G=This sentence is not provable in T — TheMadFool

If G was provable in T then this requires a sequence of inference steps in T
that prove that they themselves do not exist.

Copyright 2023 PL Olcott

Q is enumerable and so is the set of finite sequences of members of Q. The set of infinite sequences of members of Q and R are not enumerable. Okay, some things are enumerable and others aren't.
— TonesInDeepFreeze

I know all that, already said it, spent years writing proofs for professors. Not asking random internet guy about the basics of analysis. Tried to ask you about 'subjective' (maybe philosophical) responses to all the symbols that swim like fish in those textbooks you mentioned. You gave a disappointing response, like you are deaf and mute to anything that isn't mere chatbot correctness. I have loved math as a meaningful 'science' of form(s) with some intuitive validity. I care about various formalisms only because they strive to mean something, capture something beyond them. The continuum is a endlessly fascinating beast that great thinkers have wrestled with for centuries. I don't know if you know or care much about mathematical history, but I love the drama. But I'll save that for others who aren't satisfied with the relatively trivial (however difficult at times ) syntactical part. — plaque flag

I said I don't have anything immediate to say about subjective impressions of mathematics. I am not thereby like a "chatbot" that is not interested in anything other than the syntactical aspects of mathematics. Specifically, the fact that I don't have anything immediate to say about "R is mostly a black and seamless sea in the darkness" does not entail that I am uninterested in intuitions involved in the construction of the real number system. And the poster quoted above had said that he had been reading my posts over time; but in some of my posts I had written that indeed I am interested in intuitions about mathematics.

1. G is provable. So G is unprovable
2. G is not provable

So, there is G in the theory T

Have I got it right? — TheMadFool

Shouldn't G be in the form of arithmetic calculus propositions for the incomplete theorem to apply?

G is a sentence in the language of arithmetic.

There are many ways to couch incompleteness proofs. Here is one outline:

Let T be a recursively axiomatized, consistent theory that is "sufficient for a certain amount of arithmetic".

We adduce a sentence G that is is true (to be more precise, it is true in the standard model for the language of arithmetic) if and only if G is not provable in T.

Then we prove that G is not provable in T. So G is a true sentence that is not provable in T. Moreover we show also that ~G is not provable in T. So T is incomplete.

We adduce a sentence G that is is true (to be more precise, it is true in the standard model for the language of arithmetic) if and only if G is not provable in T.

Then we prove that G is not provable in T. So G is a true sentence that is not provable in T. Moreover we show also that ~G is not provable in T. So T is incomplete. — TonesInDeepFreeze

Could you demonstrate and prove the provability and unprovability of G in real arithmetic sentences in T?

We adduce a sentence G that is is true (to be more precise, it is true in the standard model for the language of arithmetic) if and only if G is not provable in T.

Then we prove that G is not provable in T. — TonesInDeepFreeze

If all it proves is that every T has the true and unprovable sentence "this sentence is true and unprovable" then it seems vacuous.

Or does it prove that every T has a "natural" example of a true and unprovable sentence, like the strengthened finite Ramsey theorem in Peano arithmetic?

Could you demonstrate and prove the provability and unprovability of G in real arithmetic sentences in T? — Corvus

If T is consistent, then T does not prove both that G is provable in T and G is not provable in G.

We prove that if T is consistent then T does not prove G and T does not prove the negation of G.

How that is all done is quite complicated, especially with the proofs of all the lemmas.

If all it proves is that every T has the true and unprovable sentence "this sentence is true and unprovable" then it seems vacuous. — Michael

"This sentence is true and unprovable" is not the sentence we prove is not provable in T.

Rather, "This sentence is not provable" is the sentence we prove is not provable in T and we prove that it is a true sentence.

Don't forget that the predicate 'provable' can be emulated in T, but the predicate 'true' cannot be emulated in T.

I don't know what you mean by "vacuous" here. G is a sentence of arithmetic. It makes a certain true claim about natural numbers. Granted, the particular claim it makes about natural numbers is probably not of interest to anyone. But that's not the point. Rather, the point is that there is no recursively axiomatized and consistent system for basic arithmetic that is complete and thus, for any given such system, there are true sentences about the natural numbers that are not provable in the system. Moreover, we can then see that there are infinitely many such true and unprovable sentences. Moreover, we then see that it is possible that some of the sentences about arithmetic that are of interest to us might be undecidable (neither provable nor disprovable) in the system. Moreover, hastened by the previous point, we do go on to show specific sentences that are of interest that are undecidable. That leads to the work showing that there is no algorithmic method for solving Diophantine equations, which is not just of interest but is basic to mathematics, even basic to high school algebra, especially for any lazy teenager like me who ever wondered, "Isn't there a step by step procedure I could use to solve any possible equation that I might be asked to solve, so I wouldn't have to think over all these problems but instead could just apply the procedure?" Moreover, we are then led to showing the undecidability of profound and fundamental questions such as the axiom of choice in ZF and the continuum hypothesis in ZFC. Moreover the techniques used in the incompleteness prove lead to the profound find that there is no solution to the halting problem, etc. And to top all of that, the P v NP problem may be the most economically valued in mathematics, as solution to it would have vast ramifications for computing and business; and incompleteness informs us that it is possible that P v NP does not have a solution (though, granted, there are a lot of people who do think it does have a not yet discovered solution).

Or does it prove that every T has a "natural" example of a true and unprovable sentence, like the strengthened finite Ramsey theorem in Peano arithmetic? — Michael

That is a good question. We know that, for example, PA and set theory are such Ts. But, putting aside the ambiguity of 'natural' and assuming a general informal sense of it, I don't know whether it holds for every qualifying T.