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Suppose 1st-order arithmetic is inconsistent along with Voevodsky http://video.ias.edu/voevodsky-80th.

It nevertheless remains true that when you have 2 apples and 2 apples, you have 4 apples. Preforming an experiment gives you the result of an experiment, which cannot be inconsistent. So there is a subset of arithmetic that is "necessarily" consistent, given the notion (maps) that arithmetic models reality. The question is, what is this "physically consistent" proper subset of arithmetic?

The second question is, what happens if the physical theory is quantum field theory, where quanta loose their individual identity or "primitive thisness"?

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    $\begingroup$ You don't need to go all the way to quantum field theory for apples to lose their individual identity: all you have to do is tweak the temperature or pressure of your surroundings... $\endgroup$ Mar 18, 2011 at 22:40
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    $\begingroup$ I think terms like """primitive thisness""" deserve more than just one set of quotation marks. $\endgroup$ Mar 18, 2011 at 23:00
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    $\begingroup$ How come this question is not closed after one hour? Where is the Moral Police today? $\endgroup$
    – user6976
    Mar 19, 2011 at 0:27
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    $\begingroup$ This question has a meta thread - tea.mathoverflow.net/discussion/987/… $\endgroup$ Mar 19, 2011 at 3:17
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    $\begingroup$ Why does being quantum mean losing arithmetic? If you add two (-1) units of charge to two (-1) units of charge, you get four (-1) units of charge, even if the units of charge are carried by electrons, which have no individual identities. $\endgroup$
    – Peter Shor
    Mar 20, 2011 at 21:44

1 Answer 1

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Presburger arithmetic which is the first order theory of natural numbers with addition has been proven to be consistent by Mojżesz Presburger. My reference for this is the wikipedia article on Presburger arithmetic.

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  • $\begingroup$ @Gil The wikipedia page you link to mentions doubly-exponential worst-case lower bounds for decidability algorithms for Presburger arithmetic. It would be interesting to know the average-case bounds... $\endgroup$
    – David Roberts
    Mar 20, 2011 at 1:08
  • $\begingroup$ @David: What does average-case mean here? You need a probability distribution on sentences in arithmetic. How do you get one? $\endgroup$
    – Peter Shor
    Mar 21, 2011 at 2:13
  • $\begingroup$ @Peter Perhaps use Kolmogorov complexity or similar, see arxiv.org/abs/1010.2067 - or posts at John Baez's blog Azimuth on algorithmic thermodynamics (johncarlosbaez.wordpress.com/category/information-and-entropy) $\endgroup$
    – David Roberts
    Mar 21, 2011 at 6:53

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