ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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.

Russell's shoes and socks

Example

Russell's illustration separates an explicit selection rule from a bare request for simultaneous choices. For a family of pairs of shoes, suppose the data includes a set $L$ meeting each pair in exactly one member—the left shoe. Then ZF constructs a choice function. For a family of pairs of indistinguishable socks, no such distinguisher is supplied; asking for a selection from every pair is an instance of choice for pairs. This example proves the first claim and identifies the second statement without asserting any model-theoretic nonimplication.

Facts & Assumptions

Given: A family $F$ of two-element sets and a set $L$ such that $S \cap L$ has exactly one element for every $S \in F$ .

[L1]

A choice function for $F$ is a function $g$ with domain $F$ such that $g (S) \in S$ for every $S \in F$ (Choice function).

[L2]

The Axiom of Choice asserts that every family of nonempty sets has a choice function (The Axiom of Choice).

Verification

technique · direct

1.1given

For each $S \in F$ , the phrase “the unique element of $S \cap L$ ” defines one element of $S$ from the supplied data.

2.1step 1.1construct

By Separation, $g = {(S, x) \in F \times ⋃ F : S \cap L = {x}}$ is a set; the uniqueness hypothesis makes it single-valued and total on $F$ .

3.1step 2.1L1

For every $S \in F$ , $g (S)$ is the unique member of $S \cap L$ , so $g (S) \in S$ . Thus $g$ is a choice function constructed in ZF from $F$ and $L$ .

4.1step 2.1L1L2∎

If the distinguisher $L$ is omitted, step 2.1 has no defining predicate to use. The assertion that an arbitrary family of pairs nevertheless has a choice function is precisely the corresponding restricted instance of [L2]. This identifies the sock question but neither assumes nor proves an independence result.

Remarks

Boundedly many indexed pairs require no choice principle: if the family is listed as $F (0), \dots, F (n - 1)$ , Every natural-number-indexed list of nonempty sets has a choice function on its family of values builds a selection one value at a time. The definition of an arbitrary finite set appears later in Finite, countably infinite, countable, uncountable ↗, so this remark uses only the indexed-family form.
The argument never uses pairwise disjointness or cardinality two. It uses only the supplied predicate selecting exactly one member of every set.

Russell's shoes and socks

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources