What are your opinions of all the theorems that rely on RH as an excluded middle?
Constructivists reject exmid, saying instead that a proof of "A or B" requires you to have in hand a proof of A or a proof of B. And nobody yet has a proof of RH nor a proof of ~RH. This is important in so-called incomplete logical systems, where some theorems are neither provable nor disprovable, and, therefore, exmid is an inadmissible axiom.
Aren't those sort of different things? I thought the whole point of provability was that it was distinct from truth.
How do you know something is true if you don't have a proof? It all depends on you views on the philosophy of mathematics. Are there "true" statements that don't have proofs? Some say yes, there are platonic ideas that are true, even if they aren't provable. Others say, "what does it even mean to say something is true, if there is no proof. What you really have is a conjecture."
Didn’t Gödel show that in most useful logical systems there are true statements that cannot be proved?
Indeed, the first incompleteness theorem tells us that any logical framework which can express Peano arithmetic must necessarily contain true (resp. false) facts for which no (resp. counter) proof can be given.
Sometimes you can prove that no proof exists about a specific sentence (that's what his incompleteness proof does), and I think you could extend this technique to construct sentences where no proof exists of whether it has a proof, etc...
(with a finite list of axioms)
I think the precise pre-condition is that the theory should be recursive, which means either a finite list of axioms _or_ a computable check to determine whether a given formula is an axiom.
This formulation misses the important aspect that whether the statement is 'true' is not absolute property (outside logical truths). We can consider truthfulness of a statement in a specific structure or in a specific theory.
E.g. a statement can be undecidable in Peano arithmetic (a theory) while true in natural numbers (a structure, model of Peano arithmetic), but that just means there is a different structure, different model of Peano arithmetic in which this statement is false.
Not quite. Any logical framework which can express Peano arithmetic must necessarily contain true facts for which no proof can be given within PA. The proof of Godel's theorem itself is a (constructive!) proof of the truth of such a statement. It's just that Godel's proof cannot be rendered in PA, but even that is contingent on the assumption that PA consistent, which also cannot be proven within PA if PA is in fact consistent. In order to prove any of these things you need to transcend PA somehow.
The latter would be an axiom. A disproof would be a proof that there is no proof, so if you’d proven that no proof exists one way or the other then you’ve proven it can’t be disproven _or_ proven.
Which means you’ve hit a branch in mathematics. You can assume it to be either true or false, and you’ll get new results based on that; both branches are equally valid.
No he showed either there are true statements that cannot be proved, OR the system is inconsistent...
https://en.wikipedia.org/wiki/Intuitionism
There is another possibility: it could be an arbitrary choice, as in the case of the parallel postulate, the axiom of choice, and non-standard models of the Peano axioms.
According to logicians, it always is an arbitrary choice. But we have a second-order notion of what "finite integer" should mean. And within that notion the idea might be false.
Here's how that plays out. Suppose that the RH cannot be proved or disproved from ZF. (It turns out that choice cannot matter for all theorems in number theory, so ZF is enough.) That means that there is a model of ZF in which RH is true. Every model of ZF contains a finite calculation for any non-trivial zero of the Riemann function. (The word "finite" is doing some heavy lifting here - it is a second order concept.) That calculation must work out the same in all models. Therefore every finite nontrivial zero has complex part 0.5. Therefore RH is actually true of the standard model of the integers. Therefore every model of ZF where RH is false, is non-standard.
The truth of RH is therefore independent of ZF. But it's true in our second order notion of what the integers are.
Sorry, you're going to have to explain that to me. The word "finite" has only one meaning AFAIK and on that definition it is definitely not the case that "every model of ZF contains a finite calculation for any non-trivial zero of the Riemann function." I don't even know what it means for ZF to "contain a calculation." The concept of calculation (at least the one that I'm familiar with) comes from computability theory, not number theory.
"finite" means what you think it means: a program that halts.
Look at Peano Axioms:
https://en.m.wikipedia.org/wiki/Peano_axioms
The Induction axiom allows us to write finite proofs of otherwise infinite facts, like every positive integer has a predecessor.
Without it, you'd be stuck with finitism (since only a finite number of numbers can be constructed in finite time, there is only a finite number of numbers.)
I think you're confusing PA with computability theory and Turing machines. "Program" is not a thing in PA.
Probability is relative to an axiom systems.
A more powerful system can prove statements that are true but not provable in a weaker system.
For example "This sentence is not provable." (Godel's statement, written informally) is true, but not provable, in first order arithmetic logic.
https://en.m.wikipedia.org/wiki/Diagonal_lemma
In constructivist mathematics, proving that a statement is not false is not a proof that it is true. Non-constructivist mathematics, on the other hand, has the axiom, for all P: P or not-P, also stated as not-not-P implies P.
The whole point of provability is that it is purely syntactic process that could be verified in finite time. Ideally it would be the same as truth, but there are some caveats.
Not according to constructivists.
I’ve heard this argument:
If RH is unprovable one way or another, then certainly no counterexample can exist to the RH otherwise you could find it and prove the RH to be false.
Hence if RH is unprovable, it must be true. I suppose this uses logic outside the logical system that RH operates in.