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Questions tagged [presburger-arithmetic]

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

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1 answer
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I want to know a lower bound for the complexity of the decision problem for $\langle \mathbb{Z}; + \rangle$. The below paper notes that Presburger arithmetic, originally $\langle \mathbb{N}; +\rangle$,...
Learner of math's user avatar
3 votes
0 answers
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Fischer and Rabin proved a superexponential bound $2^{2^{cn}}$ for the worst-case length of a proof of a proposition of length $n$ in Presburger arithmetic. The result is in Michael J. Fischer and ...
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Given a function $f: \{1,\ldots,m\} \times \{1,\ldots,n\} \to \{1,\ldots,n\}$ where $m,n$ are strictly positive natural numbers. Is there a Presburger arithmetic formula that represents it?
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5 votes
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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 ...
jpt4's user avatar
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The slides I am reading don't give the axioms of LIA (linear integer arithmetic) instead only give it's signature. I looked on the internet and can't find the axioms of LIA. For the following ...
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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 ...
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1 vote
1 answer
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The signature of Presburger arithmetic contains only the addition operation and equality, omitting the multiplication operation entirely. Is it possible to state and prove theorems in Presburger ...
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7 votes
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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 &...
Noah Schweber's user avatar
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1 answer
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It’s my understanding that Julius Büchi showed that $WS1S$, the weak monadic second-order theory of one successor is decidable by a finite-state automaton, and that this implies that Presburger ...
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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 ...
Rob's user avatar
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1 vote
1 answer
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I always work with these two and use it with no distinction, but I am probably wrong. I mean, are their signatures the same? And their axioms? I know that, for instance, the quantifier elimination ...
Theo Deep's user avatar
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Context: Presburger arithmetic is the theory $\tau$ of structure $$ A = (\mathbb{N},0,1,+,\{c|\cdot\}_{c\in\mathbb{N}})$$ where for each integer $c > 1$, the unary predicate c|n holds if and only ...
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Let $Q(\vec x, \vec y) \subseteq \mathbb Z^n \times \mathbb Z^n$ be a relation which can be represented in presburger arithmetic. Let $R(\vec x, \vec y) \equiv Q \cup (Q \circ Q) \cup (Q \circ Q \...
Siddharth Bhat's user avatar
1 vote
1 answer
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Here it says planarity is definable in first order. http://jgaa.info/accepted/recent/Brandenburg.pdf Here it says planarity testing of graphs is not a first order property. Refer https://simons....
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8 votes
3 answers
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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 ...
Eli's user avatar
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