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6:38 AM
@JadeVanadium What if we have only Σ1-Ord-Rep (for definable functions), but also have cofinal cardinals?
 
 
2 hours later…
8:46 AM
Does it satisfy "ZFC is arithmetically sound"? — user21820 1 hour ago
 
 
5 hours later…
1:50 PM
@user21820 Isn't it the case that ZFC+"there's a transitive model of ZFC" proves ZFC is arithmetically sound?
 
2:04 PM
Every transitive model of ZFC will include the real set theoretic ordinal ω, and the definition of ω in the model agrees with the external definition of ω. Assuming there exists a transitive model of ZFC, then any arithmetical formula which ZFC proves will be true about the ω of that model, which is the real ω, so it will simply be true. In the sense that ℕ=ω, everything which ZFC proves about ℕ will be actually true about ℕ, which is just to say that ZFC is arithmetically sound.
ZFC also comes pretty close to proving itself arithmetically sound, similar to how PA comes very close to proving itself consistent. ZFC proves that every finite fragment of ZFC admits a standard model. More strongly, if we merely weaken Replacement to Σn Replacement for some standard finite n, then ZFC will prove that that weakened theory has a standard model (we keep the full Specification schema in this case).
In the same way as before, ZFC proves that each of these weakened theories are arithmetically sound, albeit only as a theorem schema for each standard finite n. Of course, if we have that every finite fragment of ZFC is arithmetically sound, then ZFC is actually arithmetically sound.
 
2:25 PM
Let M be a transitive model of ZFC, and let φ be some arithmetical formula, then
ZFC⊢[ω⊨φ] implies M⊨[ω⊨φ] implies ω⊨φ
where the final implication holds since the ω of M is the real ω. More formally, we define:
x∈Ord iff [∀(y∈x), ∀(z∈y), z∈x & ∀(w∈z),w∈y]
x∈Lim iff [x∈Ord & x≠∅ & ∀(y∈x), ∃(z∈x), y∈z]
x∈ω iff [x∈Ord & x∉Lim & ∀(y∈x), y∉Lim]
We easily show that the condition x∈ω, as defined above, is a Δ0 property, which is absolute for transitive sets. That is, if M is a transitive set and ψ is a Δ0 formula in the language of sets, then ψ⇔[M⊨ψ]. This means that x∈ω⇔M⊨[x∈ω], for all x∈M. We e
The rest of the argument follows by the same logic, since the satisfaction operator ⊨ is itself Δ0. That is to say, the formula "ω⊨φ" is itself a Δ0 formula, and since M is transitive then [ω⊨φ] ⇔ M⊨[ω⊨φ]
The argument is slightly more obvious, if you talk about V[ω] instead of ω. The same basic argument works though. The V[ω] of M is the real V[ω], and so for any formula φ in the language of sets, we will have (V[ω]⊨φ) ⇔ M⊨(V[ω]⊨φ).
 
@JadeVanadium The minimal model of ZFC satisfies "There is no transitive model of ZFC."...
 
I don't think that really changes anything.
If ZFC is arithmetically sound, then every transitive model of ZFC will obey the assertion "ZFC is arithmetically sound".
 
2:42 PM
Hmm.. I'm a bit lazy to think..
I'll just take your word for it that "arithmetically sound" is absolute. It sounds like it, but as I said I'm lazy to think..
XD
Got to go!
XD
 
Oh alright, see ya hahah
 

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