Questions tagged [presburger-arithmetic]

Presburger arithmetic is the first-order theory of the natural numbers with addition.

Presburger arithmetic is the first-order theory of the natural numbers with addition, named in honor of Mojżesz Presburger, who introduced it in 1929. The signature of Presburger arithmetic contains only the addition operation and equality, omitting the multiplication operation entirely. The axioms include a schema of induction.

Presburger arithmetic is much weaker than Peano arithmetic, which includes both addition and multiplication operations. Unlike Peano arithmetic, Presburger arithmetic is a decidable theory. This means it is possible to algorithmically determine, for any sentence in the language of Presburger arithmetic, whether that sentence is provable from the axioms of Presburger arithmetic. The asymptotic running-time computational complexity of this decision problem is at least doubly exponential, however.

Presburger arithmetic cannot formalize concepts such as divisibility or primality. Generally, any number concept leading to multiplication cannot be defined in Presburger arithmetic, since that leads to incompleteness and undecidability. However, it can formulate individual instances of divisibility; for example, it proves "$\forall x \exists y \big( (y + y = x) \lor (y + y + 1 = x) \big) $". This states that every number is either even or odd.

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Why is it impossible to define multiplication in Presburger arithmetic?

Peano arithmetic defines multiplication recursivly as: $$\begin{gather}a\cdot0=a\\a\cdot S(b)=a+(a\cdot b)\end{gather}$$ Why is this not possible in Presburger arithmetic?
FUZxxl
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Presburger arithmetic

In discovering that Presburger's arithmetic is one of the weaker systems in PA that does not violate Godel's first incompleteness theorem. Upon reading the wiki article, it said that Presburger proved that his arithmetic is decidable, complete, and…
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Can Robinson's Q prove Presburger arithmetic consistent?

I made an assertion in What are some examples of theories stronger than Presburger Arithmetic but weaker than Peano Arithmetic? that Q has higher consistency strength than Pres, Presburger arithmetic; i.e., Q proves the consistency sentence for…
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Are there "interesting" theorems in Peano arithmetic, that only use the addition operation?

More precisely, are there "interesting" theorems of Presburger arithmetic, other than the following four well-known "interesting" ones? The commutativity of addition. The theorem stating there are no two consecutive even numbers. The theorem…
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Is Presburger arithmetic in "Ramsey logic" complete?

Let $\Sigma=\{+,<,0,1\}$ be the usual language of Presburger arithmetic - so we have addition but no multiplication (the additional symbols for $<,0,1$ are unnecessary but convenient). Given a "reasonable" logic $\mathcal{L}$, let…
Noah Schweber
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Nonstandard models of Presburger Arithmetic

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.…
rkrapf
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Is it not possible to state the Collatz Conjecture in a decidable theory?

I believe that the answer is no (evidenced by the 80+ year "open" status of the conjecture), but I am self-taught in formal logic and decidability theory, so I would like to present my argument for review. Presburger arithmetic, the first-order…
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Pairing in Presburger arithmetic

Is it possible to define pairing function (and the inverses) in Presburger arithmetic? I would guess no but I can't locate a reference nor construct a proof to one way or another.
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What modifications to the axioms of primitive recursion would restrict its expressivity to that of Presburger arithmetic?

Inspired by reading through this page: https://golem.ph.utexas.edu/category/2019/08/turing_categories.html https://en.wikipedia.org/wiki/Primitive_recursive_function Among the primitive recursive functions one can define are addition and…
jpt4
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How much arithmetic is required to formalize quantifier elimination in Presburger arithmetic?

As we know, Presburger arithmetic can be proved decidable by demonstrating that it admits quantifier elimination, i.e. that there is an algorithm that reduces any sentence in the language to some equivalent quantifier-free sentence (which in turn…
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Systems of arithmetic models

Presburger Arithmetic is decidable theory but weaker than Peano Arithmetic. Are there systems in some sense that are: stronger than Presburger but weaker than Peano and remain decidable? weaker than Peano but stronger than Presburger and remain…
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Showing that Presburger arithmetic is decidable by deciding if $\mathbb N \models \varphi$, but does it give provability in the axioms?

Here Presburger arithemtic is given by a set of axioms over the signature with binary operation $+$ and two constants $0$ and $1$. Similarly in Presburgers original paper he gives the arithmetic in terms of an axiomatic system. With this…
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Why doesn't the decision problem for Presburger arithmetic demonstrate that $\mathsf{P} \neq \mathsf{NP}$

From Wikipedia's article on Presburger arithmetic: Then Fischer and Rabin (1974) proved that any decision algorithm for Presburger arithmetic has a worst-case runtime of at least $2^{2^{cn}}$, for some constant $c>0$. Hence, the decision problem…
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Why is it impossible to define multiplication in Presburger arithmetic yet possible to define exponentiation in Peano Arithmethic?

Hello my question is related to Why is it impossible to define multiplication in Presburger arithmetic? and to How is exponentiation defined in Peano arithmetic?. I would have preferred to add it as a comment to one of the above discussions but I…
Motorhead
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Presburger arithmetic is consistent, but relative to what?

In the Wikipedia article for (the first-order theory of) Presburger arithmetic, it is stated (among other properties) that Presburger arithmetic is consistent. What meta-theory does he rely on in order to prove this is consistent? Is that…
user923098
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