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Presburger Arithmetic is a decidable theory but if multiplication is added to it would that theory remain decidable?

UPDATE: I began to write out the axioms that would distinguish Presburger Arithmetic from Peano Arithmetic and realized that adding the Peano axioms which give semantics for multiplication (E.g., commutativity of multiplication, the distributive law -- essentially, all of the peano axioms which include the multiplication symbol -- etc) to Presburger Arithmetic result in Peano Arithmetic. And the decidability of Peano Arithmetic is already widely known due to Gödel. So I've decided to change my question: why does a definition of multiplication in Presburger Arithmetic result in an undecidable theory? In other words, why would the addition of the multiplication symbol and axioms for multiplication result in an undecidable theory?

Rodrigo de Azevedo
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Francesco Gramano
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    See George Boolos & John Burgess & Richard Jeffrey, Computability and Logic (5th ed - 2007), page 295 : "Arithmetic is not decidable: the set of code numbers of sentences of the language L of arithmetic that are true in the standard interpretation is not recursive. [...] In contrast to the undecidability of arithmetic stand Presburger’s theorem, to the effect that arithmetic without multiplication is decidable" – Mauro ALLEGRANZA May 05 '15 at 07:41
  • I'm a bit confused by the question; when you talk about validity of a formula in a certain signature, are you talking about what follows from the logical axioms alone, rather than from axioms of some theory of arithmetic? – Trevor Wilson May 05 '15 at 07:46
  • Has my edit made it more clear, @TrevorWilson? – Francesco Gramano May 05 '15 at 07:53
  • Yes, but now I am wondering: when you say "multiplication is added to it" do you just mean that you are considering the same axioms as forming a theory in a larger signature, or are you also adding some non-logical axioms that pertain to multiplication? – Trevor Wilson May 05 '15 at 07:56
  • @TrevorWilson Could you please explain the difference between those two statements? They are honestly indistinguishable to me. – Francesco Gramano May 05 '15 at 08:04
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    Let me rephrase my question. A theory (such as Presburger arithmetic) is a set of axioms, but multiplication is not an axiom (in this context, I am thinking of multiplication as a predicate symbol with no semantic meaning.) So I don't know what it means to add multiplication to a theory. If we don't add any axioms, then we don't change what is provable in the theory, so we don't change whether the theory is decidable. – Trevor Wilson May 05 '15 at 08:12
  • The symbol for multiplication would have to be added as well as axioms for its semantics. – Francesco Gramano May 05 '15 at 08:14
  • In that case, I think the important question becomes: which axioms for its semantics? – Trevor Wilson May 05 '15 at 08:16
  • Axioms 3,4,5, and 7 from this list (I'm not sure if 12 is necessary) – Francesco Gramano May 05 '15 at 08:23
  • I think you should add that information to your question. I don't know the answer, but I bet someone else here will. It might also be good to say in the question how your theory differs from $\mathsf{PA}$ as well as how it differs from Presburger arithmetic. – Trevor Wilson May 05 '15 at 08:27

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