end{array}) .$

given

2.1

The signed-permutation formula [L2] leaves only the diagonal term, so $\det A = 2 \cdot 3 \cdot (- 2) = - 12$ .

L2step 1.1

3.1

By [L1], the content is $∣ - 12 ∣ = 12$ . Directly, the first two vectors span a base of area $6$ in the horizontal plane and the third has perpendicular height $2$ , again giving $12$ .

L1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

A. Leibman, Multidimensional Real Analysis, Lemma 5.5.4 (standard reference, not scraped)