ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)

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 one-bit Boolean-valued name

Example

In the complete Boolean algebra $B = P ({0, 1})$ , let $b = {0}$ , let e be the empty name and let $t = {⟨ e, b ⟩}$ . Then

$∥ e \in t ∥ = b, ∥ t = e ∥ = \neg b = {1}, ∥ t = t ∥ = 1_{B} .$

Valuation using the principal Boolean ultrafilter $U_{0} = {a \subseteq {0, 1} : 0 \in a}$ gives $t_{U_{0}} = {\emptyset}$ ; using $U_{1}$ gives $t_{U_{1}} = \emptyset$ .

Facts & Assumptions

Given: ZF. Calculated all three Boolean values directly in the four-element algebra and both principal valuations; no generic truth theorem or Boolean completion is consumed.

[F1]

Well-definedness of Boolean-valued semantics: The atomic clauses give well-defined joins and meets, including the empty ones.

[F2]

Valuation of names and M[G]: Valuation is defined recursively by retaining exactly the subnames whose coefficients lie in the evaluating set; in particular, the empty name evaluates to empty.

Verification

1.1

In this algebra join is union, meet is intersection and complement is relative to ${0, 1}$ . Every family has these bounds, so the algebra is complete. The empty atomic clauses give $E (e, e) = 1_{B}$ and $I (e, e) = 0_{B}$ . Thus $I (e, t) = b \land 1_{B} = b$ , while $E (t, e) = (\neg b \lor 0_{B}) \land 1_{B} = \neg b$ .

2.1

The equality clause for t with itself has in both factors the single term $\neg b \lor I (e, t) = \neg b \lor b = 1_{B}$ , so $E (t, t) = 1_{B}$ . The only subname of t is e, whose valuation is empty. Since $b \in U_{0}$ and $b \notin U_{1}$ , the valuation rule gives respectively the singleton of empty and the empty set. These U_i are filters deciding each subset by whether it contains i; no ultrafilter-extension principle or generic truth theorem is used.

F1F2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Geschke Definition 6.27 pp26–27; local finite Boolean calculation (standard reference, not scraped)