CounterexampleConstruction: 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.

Abelian groups do not satisfy AB5*

Statement refuted

The abelian category $A b$ satisfies AB5*.

Facts & Assumptions

Given: The abelian group $A = \prod_{n \geq 1} Z$ , the tail subgroups $T_{n} = {(a_{k}) : a_{1} = \dots = a_{n} = 0}$ , and the direct sum $B = ⨁_{n \geq 1} Z \leq A$ .

[L1]

AB5* is the decreasing-family identity $(⋀_{i} B_{i}) \lor C = ⋀_{i} (B_{i} \lor C)$ (The axioms AB5 and AB5*).

Counterexample

1.1

The family $(T_{n})$ is decreasing, and $⋂_{n \geq 1} T_{n} = 0$ because a sequence whose every coordinate eventually vanishes from the front has all coordinates zero. Also $B \neq A$ , since the constant sequence $(1, 1, 1, \dots)$ lies in $A$ but not in the direct sum $B$ .

givenalgebra

2.1

For every $n$ , the subgroup $T_{n} + B$ is all of $A$ : given $a = (a_{k})_{k \geq 1} \in A$ , write $a = b + t$ where $b$ has the same first $n$ coordinates as $a$ and all later coordinates $0$ , while $t$ has first $n$ coordinates $0$ and later coordinates equal to those of $a$ . Then $b \in B$ and $t \in T_{n}$ . Hence $(⋂_{n \geq 1} T_{n}) + B = B \neq A = ⋂_{n \geq 1} (T_{n} + B) .$ So the AB5* identity [L1] fails in $A b$ .

L1step 1.1∎

Depends on

The axioms AB5 and AB5*

Used by

The opposite of abelian groups does not satisfy AB5 Counterexample

Dependency tree · two levels

7 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

Charles A. Weibel, An Introduction to Homological Algebra, Appendix A.4 (standard reference, not scraped)