# Questions tagged [peano-axioms]

For questions on Peano axioms, a set of axioms for the natural numbers.

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### Unconventional models and Peano Arithmetic

I'm trying to show that $\mathbb{Z}[x]^+ \models \mathsf{PA}^-$. What are the initial segments of this model?
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### Can An Axiom Schema be Independent?

Consider the following theory: Ring Theory (RT) + $\forall x(Sx=x+1)$ + first order induction (Ind). The finite rings $Z/nZ$ are models of this theory. Now consider RT + $\forall x(Sx=x+1)$ + Not(Ind)....
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### How does (ZFC-Infinity+"There is no infinite set") compare with PA?

How does (ZFC-Infinity+"There is no infinite set") compare with (first order) PA? Intuitively, neither theory should be more powerful than the other.
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### Gödel, Escher, Bach: $b$ is a power of $10$.

I’d like to verify if my formula correctly expresses that a number is a power of $10$, using the $\sf{TNT}$ language provided by Hofstadter in his famous book Gödel, Escher, Bach: An Eternal ...
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### Can we apply a successor operation in Peano's arithmetic infinitely many times, is the successor well defined?

If we look at the axioms of Peano arithmetic, e.g. http://mathworld.wolfram.com/PeanosAxioms.html, they contain an axiom: If $a$ is a number, the successor of $a$ is a number. However, the axioms do ...
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### Sequent calculus and first incompletness theorem

Wikipedia says that sequent calculus is sound and complete (http://en.wikipedia.org/wiki/Sequent_calculus#Properties_of_the_system_LK). Godel first incompleteness theorem says that system capable of ...
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### Are all models of peano arithmetics descibed using first order logic non standard?

It is known that there are non-standard models of Peano Arithmetics when it is described using first order logic. My question is if there is standard model (one which does not contains non-standard ...
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### Why is Peano arithmetic undecidable?

I read that Presburger arithmetic is decidable while Peano arithmetic is undecidable. Peano arithmetic extends Presburger arithmetic just with the addition of the multiplication operator. Can someone ...
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