ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The subobject lattice of a two-dimensional vector space over F_2 is the diamond M_3

Example

Let $V = F_{2}^{2}$ . Its subspaces are $0$ , the three lines through the origin, and $V$ itself. So the subobject lattice of $V$ in $F_{2} - M o d$ is exactly the diamond $M_{3}$ .

Facts & Assumptions

Given: The vector space $V = F_{2}^{2}$ .

[L1]

Subobjects form a lattice in every abelian category (The subobjects of an object in an abelian category form a lattice).

[L2]

The A-page counterexample identifies the same diamond pattern as non-distributive (A subobject lattice of an abelian category need not be distributive).

Verification

technique · direct

1.1

Over $F_{2}$ , the nonzero vectors of $V$ are $(1, 0)$ , $(0, 1)$ , and $(1, 1)$ , and each spans a distinct one-dimensional subspace. Any two distinct lines meet only in $0$ , and because they are not equal each pair spans all of $V$ . So the subspace lattice has exactly the five elements $0$ , the three lines, and $V$ .

L1algebra

2.1

This is the same five-element diamond described in [L2]. Hence the subobject lattice of a very familiar module can already be modular without being distributive.

L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Daniel Murfet, Abelian Categories, Section 4.2 (standard reference, not scraped)