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26 votes
3 answers
7k views

Presburger Arithmetic

Presburger arithmetic apparently proves its own consistency. Does anyone have a reference to an exposition of this? It's not clear to me how to encode the statement "Presburger arithmetic is ...
Brendan Cordy's user avatar
18 votes
3 answers
1k views

Computable nonstandard models for weak systems of arithmetic

By Tennenbaum's theorem, PA itself does not have any computable nonstandard models. The integer polynomials which are 0 or have a positive leading coefficient form a computable nonstandard model of ...
user avatar
16 votes
2 answers
713 views

Is (Z,+,0,1,P2,P3) decidable?

Is Presburger arithmetic, augmented with two unary predicates P2, P3, for powers of 2 and powers of 3 respectively, decidable? I know that adding just one of P2, P3 to Presburger keeps it decidable, ...
ikp's user avatar
  • 428
14 votes
3 answers
2k views

Which recursively-defined predicates can be expressed in Presburger Arithmetic?

In Presburger Arithmetic there is no predicate that can express divisibility, else Presburger Arithmetic would be as expressive as Peano Arithmetic. Divisibility can be defined recursively, for ...
12 votes
0 answers
249 views
+50

Is there a decidable theory of arithmetic with a non-collapsing quantifier hierarchy?

This question is very close to this old MSE question of mine, which is still unanswered. Is there an (ideally reasonably-natural!) expansion of the structure $(\mathbb{N};+)$ in a finite language ...
Noah Schweber's user avatar
7 votes
1 answer
334 views

Why include $0$ and $1$ in the signature of Presburger arithmetic?

I recently became interested in some problems concerning decidability of extensions of Presburger arithmetic. However, being a number-theorist rather than a logician, I am confused by some notational ...
Jakub Konieczny's user avatar