# 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 [1] 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?

## 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. ${\rm Con}(PA(X) +\Phi(X))$;

2. If $(M,X)\models \Phi(X)$, then $X$ is not definable in $M$.

Problem [2]: 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 $(M,X)$?

## Elementary submodels

If $S$ is a partial inductive satisfaction class for a model $M$, then let ${M_S}$ be the $PA$-reduct of the smallest elementary submodel of $(M,S)$. If $M_S$ is not $\omega$, then the restriction of $S$ to $M_S$ is a partial inductive satisfaction class for $M_S$; hence $M_S$ is recursively saturated.

It is not hard to prove that if a recursively saturated $M'\prec M$ is cofinal in $M$, then, there is a partial inductive satisfaction class $S$ such that $M'=M_S$. It follows from a result on strange cuts in [3] that every countable recursively saturated $M$ that is not arithmetically saturated has a recursively saturated elementary cut $M'$ which is not $M_S$ for any $S$.

Problem: Let $M\models PA$ be countable and recursively saturated. For which recursively $M'\prec M$ do there exist partial inductive satisfaction classes $S$ such that $M'=M_S$?

## References

- Matt Kaufmann and James H. Schmerl.
*Remarks on weak notions of saturation in models of Peano arithmetic.*J. Symbolic Logic 52(1):129--148, 1987. www DOI MR bibtex - Roman Kossak.
*Four problems concerning recursively saturated models of arithmetic.*Notre Dame J. Formal Logic 36(4):519--530, 1995. (Special Issue: Models of arithmetic) www DOI MR bibtex - Roman Kossak.
*Models with the $\omega$-property.*J. Symbolic Logic 54(1):177--189, 1989. www DOI MR bibtex