ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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.

Empty and single-branch tree codes

Example

In ZF, the empty tree has empty body. For each $x \in N^{N}$ the prefix tree $T_{x} = {x ↾ n : n \in N}$ has body ${x}$ . In particular, for $z (n) = 0$ , the tree ${\emptyset, (0), (0, 0), \dots}$ has body exactly the constant-zero sequence. These bodies are closed in the cylinder topology.

Facts & Assumptions

[F1]

A branch has every finite prefix in its tree, including the empty prefix; see Trees and their bodies.

Verification

Given: $x \in N^{N}$ , $T_{x}$ as above, and the empty tree.

1.1

Every putative branch of $\emptyset$ would have to satisfy $y ↾ 0 = \emptyset \in \emptyset$ , which is false. Hence $[\emptyset] = \emptyset$ , a closed set.

1.2

Restricting $x ↾ n$ to $m \leq n$ gives $x ↾ m \in T_{x}$ , so $T_{x}$ is a tree. The point $x$ lies in its body. If $y \in [T_{x}]$ , its prefix of length $k + 1$ equals the unique member $x ↾ (k + 1)$ of $T_{x}$ of that length. Consequently $y (k) = x (k)$ for every $k$ , giving $[T_{x}] = {x}$ .

F1given

2.1

For $z (n) = 0$ , step 1.2 gives $y (k) = 0$ at every coordinate, the claimed concrete calculation. If $y \neq x$ , choose one differing coordinate $k$ ; its cylinder of length $k + 1$ excludes $x$ . Thus the complement of ${x}$ is open, establishing the closedness assertion. QED.

step 1.2

Depends on

Trees and their bodies

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

Application of Lemma 1.14(ii) (standard reference, not scraped)