ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

Stems, direct extensions, and the generic sequence

Example

Let $M$ be a transitive model of ZFC containing a normal measure $U$ on an uncountable cardinal $κ$ , let $G$ be $M$ -generic for the corresponding Prikry forcing, choose $α < β < γ < κ$ , and put

$A_{δ} = {ξ < κ : δ < ξ} .$

Then

$p = (⟨ α ⟩, A_{α}), q = (⟨ α, β ⟩, A_{β}), r = (⟨ γ ⟩, A_{γ})$

are Prikry conditions. The condition $q$ extends $p$ but is not a direct extension; by contrast

$p^{*} = (⟨ α ⟩, A_{β})$

is a direct extension of $p$ . The conditions $p$ and $r$ are incompatible. For a generic filter, the dense requirements on stem length and final height make the union of its stems an increasing cofinal $ω$ -sequence in $κ$ .

Facts & Assumptions

Given: $M, U, κ, G, α, β, γ$ are as above, and conditions are ordered stronger-below.

[F1]

Prikry forcing and its direct-extension order: Conditions have finite strictly increasing stems, upper parts in $U$ , and extensions end-extend the old stem using points from its upper part; direct extensions keep the stem fixed.

[F2]

The Prikry generic sequence changes cofinality to omega: In $M [G]$ , the union of the stems in $G$ is a strictly increasing sequence of order type $ω$ cofinal in $κ$ .

[F3]

Dense open sets and generic filters over a model: An $M$ -generic filter meets every dense subset of the forcing that belongs to $M$ .

Verification

1.1

Every tail $A_{δ}$ belongs to $U$ : its complement is the union of fewer than $κ$ singletons, while $U$ is nonprincipal and $κ$ -complete. Its minimum is $δ + 1$ . Hence $p, q, r, p^{*}$ satisfy the upper-part inequality in F1; in particular, $α < α + 1$ , $β < β + 1$ , and $γ < γ + 1$ .

1.2

If a condition extended both $p$ and $r$ , its stem would end-extend both one-entry stems. Its first entry would then have to be both $α$ and $γ$ , contrary to $α < γ$ . Thus $p ⊥ r$ . Notice that shrinking either upper part cannot repair this disagreement at the first stem entry.

1.3

For $n < ω$ , write $D_{n} = {(s, A) : ∣ s ∣ \geq n}$ . From a stem of length $m < n$ , choose successively $n - m$ increasing points of its upper part and then shrink above the last chosen point; F1 shows that the resulting condition lies in $D_{n}$ . Thus $D_{n}$ is dense. For $η < κ$ , let $E_{η}$ consist of conditions with nonempty stem and last entry above $η$ . Given $(s, A)$ , the measure-one set $A$ is unbounded, so choose $ξ \in A$ above both $η$ and every entry of $s$ , append $ξ$ , and shrink the upper part to $A \cap A_{ξ}$ . This gives an extension in $E_{η}$ , so $E_{η}$ is dense.

2.1

The stem $⟨ α, β ⟩$ end-extends $⟨ α ⟩$ , its new entry $β$ lies in $A_{α}$ , and $A_{β} \subseteq A_{α}$ . Thus $q \leq p$ . Their stems differ, so $q ̸ \leq^{*} p$ . On the other hand $p^{*} \leq p$ , $A_{β} \subseteq A_{α}$ , and the stems of $p^{*}$ and $p$ agree, so $p^{*} \leq^{*} p$ .

F1step 1.1

3.1

Each $D_{n}$ and $E_{η}$ is a member of $M$ , because it is defined there from the ground forcing and the displayed ground parameters. By F3, $G$ meets every one of them. Meeting all $D_{n}$ makes the compatible stems have union of domain $ω$ , and meeting all $E_{η}$ makes that union unbounded in $κ$ . Since extensions only end-extend strictly increasing stems, the union is a strictly increasing cofinal $ω$ -sequence, exactly as F2 asserts.

F2F3step 1.3∎

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