http://modelsofpa.info/index.php?title=Nonstandard_satisfaction_classes&feed=atom&action=historyNonstandard satisfaction classes - Revision history2022-01-26T09:19:50ZRevision history for this page on the wikiMediaWiki 1.24.4http://modelsofpa.info/index.php?title=Nonstandard_satisfaction_classes&diff=190&oldid=prevRkossak: 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 "== 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..."</p>
<p><b>New page</b></p><div>== Fullness of $M$ ==<br />
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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 />
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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 />
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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}<br />
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Kaufmann and Schmerl <cite> kaufmannschmerl1987:remarks </cite> showed that there are completions $T$ of $PA$, such that for every $M\models T$, ${\rm Full}(M)$ contains no definable nonstandard elements. <br />
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Problem: Suppose ${\rm Full}(M)=M$, does $M$ have a full inductive satisfaction class?<br />
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== A converse to Tarski? ==<br />
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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 />
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$FS(X)$ is an example of a formula $\Phi(X)$ such that <br />
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1. ${\rm Con}(PA(X) +\Phi(X))$;<br />
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2. If $(M,X)\models \Phi(X)$, then $X$ is not definable in $M$.<br />
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Problem <cite> kossak1995:four </cite>: 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 />
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== Elementary submodels ==<br />
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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 />
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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 <cite> kossak1989:models </cite> 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 />
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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 />
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{{References}}</div>Rkossak