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What is a simple (the simplest?) axiom which can be added to the usual PA axioms so that the new "non-standard PA theory" no longer has the Standard Model as one of its models? Assuming of course that the new axiom is consistent with the original axioms.

EDIT: I intend a first order theory. If possible by adding a single new axiom. If absolutely necessary, then an RE axiom schema.

Tommy R. Jensen
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    You mean adding a constant $\gamma$ and infinitely many axioms $\gamma\neq 0$, $\gamma\neq 1$, ... – user10354138 Jul 18 '20 at 11:38
  • I see what you are saying, and it would do the job alright. Except it would obviously be more than one axiom. Is it clear that we cannot do better? – Tommy R. Jensen Jul 18 '20 at 12:14
  • @user10354138 That axiom explicitly contradicts one of the PA axioms. So there is no model for your theory at all, standard or not.' – David C. Ullrich Jul 18 '20 at 12:41
  • That is interesting to me. I started reading a paper in which the theory in question is required to have only non-standard models of PA. It seemed an aside point initially. But I began wondering how to achieve it. On second thought I can believe that it might be just as impossible as to construct a theory which has only the standard model. Because the new attempted axiom meaning "the theory does not have the standard model" has a negation resembling "the theory only has the standard model". Probably my question is simply too naive. – Tommy R. Jensen Jul 18 '20 at 13:32
  • I am not an expert, but the approach user10354138 gives is the usual one to force the model to be nonstandard. There is a proof that first order logic cannot force there to only be the standard model. You would like the induction axiom to say "all numbers are standard" but that is a second order statement. – Ross Millikan Jul 18 '20 at 13:42
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    It's not very simple, but does $\lnot \mathrm{Con}(\mathsf{PA})$ work? – Nate Eldredge Jul 18 '20 at 14:32
  • @Ross Millikan : I believe you may force the standard model by adding every statement that is true in the standard model as an axiom to make a first order theory. But it is "impossible", because we cannot effectively determine which statements are true, The approach mentioned by user10354138 is a straightforward axiom schema that forces a nonstandard model. This is not a contradiction, because the negation of the conjunction of axioms in the schema is not itself a schema, since it becomes a logical disjunction. – Tommy R. Jensen Jul 18 '20 at 15:20
  • @Nate Eldredge : Adding $\neg$Con() as an axiom to PA might kill off some models. But are you certain that it is consistent with PA and that it rules out the standard model in particular? – Tommy R. Jensen Jul 18 '20 at 16:04
  • @TommyR.Jensen: I believe so, see https://math.stackexchange.com/questions/2305133/is-conpa-true-in-standard-model – Nate Eldredge Jul 18 '20 at 16:08
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    @TommyR.Jensen: And $\lnot \mathrm{Con}(\mathsf{PA})$ is consistent with $\mathsf{PA}$, assuming $\mathsf{PA}$ is itself consistent. If it were not, then $\mathsf{PA}$ would prove $\mathrm{Con}(\mathsf{PA})$, which by Godel's second incompleteness theorem it cannot do. – Nate Eldredge Jul 18 '20 at 16:12
  • @TommyR.Jensen: I don't believe that is sufficient. There could be more elements than the standard ones. There are constraints on what is true about them-they are all greater than all the naturals so every statement that is true for all $n$ large enough is true of all of them. – Ross Millikan Jul 18 '20 at 18:21
  • @RossMillikan: Indeed, doesn't upward Lowenheim-Skolem guarantee that there are other models (of arbitrarily high cardinality) which satisfy exactly the same first-order statements as the standard one? – Nate Eldredge Jul 18 '20 at 21:29
  • @NateEldredge: Yes. I believe one can also get countable models that satisfy all the same first order statements as standard "true arithmetic". – Ross Millikan Jul 18 '20 at 21:43
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    @NateEldredge I was talking about a comment that seems to have disappeared, where it was asserted that PA plus the one axiom "there exists a non-zero non-asuccessor" added had the required property. – David C. Ullrich Jul 18 '20 at 21:43
  • @DavidC.Ullrich: Ah, I see. I deleted my comment. – Nate Eldredge Jul 18 '20 at 21:44

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Since the new axiom must be consistent but not hold in the standard model, it must be undecidable in $\sf PA$. But for most (all?) of the statements which we have proved undecidable in $\sf PA$, we know whether or not they hold in the standard model. We can therefore obtain a nonstandard theory by adding the statement or its negation to $\sf PA$.

So answering the question simply requires us to find the simplest statement which is known to be undecidable in $\sf PA$. Of course Gödel sentence was the first known example of such, but since then various simpler undecidable sentences have been found. The most famous are Goodstein's theorem (proved undecidable by Kirby and Paris), and the Strengthened Finite Ramsey Theorem (proved undecidable by Paris and Harrington). Both these statements are known to be true in the standard model, so we obtain a non-standard theory by adding the negation of one of them as an axiom.

Oscar Cunningham
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