False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-01

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 wedge product is not commutative

Statement

False claim: the wedge product is commutative.

Facts & Assumptions

Given: The standard coordinate $1$ -forms $d x, d y$ on $R^{2}$ .

[L1]

The wedge product is graded commutative, so for $1$ -forms $η, θ$ one has $η \land θ = - θ \land η$ (The wedge product is associative and graded commutative).

Refutation

technique · direct

1.1

Evaluating on $(\partial_{x}, \partial_{y})$ gives $(d x \land d y) (\partial_{x}, \partial_{y}) = 1$ , so $d x \land d y \neq 0$ .

givenalgebra

2.1

By [L1], $d y \land d x = - d x \land d y$ . Since step 1.1 shows $d x \land d y$ is nonzero, it follows that $d y \land d x \neq d x \land d y$ .

L1step 1.1

3.1

Therefore the wedge product is not commutative.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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)