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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 don't have commenting powers yet :-( Anyway, when I look at the answer to how exponentiation is defined, using sequences and the Chinese remainder theorem, I assume that Presburger is simply not powerful enough to play the same trick to define $\times$ in terms of +?

thanks

Rodrigo de Azevedo
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Motorhead
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  • The proof that exponentiation is Diophantine is weirdly technical and off-putting, but it is the reason. – Thomas Andrews Jan 25 '15 at 01:58
  • Specifically, there is a polynomial $f(x,y,z,w_1,w_2,\dots,w_n)$ such that for any $(x,y,z)\in\mathbb N$, $\exists w_1,w_2,\dots,w_n$ yielding a solution to $f$ if and only if $x^y=z$. That's a deep and weird result. Note that a polynomial can be expressed in terms of sum and product only: $x^3+3xy+z=x\cdot x\cdot x + 3\cdot x\cdot y + z$. – Thomas Andrews Jan 25 '15 at 02:01
  • Once you've defined ore assumed $+$ and $\times$, you can define exponentiation without encodings or the like using this result. @CarlMummert – Thomas Andrews Jan 25 '15 at 02:08
  • @CarlMummert Having read the proof that exponentiation is Diophantine, I saw nowhere in it that it assumed a particular model. Maybe I'm wrong. – Thomas Andrews Jan 25 '15 at 02:12
  • Sigh, no, you just need to prove that the triples $(x,y,z)$ satisfy the recursive definition of triples satisfying it, including that it is single-valued. You don't need to already have exponentiation defined in the structure. @CarlMummert – Thomas Andrews Jan 25 '15 at 02:17
  • Give my $P$, if I write $R(x,y,z)$ for the relationship, then if I can show via Peano that: $R(x,0,z)\iff z=1$, $R(x,Sy,z)\iff \exists w(z=xw\land R(x,y,w))$ then $R$ defines exponentiation in all models, and we can show via induction on $y$ that $\forall x,y(\exists_1 z,R(x,y,z))$. @CarlMummert – Thomas Andrews Jan 25 '15 at 02:23
  • The definability, indeed of any recursive function, was shown some $40$ years earlier than the solution of Hilbert's Tenth Problem. – André Nicolas Jan 25 '15 at 05:23

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Yes, Presburger arithmetic is not strong enough to quantify over sequences, which is the reason it can't define multiplication.

Carl Mummert
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