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user57888
user57888
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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.

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 there exists such an $A$.

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.

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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user57888
user57888
  • 1.2k
  • 1
  • 11
  • 14

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 there exists such an $A$.

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.