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.