http://modelsofpa.info/index.php?title=Nonstandard_satisfaction_classes&feed=atom&action=history Nonstandard satisfaction classes - Revision history 2022-01-26T09:19:50Z Revision history for this page on the wiki MediaWiki 1.24.4 http://modelsofpa.info/index.php?title=Nonstandard_satisfaction_classes&diff=190&oldid=prev Rkossak: Created page with "== 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 eith..." 2013-01-23T19:39:31Z <p>Created page with &quot;== 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 eith...&quot;</p> <p><b>New page</b></p><div>== Fullness of $M$ ==<br /> <br /> 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 $\lnot\varphi\in S$.<br /> <br /> 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.<br /> <br /> It is easy to see that ${\rm Full}(M)$ is a cut of $M$ and, if ${\rm Full}(M)&gt;\omega$ 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}<br /> <br /> Kaufmann and Schmerl &lt;cite&gt; kaufmannschmerl1987:remarks &lt;/cite&gt; showed that there are completions $T$ of $PA$, such that for every $M\models T$, ${\rm Full}(M)$ contains no definable nonstandard elements. <br /> <br /> Problem: Suppose ${\rm Full}(M)=M$, does $M$ have a full inductive satisfaction class?<br /> <br /> == A converse to Tarski? ==<br /> <br /> 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.<br /> <br /> <br /> $FS(X)$ is an example of a formula $\Phi(X)$ such that <br /> <br /> 1. ${\rm Con}(PA(X) +\Phi(X))$;<br /> <br /> <br /> 2. If $(M,X)\models \Phi(X)$, then $X$ is not definable in $M$.<br /> <br /> Problem &lt;cite&gt; kossak1995:four &lt;/cite&gt;: 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 $(M,X)$?<br /> <br /> <br /> == Elementary submodels ==<br /> <br /> If $S$ is a partial inductive satisfaction class for a model $M$, then let ${M_S}$ be the $PA$-reduct of the<br /> 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. <br /> <br /> 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 &lt;cite&gt; kossak1989:models &lt;/cite&gt; 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 $S$.<br /> <br /> 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 $M'=M_S$?<br /> <br /> <br /> <br /> {{References}}</div> Rkossak