Nonstandard satisfaction classes.
Fullness of M
A satisfaction class for a model M is e-full, if S decides all If for every \Sigma_e sentence (in the sense of M) \varphi with parameters either \varphi\in S or \lnot\varphi\in S.
Fullness of M, {\rm Full}(M)\} is the set of those e\in M M for which M has an e-full inductive partial satisfaction class.
It is easy to see that {\rm Full}(M) is a cut of M and, if {\rm Full}(M)>\omega then M is recursively saturated. Also if M is countable and {\rm Full}(M) contains an element greater than all definable elements of M, then {\rm Full}(M)=M.\end{enumerate} \end{prop}
Kaufmann and Schmerl showed that there are completions T of PA, such that for every M\models T, {\rm Full}(M) contains no definable nonstandard elements.
Problem: Suppose {\rm Full}(M)=M, does M have a full inductive satisfaction class?
Reference: Kaufmann, Matt; Schmerl, James H. Remarks on weak notions of saturation in models of Peano arithmetic. J. Symbolic Logic 52 (1987), no. 1, 129–148.
== A converse to Tarski?
==
Let FS(X) be a formula of the language of PA with a additional predicate symbol X expressing that X is a full satisfaction class.
FS(X) is an example of a formula \Phi(X) such that
1. {\rm Con}( A(X) +\Phi(X));
2. If (M,X)\models \Phi(X), then X is not definable in M.
\end{enumerate}
Problem: Suppose \Phi(X) satisfies 1. and 2. above. Is it true that for every M and X\subseteq M, if (M,X)\models \Phi(X), then there is a nonstandard satisfaction class definable in (M,X)?
Reference: Kossak, Roman Four problems concerning recursively saturated models of arithmetic. Special Issue: Models of arithmetic. Notre Dame J. Formal Logic 36 (1995), no. 4, 519–530.