[gedpeck]

You know more than me on logic so I defer to your expertise.

https://math.stackexchange.com/questions/4753432/g%C3%B6dels...

Andreas Blass in the comments says that the Incompleteness results don’t apply to the second order Axioms (tabling about PA_2 here and not Z_2) and that the second order axioms are not computably enumerable. Maybe that’s the correct concept I was remembering from mathematical logic class. Don’t know if computably enumerable is the same as recursively enumerable but given what you’ve said I’m guessing they are different notions.

Consider the standard model of ZFC. Assume ZFC is consistent. Within this model there is one model of PA_2. Collect all true statements in this model of PA_2. Call that Super PA. That’s now my axiomatic system. I now have an axiomatic system that proves all true statements of arithmetic. Surely this set of axioms is not recursively enumerable.

Full semantics, on the other hand, are categorical (there is only one model), but this has nothing to do with the axioms not being recursively enumerable -- it is just because we use a different notion of model.

If those axioms were recursively enumerable then the Incompleteness theorems would apply, right?

What Noah Schweber says here seems pertinent:

https://math.stackexchange.com/questions/4972693/is-second-o...

[semolinapudding]

There's a bit of a definition issue at play here. When Andreas Blass and Noah Schweber say that there is no proof system for PA_2, they mean that there is no effective proof system that is complete for the full semantics. If you subscribe to their definition of a proof system, you end up saying that there is no such thing as a proof in PA_2, and thus that incompleteness is meaningless -- which I personally find a bit silly.

On the other hand, proof theorists and computer scientists are perfectly happy to use proof systems for second order logic which are not complete. In that case, there are many effective proof systems, and given that the axioms of PA_2 are recursively enumerable (they are in finite number!), Gödel's incompleteness will apply.

If you are still not convinced, I encourage you to decide on a formal definition of what you call PA_2, and what you call a proof in that system. If your proof system is effective, and your axioms are recursively enumerable, then the incompleteness theorem will apply.

[gedpeck]

Thanks for the reply.

[Tainnor]

> and that the second order axioms are not computably enumerable

He says that true sentences in second order logic aren't computably enumerable, he's not talking about the axioms.

> Don’t know if computably enumerable is the same as recursively enumerable

I've only ever seen them used as synonyms.

> Collect all true statements in this model of PA_2. Call that Super PA. That’s now my axiomatic system. I now have an axiomatic system that proves all true statements of arithmetic. Surely this set of axioms is not recursively enumerable.

What you call "Super PA" is called "the theory of PA". Its axioms are indeed not computably enumerable. That doesn't mean that the axioms of PA themselves aren't computably enumerable. And this much is true both for first and second order logic.

(edit: in fact, the set of Peano axioms isn't just computably enumerable, it's decidable - otherwise, it would be impossible to decide whether a proof is valid. This is at least true for FOL, but I do think it's also valid for SOL)

[gedpeck]

Thanks for the reply.

...it's decidable - otherwise, it would be impossible to decide whether a proof is valid.

Isn't that the whole issue with PA_2 vs. PA? In PA_2 with "full semantics" there is no effective procedure for determining if a statement is an axiom. In my mind this is what I mean by the incompleteness results not applying to PA_2. They do apply to Z_2 since that is an effective (computable?) system.

But Z2 is usually studied with first-order semantics, and in that context it is an effective theory of arithmetic subject to the incompleteness theorems. In particular, Z2 includes every axiom of PA, and it does include the second-order induction axiom, and it is still incomplete.

Therefore, the well-known categoricity proof must not rely solely on the second-order induction axiom. It also relies on a change to an entirely different semantics, apart from the choice of axioms. It is only in the context of these special "full" semantics that PA with the second-order induction axiom becomes categorical.

https://math.stackexchange.com/questions/617124/peano-arithm...

Thanks for the knoweledge. I'm going to read up more on this stuff.