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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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Every prime ideal on the power set of omega is principal in the tail-flip symmetric model

Statement

In Ftf, every prime ideal of the internal Boolean algebra P(ω) is principal.

Facts & Assumptions

Given: A prime proper ideal I of P(ω) in Ftf, represented by an HS name I˙.

[F1]

The tail-complement automorphism fixes finitely supported names says that, after choosing m with Hmsym(I˙), a coordinate-m tail flip can be chosen to fix any given finite condition and the name I˙ while complementing Sm modulo a finite set.

[F2]

Boolean ideals, filters, prime ideals and ultrafilters gives downward and finite-union closure, propriety, and the prime implication ABIAI or BI.

[F4]

Forcing theorem supplies the truth lemma used to obtain one condition forcing the actual prime-ideal decision.

[F5]

Symmetry lemma for forcing automorphisms transports forced formulas and their names under a forcing automorphism.

Proof

technique · contradiction, with both alternatives forced by primality
1.1

Suppose for contradiction that I is not principal. For each k<ω, primality applied to {k}(ω{k})=I puts one of the two factors in I. If ω{k}I, downward closure and propriety give I={Aω:kA}, the principal prime ideal generated by that coatom. Thus {k}I for every k, and finite-union closure puts every finite subset of ω in I.

F2assume-contra
1.2

Since I˙ is HS, choose m with Hmsym(I˙), as in F1. Put A=Sm, which belongs to Ftf because its canonical name is supported by Hm+1. Since A(ωA)=, F2 yields either AI or ωAI. Let C denote the member selected by these two exhaustive cases, and let C˙ be the corresponding canonical name.

F1F2
2.1

By F4 choose a finite Q in the actual generic which forces C˙I˙. Apply F1 with the support bound m, condition Q, and coordinate m. Its tail automorphism π fixes both Q and I˙. By F5, Q=πQ forces πC˙I˙. Hence both C and πC belong to I in the actual extension.

F1F4F5step 1.2
3.1

If C=A, then F1 gives πC(ωA) finite. If C=ωA, automorphisms commute with Boolean complementation and F1 gives πCA finite. Thus in either case πC differs finitely from ωC. By step 1.1 the finite difference belongs to I; since ωCπC(πC(ωC)), F2 puts ωC in I.

F1F2F3step 1.1step 2.1
4.1

Step 2.1 gives CI and step 3.1 gives its complement in I. Finite-union closure then puts ω in I, contradicting propriety. Therefore the assumption in step 1.1 was false, and every prime ideal is principal. Both decisions in step 1.2 lead to the same contradiction, and every finite-modification use was derived from the finite singletons rather than assumed.

F2step 1.1step 1.2step 2.1step 3.1discharge-contradiction

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