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.

The cross product is recovered from $⋆ (u \land v)$ in $R^{3}$

Example

For $u = (1, 0, 2)$ and $v = (0, 3, 1)$ in oriented Euclidean $R^{3}$ ,

$u \land v = (e_{1} + 2 e_{3}) \land (3 e_{2} + e_{3}) = 3 e_{1} \land e_{2} + e_{1} \land e_{3} - 6 e_{2} \land e_{3},$

and applying the Hodge star gives $3 e_{3} - e_{2} - 6 e_{1}$ , that is $(- 6, - 1, 3)$ , which is exactly the coordinate cross product $u \times v = (0 \cdot 1 - 2 \cdot 3, 2 \cdot 0 - 1 \cdot 1, 1 \cdot 3 - 0 \cdot 0) = (- 6, - 1, 3)$ .

Facts & Assumptions

Given: The vectors $u = (1, 0, 2)$ , $v = (0, 3, 1)$ and the standard oriented orthonormal basis of $R^{3}$ .

[L1]

In oriented Euclidean three-space, the cross product is $u \times v = ⋆ (u \land v)$ (In oriented Euclidean three-space, the cross product is $⋆ (u \land v)$ ).

Verification

technique · direct

1.1

Expanding the wedge gives $u \land v = (e_{1} + 2 e_{3}) \land (3 e_{2} + e_{3}) = 3 e_{1} \land e_{2} + e_{1} \land e_{3} + 6 e_{3} \land e_{2} = 3 e_{1} \land e_{2} + e_{1} \land e_{3} - 6 e_{2} \land e_{3}$ .

algebra

1.2

The coordinate cross product is $u \times v = (u_{2} v_{3} - u_{3} v_{2}, u_{3} v_{1} - u_{1} v_{3}, u_{1} v_{2} - u_{2} v_{1}) = (- 6, - 1, 3)$ .

algebra

2.1

Applying the star to step 1.1, with $⋆ (e_{1} \land e_{2}) = e_{3}$ , $⋆ (e_{1} \land e_{3}) = - e_{2}$ , $⋆ (e_{2} \land e_{3}) = e_{1}$ , gives $⋆ (u \land v) = 3 e_{3} - e_{2} - 6 e_{1} = (- 6, - 1, 3)$ .

L1step 1.1algebra

3.1

Steps 2.1 and 1.2 produce the same vector, in agreement with the identity of [L1].

L1step 1.2step 2.1∎

Depends on

In oriented Euclidean three-space, the cross product is $\star(u\wedge v)$

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

Reyer Sjamaar, Manifolds and Differential Forms, §2.5 (standard reference, not scraped)

The cross product is recovered from ⋆(u∧v)⋆(u∧v) in R3R3