ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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 commutative monoid as a one-object strict monoidal category

Example

Let $(M, \cdot, e)$ be a commutative monoid. Make a one-object category $C_{M}$ whose unique object is $*$ and whose endomorphisms are the elements of $M$ , with composition given by $\cdot$ . Define tensor on objects by $* \otimes * = *$ and on morphisms again by $\cdot$ .

Facts & Assumptions

Given: A commutative monoid $(M, \cdot, e)$ .

[L1]

A monoid yields a one-object category whose endomorphism monoid is the original monoid (A monoid is a one-object category, and a group is a one-object category in which every morphism is invertible).

[L2]

In a strict one-object monoidal category, the endomorphism monoid is commutative (The endomorphisms of the tensor unit form a commutative monoid).

Verification

technique · direct

1.1L1

By [L1], $C_{M}$ is a category with identity $e$ and composition given by $\cdot$ .

2.1step 1.1algebra

Because $M$ is commutative, $((m^{'} \otimes n^{'}) \circ (m \otimes n)) = (m^{'} \cdot n^{'}) \cdot (m \cdot n) = (m^{'} \cdot m) \cdot (n^{'} \cdot n) = ((m^{'} \circ m) \otimes (n^{'} \circ n))$ , so tensor is a bifunctor. Associativity and unit are literal equalities because there is only one object and the tensor on morphisms is exactly the monoid product.

3.1step 2.1L2∎

Thus $C_{M}$ is a one-object strict monoidal category. The theorem in [L2] explains why commutativity is the right hypothesis for this strict example.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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