Nonstandard models of Presburger Arithmetic

Question

I have a question about nonstandard models of Presburger Arithmetic. I read that an example of a nonstandard model is the set of polynomials with rational coefficients with positive leading coefficient and a (positive) integer constant coefficient. It is obvious that all axioms other than the induction axiom schema are satified; but how can one prove that the induction axiom holds?

Well, using only addition, the polynomial $a_0+a_1x+\dots+a_nx^n$ is the same as the $(n+1)$-tuple $(a_0,a_1,\dots,a_n)$. The key is probably that the formulas also contain only addition.. As a first idea, I would try formula induction for the induction axiom. — Berci, Feb 18 '13 at 17:01
And one less searching window for the rest who might be interested in the question: http://en.wikipedia.org/wiki/Presburger_arithmetic — Berci, Feb 18 '13 at 17:12
@boumol: I found it on http://www.cs.toronto.edu/~sacook/csc438h/notes/page39.ps — rkrapf, Feb 18 '13 at 22:24
@Berci: Thanks, that's the idea I had too, but it is not that trivial... — rkrapf, Feb 18 '13 at 22:27

hmakholm left over Monica · Answer 1 · 2013-02-19T12:54:53.053

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As you describe it,

the set of polynomials with rational coefficients with positive leading coefficient and a (positive) integer constant coefficient

is not a model of Presburger arithmetic. Presburger arithmetic proves $$ \forall a : a=0 \lor a=1 \lor \exists b: a=b+1+1 $$ but this is not true in your model for $a=X+1$ (where $X$ is the formal variable in the polynomial).

I think that you might get a model if you allow the constant term to be negative when it is not the leading coefficient. In that case $X$ would intuitively model "infinity", and $X^2$ an even larger infinity and so forth. Then a more-or-less standard compactness argument shows that Presburger arithmetic must have a model containing at least all of these elements, but don't ask me for proof that there's enough of them to constitute a model.

A proof will probably not attempt to verify each axiom directly, but will instead prove that the model is elementarily equivalent to the ordinary naturals, relative to the restricted language of Presburger Arithmetic. Some form of quantifier elimination may work.

edited Feb 19 '13 at 12:54

answered Feb 19 '13 at 02:29

hmakholm left over Monica

291,611

Thanks, yes that's right, the constant term can be negative if it is not the leading coefficient. Can you explain what you mean in the last paragraph? Could it work to combine formula induction with quantifier elimination? – rkrapf Feb 19 '13 at 15:45
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@rkrapf: I don't expect that induction over formulas is going to to be helpful here. Rather I would look for a proof elementary equivalence by considering arbitrary formulas in prenex form where the quantifier-free part is a disjunctive normal form. Then each disjunct consists of a system of linear equations with integer coefficients, combined with a zero or more disequalities that restrict the solutions. – hmakholm left over Monica Feb 19 '13 at 19:04
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... and it is then possible that one can bring some linear algebra to bear on proving that every such prenex-CNF sentence that is true in $\mathbb N$ is also true in the polynomial model. This will then be the case in particular for instances of the induction axiom. – hmakholm left over Monica Feb 19 '13 at 19:27
Thank you, I will try it that way. – rkrapf Feb 19 '13 at 21:02

score 2 · Answer 2 · edited Dec 06 '23 at 10:40

A simple model of Presburger's arithmetic is a subset of complex numbers

$$ M=\{a+bi: a \in Z \land b \in Q \land b \geq 0 \land (b=0 \implies a\geq 0) \} $$

with addition and two constants (0+0i) and (1+0i).

See editor's remarks to the paper of Th. Skolem published in 1934 in Fundamenta Mathematice. pp.150-161 (available here) The editor (presumably A. Tarski) attributes this remark to S. Jaśkowski.

score 0 · Answer 3 · answered Mar 27 '23 at 05:42

Poizat's "A Course in Model Theory" (chapter 7.3) provides a more useable alternative axiomatization for $(\mathbb{N},+)$.

Addition is commutative
Addition is associative
$0$ is the additive identity
$\forall x: x+1 \neq x$
$\forall x: \forall y: x+y=1 \implies x = 0 \vee x = 1$
$\forall x: \forall y: \exists z: x = y+z \vee y = x+z$
$\forall x: \forall y: \forall u: \forall v: (x = y+u \wedge y = x+v \implies x = y)$

And finally, for each $n$: $$\forall x: \exists y: \exists r: x = ny+r \wedge r < n$$

This set appears much easier to check for any proposed nonstandard model of Presburger arithmetic.

Nonstandard models of Presburger Arithmetic

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