That's a very modest claim.

That there is a recursive axiomatization of a theory T entails that it is computable whether any given string is or is not an axiom.That follows from the fact that any recursive relation is computable.

By the way, I don't know what would be the value in telling others that I said it, since I am not in any way an authoritative expert in logic of mathematics. The only thing I really know very much about is jazz.

I got this reply, which is really interesting for me to understand:

The real key to Gödel is that the axioms are recursively enumerable, not countable. We can show there exist maximal consistent subsets of the countable set of all statements, and take those as axioms. Then we can show that maximality implies completeness. It’s just not useful for human or computer-read proofs, because there is no way to algorithmically prove each step is allowed.

What do you think @jgill, and @TonesInDeepFreeze?

Here, 'computable' is to be taken as 'Turing machine computable'.

"If R is recursive then R is Turing machine computable".

That has a long and complicated proof, but it's not controversial that it is proven.

Note that this does not even need to assume the Church-Turning thesis, as its converse ("If R is recursive then R is computable (in an intuitive informal sense of 'computable')") is not controversial.

This user has been deleted and all their posts removed.

This user has been deleted and all their posts removed.

an axiom I understand here is an expression, like Godel's sentence, such that neither it nor its negation is provable, yet is also provably true, being proved meta-mathematically — tim wood

That is incorrect. You have conflated "formula F is independent of axiom set S" with "formula F is an axiom in the axiom set S".

Axiomatization is a relation between a set of sentences (called 'the axioms') and the set of sentences (called 'the theorems') that are provable from those axioms. I.e.:

A set of sentence S is an axiomatization of a set of sentences T iff every member of T is provable from some finite subset of S.

A theory is a set of sentence closed under deduction (provability).

A recursive axiomatization S of a theory T is a recursive set of sentences S such that every member of T is provable from a finite subset of S.

Trivially, any axiom in S is provable from S. (Trivially, any axiom is provable from itself.)

A theory T is recursively axiomatizable iff there is a recursive set S that is an axiomatization of T.

/

Here's the other question asked just for fun:

https://math.stackexchange.com/questions/4142390/completeness-without-uncountability

recursively enumerable is not the same as recursive, and implies non-recursiveness. — tim wood

That is incorrect.

It is an easy theorem that R is recursive iff (R is recursively enumerable & ~ R is recursively enumerable).

A recursive enumeration does not itself provide a decision procedure for R, but that does not entail that there does not exist a decision procedure for R.

What do you think — Shawn

I would have to think it through. I am simply not caught up in the StackExchange thread.

The very first sentence in the very first reply to you in that thread:

"What does it mean for a theorem to be complete or uncountable?"

Please stop using the phrases "theorem is complete" or "theorem is countable". They make no sense, and all they do is make people have to stop you in your tracks before we can even answer you about anything else you ask or say.

This user has been deleted and all their posts removed.

a recursive set is not the same thing as a recursively enumerable set, — tim wood

In the exact sense that the set of recursive sets is a proper subset of the set of recursively enumerable sets.

closure — tim wood

A theory is closed under deduction.

But the clause you have in mind is recursive axiomatizability. (I don't think that is usually referred to as 'closure'.)

At least for now, I've decided not to post there. There are already too many confusions in the discussions, and I'm not up to sorting through them with the participants. Also, I don't like the interface and layout of the post and thread formatting there.

Can you explain this, as I'm quite interested? — Shawn

Well, Shawn, you asked for it. Sorry, buddy, it's convoluted and I doubt comprehensible. I don't usually go to professional math sites like StackExchange, preferring to do the job myself, but I don't think it would make any difference for this particular problem. So I'll put it up here. And one of the math people on the forum might have a suggestion:

