# Can there be two different math?

As per usual, let PA denote Peano Arithmetic and ZFC denote Zermelo-Fraenkel set theory with choice. Furthermore, ZFC 'validates' PA, in the sense that it proves that the PA axioms hold for the standard model of the natural numbers. Arguably, PA also validates ZFC, albeit in a weaker sense. In particular, note that, if ZFC is consistent, then the sentences it proves about the natural numbers do not contradict the theorems of PA.

However, I'm wondering: can we find theories PA' and ZFC' that contradict the more usual PA and ZFC, but which validate each other in much the same way that PA and ZFC validate each other? Obviously the answer is 'yes', but what if we further require that PA and PA' agree on on all $\Pi_1$ sentences and all $\Sigma_1$ sentences, and similarly that ZFC and ZFC' agree on all $\Pi_1$ sentences and all $\Sigma_1$ sentences in the language of arithmetic?

Then we'd have no 'trial-and-error' method for choosing between the foundation (PA,ZFC) versus the foundation (PA',ZFC'), since it is only $\Pi_1$ and $\Sigma_1$ sentences in the language of arithmetic that can actually be checked in a mechanical fashion. More precisely, the $\Pi_1$ sentences of arithmetic are precisely those that can be mechanically falsified (if false), while $\Sigma_1$ sentences are those that can be mechanically verified (if true).

So is this possible?

And if so, how do we decide which is the correct, true math? (And, is this even a coherent idea?)

For instance, it is known that the twin-prime conjecture is $\Pi_2$. What if the PA/ZFC approach to math proves that the twin-prime conjecture is true, while the PA'/ZFC' approach proves that its false? What then?

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I don't see how you can say that $\sf ZFC$ and $\sf PA$ validate each other where $\sf ZFC$ is much much stronger. – Asaf Karagila Aug 12 '13 at 1:43
@AsafKaragila, if ZFC is consistent, then it doesn't contradict PA (interpreted as talking about $\mathbb{N}$); so, I regard that as a form of validation. – goblin Aug 12 '13 at 1:46
Sure, ZFC "validates" PA in proving its consistency (via a model), but in what sense does PA "validate" ZFC? – hardmath Aug 12 '13 at 1:49
I would like to thank my lawyer, @hardmath, for taking care of the follow up while I was typing an answer somewhere else. :-) – Asaf Karagila Aug 12 '13 at 1:50

Are you thinking to a sort of "crucial experiment" like those used in empirical sciences in order to choose between two competing theories ?

You are assuming the union of the sets of $\Pi_1$ sentences and $\Sigma_1$ sentences as a sort of "empirical basis" of mathematical facts that forms the "evidence" against which compare a couple of competing theories (i.e. PA and PA').

If both theories agree on this basis, how can we decide between them ?

Assuming that PA and PA' are consistent, we know that they have models. If $\mathcal M$ is a model of PA (PA'), the (closure of) theorems of PA (PA') are sentences that express "true facts" about the model; is it right ?

Assume that we can find a $\Pi_2$ formula that is proved by (e.g.) PA and disproved by PA'.

But then, what are the evidence for/against it ? If the mathematical community does not share an "intuition" for or against the twin-prime conjecture, the only way left to affirm (or deny) a mathematical fact is through a proof (accepted by the math community) of it.

Is this case different from that of euclidean vs non-euclidean geometry ?

Leaving aside the analogies, what PA' will looks like ? Assuming first-order logic with individual constant $0$, function symbol $'$ and two predicate symbols : $+$ and $.$, in order to obtain PA' it is necessary to add to the usual FO version of Peano Axioms, some new axiom that is not provable in PA itself.

This must be a statement undecidable on the basis of (FO version of) PA's axioms, otherwise PA' will be inconsistent (or PA itself).

(Note : is it true that this statement is like parallel postulate of euclidean geomtery ?, i.e. is undecidable on the basis of the remaining axioms)

Now we want that PA' behave like PA regarding the set of $\Pi_1$ and $\Sigma_1$ sentences : is it possible ?

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I think you have understood my concerns perfectly. But, what does your last sentence mean? It seems that you are arguing that neither PA nor PA' are more correct. – goblin Dec 20 '13 at 10:34
At the end of 19th century mathematician proved the relativ consistency of eucl and non-eucl geometries. So, which is true ? Both have models. If we consider geometry as a description of physical space, may we say that Einstein's relativity gives support to the interpretation of non-eucl geometry as a "true" description of it ? If so, we do not say that eucl geometry is false : may we say that physical space is not the "intended model" of it ? – Mauro ALLEGRANZA Dec 20 '13 at 10:43
I'm not sure I agree with the analogy. The shape of space in any given region is, in principle at least, testable by observation. How can we test between PA and PA'? – goblin Dec 20 '13 at 10:47
I'm trying to follow your suggestion: to use $\Pi_1$ sentences and $\Sigma_1$ sentences as "empirical basis" of mathematical facts that are the "evidence" for "testing" PA and PA'. If we cannot use it to choose between them, and assuming both consistent, we can say that PA has a model $\mathcal M$ and PA' a model $\mathcal M'$. "Our" numbers are the intended model of PA, but if PA disproves the twin-prime conjecture, what we must do? Stay with the conjecture, and so conclude that "our" numbers are nor more the intended model of PA, or stay with PA, and conclude that the conjecture is false? – Mauro ALLEGRANZA Dec 20 '13 at 11:03