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Let $M$ be a model of ${\sf ZFC}$. A satisfaction class $S$ for $M$ is subset of $M$'s ordered pairs which satisfies in $M$ the standard Tarskian compositional axioms. E.g.:

$M\vDash \forall \phi, \psi\in\mathcal L_\in\forall a:\omega \to V(\langle\phi\wedge \psi, a\rangle\in S\leftrightarrow \langle\phi, a\rangle\in S \wedge \langle \psi, a\rangle\in S)$

Question: Is there a model $M$ of ${\sf ZFC}$ with two satisfaction classes $S$ and $S'$ such that for some formula $\phi$ and assignment function $a$ in $M$, $\{x: \langle\phi, a[0/x]\rangle\in S\} = S'$ (where $a[0/x]$ is the assignment function according to $M$ which is just like $a$ except that it assigns $x$ to $0$)?

Obviously, $M$ will have to be non-standard, as will $\phi$. Similarly, by Tarski's theorem on the undefinability of truth, $S$ and $S'$ can't agree on pairs of formulas and assignment functions. And as Joel notes below, a simple induction using a predicate for $S$ would show that $S$ and $S'$ do agree on formulas and assignment functions. So $M$ will not be a model of ${\sf ZFC}$ with Separation extended to a predicate for $S$.

I'm primarily interested in models of ${\sf ZFC}$, but I'd also be interested in models of ${\sf PA}$.

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  • $\begingroup$ What is a satisfaction class? $\endgroup$ – Emil Jeřábek supports Monica Apr 22 '16 at 19:57
  • $\begingroup$ @EmilJeřábek Added a definition. Does that help? $\endgroup$ – Sam Roberts Apr 22 '16 at 20:06
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In case of Peano Arithmetic the answer is yes (emphatically yes), if I understand your question correctly. This follows from Theorem 3.3 in Smith's "Nonstandard Definability" APAL 42 (1989) pp. 21-43 which says that for any recursively saturated countable model $M$ of PA and any set $A \subseteq M$ the set $A$ may be presented in a form $\{x \in M \mid (\phi, x) \in S \}$ for some satisfaction class $S$ in $M$ iff the expansion $(M,A)$ is recursively saturated (I'm slightly confused with the $0$ in your notation and I know I am a bit sloppy with writing of a pair (a formula, an element) being in a satisfaction class rather than (a formula, an assignment), but I hope this is what you looked for).

Now take any $A$ in the model $M$ which is a recursively saturated satisfaction class (being a satisfaction class can be finitely axiomatised and being recursively saturated is axiomatised by a scheme, so by resplendency of countable recursively saturated models and the fact that the theory of satisfaction class without induction (or replacement in your case) is conservative over PA there exists such an $A$ in $M$).

So we conclude that: in any countable recursively saturated model $M$ of PA there exists a class $S$ which defines another satisfaction class or even a chain of such classes.

I haven't really checked whether this proof carries over to ZFC, but I would be extremely surprised if it were not the case. So I'm 99 % sure the answer to your question is very strong yes.

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  • $\begingroup$ Thanks for this! The proof is out of my comfort zone, so I'll leave it open for another day and accept if no one objects. (I take assignment functions to be maps from $\omega$ to $V$. Effectively, this gives us an assignment to the variables, since we can take it to assign $a(0)$ to the least variable, $a(1)$ to the next, etc (relative to some salient well-ordering of the variables)). $\endgroup$ – Sam Roberts Apr 23 '16 at 19:25
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A model of ZFC can indeed have two satisfaction classes. This is a result going back to Krajewski 1974, and you can find an account of it and some and some other related matters in my paper, J. D. Hamkins, R. Yang, Satisfaction is not absolute.

Meanwhile, your context is asking for much more than just two different satisfaction classes. You have a model $M$ of $\text{ZFC}$ with a satisfaction class $S$ and another $S'$ that is definable relative to it.

If you mean that $M$ is a model of $\text{ZFC}(S)$, that is, in the language with $S$ as a predicate, such as in GBC, then your situation is impossible. The reason is that one can never have two different satisfaction classes in a context where we have induction relative to the two classes, because you can prove by induction on formulas that they would agree. That is, if you look at the least formula $\varphi$ at which the two classes disagree (possibly nonstandard), then this will immediately violate the Tarski conditions for that formula. So you don't even need full $\text{ZFC}(S)$, but rather only induction on the natural numbers relative to $S$.

Without the $\text{ZFC}(S)$ or at least the induction assumption on $M$ relative to $S$, then this seems to become an interesting, subtle question.

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  • $\begingroup$ Thanks! Yes, uncommonly for me, I do have the `interesting, subtle, question' in mind. Will edit to make that a bit clearer. $\endgroup$ – Sam Roberts Apr 22 '16 at 20:15

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