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Q: Is Presburger arithmetic in stronger logics still complete?

Noah SchweberOriginally asked at MSE: Let $\Sigma=\{+,<,0,1\}$ be the usual language of Presburger arithmetic. Given a "reasonable" logic $\mathcal{L}$, let $\mathbb{Pres}(\mathcal{L})$ be the $\mathcal{L}$-theory consisting of Axioms 1-4 of the usual presentation of Presburger arithmetic, and for each $\ma...

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

Jakub KoniecznyI 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 questions which are probably very basic. Presburger arithmetic is the first order theory of the i...

3
Q: Explicit superexponential growth for Presburger Arithmetic

Mikhail KatzFischer 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 Michael O. Rabin, Super-Exponential Complexity of Presburger Arithmetic, Proceedings of the SIAM-A...


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