Start with a sequence of complex functions: $\left\{ {{f}_{k}} \right\}_{k=2}^{\infty }$,
And create a second sequence: ${{F}_{k-1,n}}(z)={{f}_{k-1}}\left( {{F}_{k,n}}(z) \right)$, ${{F}_{n,n}}(z)=z$ Which can be written as: ${{F}_{1,n}}(z)=z+{{\rho }_{1}}{{\varphi }_{1}}\left( {{F}_{2,n}}(z) \right)+{{\rho }_{2}}{{\varphi }_{2}}\left( {{F}_{3,n}}(z) \right)+\cdots +{{\rho }_{n-1}}{{\varphi }_{n-1}}\left( {{F}_{n,n}}(z) \right)$

${{\varphi }_{k}}(z)=-zQ(z)/\left( 1+{{\rho }_{k}}Q(z) \right),Q(z)=xCos(y)+iySin(x)$


Then show $\underset{n\to \infty }{\mathop{\lim }}\,{{F}_{1,n}}(z)$ exists.

There have been discussions on this thread about recursive objects. Here's a Lulu from actual math, not about math. :cool:

This dynamical system is associated with an image I will post.

Reproductive Universe

I forgot to mention, one way a theorem could be considered "complete" is when the theorem is "sharp", which means the hypotheses cannot be reduced and the theorem still reach the conclusion. Numerical analysis is an area where one used to see this idea play out.

@TonesInDeepFreeze,

I've been wondering if the question posted over at StackExchange, is generalizable?


What do you think?

I'm also trying to figure out whether the complexity class is determinate for the first question I asked on StackExchange.

https://math.stackexchange.com/questions/4144858/finding-the-shortest-proof-how-complex-is-it

First, for your questions, these still need to be clearly settled:

(1) Given a recursively axiomatizable theory with finitely many axioms, and with 'length of proof' defined as the sum of the lengths of the formulas that are the lines in the proof, is there an algorithm to determine. for any given theorem, the least length of a proof?

(2) Given a recursively axiomatizable theory with infinitely many axioms, and with 'length of proof' defined as the sum of the lengths of the formulas that are the lines in the proof, is there an algorithm to determine, for any given theorem, the least length of a proof?

(3) Given a recursively axiomatizable theory with finitely many axioms, and with 'length of proof' defined as the number of lines in the proof, is there an algorithm to determine, for any given theorem, the least length of a proof?

(4) Given a recursively axiomatizable theory with infinitely many axioms, and with 'length of proof' defined as the number of lines in the proof, is there an algorithm to determine, for any given theorem, the least length of a proof?

Should I post this on StackExchange?

How would you go about this?

Those are my questions to sort out the discussion at StackExchange. You can post it if you want, but I probably won't follow up there or here, as I explained.

What are your responses to those questions, just wondering?

I don't know, because I got confused by what different people said at StackExchange.

https://math.stackexchange.com/questions/1002618/how-to-find-the-shortest-proof-of-a-provable-theorem?

https://thephilosophyforum.com/discussion/comment/539584

In all cases, it seems to me that, since we are concerned with finding the shortest proof, we only need to consider the finite part of the signature that occurs in the theorem, and we don't need to consider alphabetic variants.

(1) I think 'Yes'. There are only finitely many strings of a given finite length from a finite set of symbols, and only finitely many axioms to check whether a string is an axiom.

(2) I think 'Yes'. There are only finitely many strings of a given finite length from a finite set of symbols, and it is algorithmic to check whether any one of them is an axiom.

(3) I didn't check the details of the argument at

https://math.stackexchange.com/questions/1002618/how-to-find-the-shortest-proof-of-a-provable-theorem

that purports to show that the answer is 'No', but the poster seems to be knowledgeable.

(4) I didn't check the details of the argument at

https://math.stackexchange.com/questions/1002618/how-to-find-the-shortest-proof-of-a-provable-theorem

that purports to show that the answer is 'No', but the poster seems to be knowledgeable. And, aside from his argument, it does seem to make sense that if there are infinitely many axioms, then there are infinitely many different possible proof lines, so the algorithm could go on indefinitely before arriving at an answer.

I kind of have this topic in mind again, and was thinking that if completeness cannot be determined for theorems, then can they be ascertained in complexity?

It would be interesting to look into this as a general proposition.

What do you think, @TonesInDeepFreeze?