ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 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.

Supervector spaces with the sign braiding

Example

Let $S V e c t$ be the category of $Z / 2$ -graded vector spaces. For homogeneous vectors $v, w$ , define

$c_{V, W} (v \otimes w) = (- 1)^{∣ v ∣ ∣ w ∣} w \otimes v .$

This is a symmetric braiding.

Facts & Assumptions

Given: $Z / 2$ -graded vector spaces and homogeneous vectors of degrees $∣ v ∣, ∣ w ∣ \in {0, 1}$ .

[L1]

A symmetric monoidal category is a braided monoidal category with involutive braiding (Symmetric monoidal category).

Verification

technique · direct

1.1

The displayed map is natural and invertible: applying it twice multiplies a pure tensor by $(- 1)^{∣ v ∣ ∣ w ∣} (- 1)^{∣ w ∣ ∣ v ∣} = (- 1)^{2 ∣ v ∣ ∣ w ∣} = 1$ , so $c_{W, V} \circ c_{V, W} = 1_{V \otimes W}$ .

givenL1algebra

2.1

On a triple tensor $u \otimes v \otimes w$ , each route around a braiding hexagon contributes the sign $(- 1)^{∣ u ∣ ∣ v ∣ + ∣ u ∣ ∣ w ∣} = (- 1)^{∣ u ∣ (∣ v ∣ + ∣ w ∣)}$ , because moving $u$ past $v \otimes w$ is the same as moving it past $v$ and then past $w$ . Thus both hexagons commute.

step 1.1algebra

3.1

By [L1], these two checks show that $S V e c t$ carries a symmetric braiding, namely the sign braiding.

L1step 1.1step 2.1∎

Depends on

Symmetric monoidal category

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories, Example 8.2.2 (standard reference, not scraped)