학술논문

End extensions of models of weak arithmetic theories.
Document Type
Journal
Author
Dimitracopoulos, Costas (GR-UATH-QH) AMS Author Profile; Paschalis, Vasileios S. (GR-UATH-DM2) AMS Author Profile
Source
Notre Dame Journal of Formal Logic (Notre Dame J. Form. Log.) (20160101), 57, no.~2, 181-193. ISSN: 0029-4527 (print).eISSN: 1939-0726.
Subject
03 Mathematical logic and foundations -- 03C Model theory
  03C62 Models of arithmetic and set theory

03 Mathematical logic and foundations -- 03H Nonstandard models
  03H15 Nonstandard models of arithmetic
Language
English
Abstract
One of the most outstanding open questions in model theory of weak systems of arithmetic is whether every countable model $M$ of $B\Sigma_1$ has a proper end extension to a model of $I\Delta_0$. A.~J. Wilkie and J.~B. Paris [in {\it Logic, methodology and philosophy of science, VIII (Moscow, 1987)}, 143--161, Stud. Logic Found. Math., 126, North-Holland, Amsterdam, 1989; MR1034559] showed that the answer is positive under the additional assumption of $\Delta_0$-fullness of $M$. $\Delta_0$-fullness is a somewhat technical condition, implied by several other more common assumptions, such as $\Pi_1$-short recursive saturation, or totality of the exponential function. Dimitracopoulos and Paschalis give a survey of the known results, and then, following a remark from the paper by Wilkie and Paris, they give a new proof that every countable model of $B\Sigma_1+\ssf{exp}$ has a proper end extension to a model of $I\Delta_0$. Their proof is a careful application of the Arithmetized Completeness Theorem formulated for the notion of tableau consistency. In the last section of the paper, the authors modify their proof to show that the assumption $\ssf{exp}$ can be replaced by any of the following: (i) $M$ is $\Pi_1$-short recursively saturated; (ii) the $\Delta_0$-hierarchy provably collapses in $I\Delta_0$, and for some nonstandard $\gamma$, $M\models \forall x \exists y [y=x^\gamma]$; (iii) the $\Delta_0$-hierarchy provably collapses in $I\Delta_0$, and there is an $a$ in $M$ such that for all $b$, $M\models b\leq a^n$, for some standard $n$.