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Is Robinson Arithmetic complete in the sense of Gödels completeness theorem? And is Robinson Arithmetic incomplete in the sense of Gödels first incompleteness theorem? If RA is both it would be a good example to explain the different notions of "completeness" used in the two theorems mentioned.

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Gödel's completeness theorem isn't about theories but about the logic used in the language of a theory, namely about the connection between the syntax and semantics of that logic. The logic used to describe Robinson's arithmetic (first order logic) is complete and at the same time RA is not complete as a theory. The same applies to first order Peano Arithmetic. The logic used in the language of PA is complete but PA is not a complete theory. Same thing applies to ZFC. Generally, every incomplete theory described in first order logic has that property.

Apostolos
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  • I thought Robinson's Arithmetic WAS complete as a theory. What am I misremembering? – Jason DeVito - on hiatus Dec 04 '10 at 16:17
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    @Jason: Robinson's arithmetic is incomplete and undecidable much like PA, but it's finitely axiomatized (en.wikipedia.org/wiki/Robinson%27s_Arithmetic). Maybe you're thinking about Presburger arithmetic which is decidable? – Apostolos Dec 04 '10 at 16:35
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    Yes - it was Presburger arithmetic I was thinking of. Thank you for the clarification. – Jason DeVito - on hiatus Dec 04 '10 at 17:14
  • @Apostolos: Hey, I've never heard of Presburger arithmetic before. Thanks for that Apostolos. But something is bugging me: a quick glance at the wikipedia page taught me that Presburger arithmetic is basically Peano arithmetic without multiplication. But isn't that strange? If you have addition already, it should be a trivial matter to define multiplication. Thus Presburger arithmetic should be equivalent to Peano arithmetic. What am I missing? Seemingly some axiom about how addition relates to multiplication or something? – Raskolnikov Dec 04 '10 at 17:14
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    @Raskolnikov What tools do you have to define multiplication in Presburger? It's a worthwhile exercise to try it and see why you can't do it. – Mark Reitblatt Dec 04 '10 at 17:23
  • Well, I'd say you have addition. If I have to explain multiplication to a kid, I'd say he has to add a certain number of times the same number. But maybe that doesn't work because it requires some second order definition or something? – Raskolnikov Dec 04 '10 at 18:07
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    @Raskolnikov Yeah, that's the idea. The number of times you have to add "x" will not be constant. If you try to write it out in the formal language, you'll get stuck. – Mark Reitblatt Dec 04 '10 at 19:05