TITLE: Which recursively-defined predicates can be expressed in Presburger Arithmetic?
QUESTION [14 upvotes]: 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 example $D(a,c) \equiv \exists b \: M(a,b,c)$, $M(a,b,c) \equiv M(a-1,b,c-b)$, $M(1,b,c) \equiv (b=c)$.  But some predicates which can be expressed in Presburger Arithmetic also have recursive definitions, for example $P(x,y,z) \equiv (x+y=z)$ versus $P(x,y,z) \equiv P(x-1,y+1,z)$, $P(0,y,z) \equiv (y=z)$.
How to tell if a predicate, defined recursively without use of multiplication, has an equivalent non-recursive definition which can be expressed in Presburger Arithmetic?

REPLY [9 votes]: In addition to the answers above, it is worth mentioning that the existential fragment of Presburger arithmetic can actually be extended by a full divisibility predicate while retaining decidability [1]. Satisfiability in the resulting fragment is in NEXP, the lower bound being NP [2].
Also, it is possible to define a family of formulas $\Phi_n(x,m)$ such that for every natural number $m$ representable using $n$-bits, $\Phi_n(x,m)$ holds if and only if $x \equiv 0 \bmod m$ [3].

[1] L. Lipshitz, “The Diophantine problem for addition and divisibility,”
Transactions of the American Mathematical Society, vol. 235, pp. 271–
283, 1976.

[2] Lechner, A.; Ouaknine, J.; Worrell, J., "On the Complexity of Linear Arithmetic with Divisibility," in Proceedings of the 30th Annual ACM/IEEE Symposium on Logic in Computer Science (LICS), pp. 667-676, 2015.

[3] Haase, C., "Subclasses of Presburger arithmetic and the weak EXP hierarchy," in Proceedings of the Joint Meeting of the Twenty-Third EACSL Annual Conference on Computer Science Logic (CSL) and the Twenty-Ninth Annual ACM/IEEE Symposium on Logic in Computer Science (LICS) (CSL-LICS '14). ACM, New York, NY, USA, Article 47, 2014.