ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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.

A wedge product detects linear dependence in concrete coordinates

Example

In $R^{3}$ , four samples pair the zero/nonzero behaviour of the wedge with dependence/independence. Dependent pairs give the zero wedge: $e_{1} \land (2 e_{1}) = 0$ , and for $u = (1, 2, 0)$ , $v = (2, 4, 0)$ one has $u \land v = 0$ because $v = 2 u$ . Independent pairs give a nonzero wedge: $e_{1} \land e_{2} \neq 0$ , and for $u = (1, 0, 0)$ , $v = (0, 1, 0)$ one has $u \land v = e_{1} \land e_{2} \neq 0$ .

Facts & Assumptions

Given: The standard basis of $R^{3}$ and the four displayed pairs.

[L1]

A decomposable wedge is nonzero exactly when its vectors are linearly independent (In a finite-dimensional vector space, a decomposable wedge is nonzero exactly when its vectors are linearly independent).

Verification

technique · direct

1.1

The pair $(e_{1}, 2 e_{1})$ is dependent and its wedge is $e_{1} \land (2 e_{1}) = 2 (e_{1} \land e_{1}) = 0$ , the zero case of [L1].

L1algebra

1.2

The pair $(e_{1}, e_{2})$ is independent and its wedge $e_{1} \land e_{2}$ is a basis vector of $Λ^{2} R^{3}$ , hence nonzero, the nonzero case of [L1].

1.3

For $u = (1, 2, 0)$ and $v = (2, 4, 0)$ , the relation $v = 2 u$ gives $u \land v = 2 (u \land u) = 0$ , matching the dependence.

L1algebra

2.1

For $u = (1, 0, 0)$ and $v = (0, 1, 0)$ , one has $u \land v = e_{1} \land e_{2}$ , which step 1.2 shows is nonzero, matching the independence.

step 1.2L1

3.1

In all four samples the wedge is zero exactly for the dependent pairs, in agreement with [L1].

step 1.1step 1.2step 1.3step 2.1∎

Depends on

In a finite-dimensional vector space, a decomposable wedge is nonzero exactly when its vectors are linearly independent

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

Keith Conrad, Exterior Powers (standard reference, not scraped)