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Internal Set Theory (IST) is a conservative extension of ZFC that adds three axioms that serve to define a predicate standard such that all numbers are either standard or not. There are finitely many standard numbers and numbers larger than these are said to be illimited or i-large. Informally speaking, illimited numbers are just too big to grasp so they don't matter.

Suppose we define a formal language in the usual way by choosing an alphabet and setting out the recursive syntactical rules that define a well-formed formula (wff). I have in mind the language of Peano arithmetic. But now we impose an additional criterion that a string needs to satisfy in order to qualify as a wff: its Gödel number must be a standard number. The intuitive justification for doing this is that strings with non-standard Gödel numbers are too large to be accessible to us and so should not matter.

Once we have the language defined, we form a theory using the usual recursive axioms for first order Peano arithmetic, closed under classical logic. We have not imposed any specific upper bound on the length of a formula, so the theory is adequate for the purposes of basic arithmetic. And we have not imposed any specific upper bound on the length of a proof, so we do not have to worry that a given sentence is unprovable just because it is too long.

Since there are finitely many standard numbers, this language contains finitely many wffs. Since the language is finite, any theory defined within it is finite also. Ordinarily, a theory that has a finite model cannot be essentially incomplete.

So, is there any interesting sense in which such a theory of arithmetic avoids Gödel's first incompleteness theorem?

I can imagine a few possible responses... We've made the definition of wff non-constructive, so we've broken the usual relation between finiteness and completeness. The collection of wffs is not a set, so there isn't a clear way to define incompleteness. We've just transplanted the incompleteness from the theory into the language. This is just a version of ultrafinitism and not particularly interesting. I'd be interested to hear what our logic experts think.

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    "illimited numbers are just too big to grasp so they don't matter" : why do they not matter :-?
    – Mikhail Katz
    Commented Jul 8 at 11:48
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    Not just numbers, incidentally; the predicate "standard" applies to sets, as well.
    – Mikhail Katz
    Commented Jul 8 at 11:48
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    "Since there are finitely many standard numbers" : this is incorrect. Where did you get such an idea?
    – Mikhail Katz
    Commented Jul 8 at 11:50
  • Your ref says IST is to clearly express Robin's nonstandard analysis via syntactic enrichment. Since it's conservative extension of ZFC it cannot avoid incompleteness theorems. Besides issues already raised, your 'a theory that has a finite model cannot be essentially incomplete' can be raised subtly, PA's standard model is NOT a finite model viewed within PA? On a more deeper insight of incompleteness, its root cause lies in the multiplication operation even through standard numbers. Thus the extended Standard predicate introduced by IST has no effect on any incompleteness results at all...
    – Double Knot
    Commented Jul 8 at 23:33
  • @DoubleKnot IST itself is incomplete for sure. But that is not the point. We can use set theory to construct theories that are complete. What I am doing is asking whether if we construct a theory of arithmetic in a particular way we can avoid Gödel's first incompleteness theorem in some interesting sense. We could avoid GFIT trivially just by imposing a specific upper bound on Gödel numbers, e.g. 10 ↑ (10 ↑ 10) but that would have unfortunate consequences since there would be cases where φ is a wff but ¬φ is not.
    – Bumble
    Commented Jul 9 at 1:49

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The basic premise of your question ("there are finitely many standard numbers") is incorrect. What is true in IST is a statement that does have a finitistic flavor, but it is a bit more involved: there is a finite set S such that each standard number belongs to S. However, the set S is itself nonstandard, so it has a nonstandard number of elements (namely, an unlimited number; as you mentioned, some authors use the term "illimited").

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  • What I had in mind is, as you say, that there exists a finite (non-standard) set of all the standard numbers. This is theorem 1.2 of Nelson's original paper. Granted we cannot say how many standard numbers there are, nor speak of a largest standard number, but we know there are not infinitely many.
    – Bumble
    Commented Jul 8 at 12:44
  • As to why i-large numbers don't matter, I suppose the intuition behind it is that in a finite universe that will last a finite amount of time and have a finite number of rational agents reasoning about a finite quantity of numbers, it is pointless to be concerned with numbers that will forever be beyond our grasp. Ditto for strings.
    – Bumble
    Commented Jul 8 at 12:45
  • "there exists a finite (non-standard) set of all the standard numbers": This is a misunderstanding. What Nelson proved is that there exists a finite set containing all standard numbers, not that it consists precisely of the standard numbers. The collection of standard numbers is not a set in any axiomatic theory of nonstandard analysis.
    – Mikhail Katz
    Commented Jul 8 at 12:56
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    "As to why i-large numbers don't matter, I suppose the intuition behind it is that in a finite universe that will last a finite amount of time" : but that finite amount of time will already include some nonstandard instances :-) Far from claiming that such "i-large numbers don't matter", Nelson (following Robinson) placed them at the basis of infinitesimal analysis. The whole point of such frameworks (whether model-theoretic or axiomatic) is to enable the practicing mathematician to use infinitesimals and unlimited numbers, with the benefit that both definitions and proofs becomes simpler...
    – Mikhail Katz
    Commented Jul 8 at 13:00
  • Currently there exist proofs using NSA that have not been matched with an "infinitesimal-frei" version.
    – Mikhail Katz
    Commented Jul 8 at 13:00

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