Can an unprovable statement be interpreted as being generally true in some cases?

Question

For example, let's say that Goldbach's conjecture turns out to be unprovable. This would mean that a program cannot devise a way to check whether any counterexample exists. This seems to mean that there in fact is no counterexample - as we can develop a program that checks until some particular number, then if it stops, then write a program that starts from that number to some number and so on. This is all finite. If an even integer that cannot be represented by the sum of two primes is found, then the process would be finite.

This all seems for me to imply that a theorem is in fact true - but I am not sure if this is right reasoning.

Edit:

My intention when I wrote "unprovable" was independent. That's my mistake.

OK. So, let's say Goldbach's conjecture is independent of PA, but we prove Goldbach's conjecture in a stronger (informally, more powerful) theory (I guess there are many...). This would mean (correct me if I am wrong here) that we select one model of PA, let's say standard one, and then prove that the conjecture is true in that model using a more powerful theory.

So as Asaf says, ZFC has no one standard model - so is that the reason why consistency of continuum hypothesis cannot be settled, while other theorems that relate to PA technically can be settled as true or not in particular model?

Edit 2:

I know that continuum hypothesis can be settled in an extension of ZFC (as either true or false for a particular model of ZFC, and it would depend on different kinds of extension) - so a better rephrase of the edit above would be: can there be any statement of PA that cannot be proven true or false for a particular model of PA using a more powerful theory? And can anyone tell me more about

Answer 1

There are two reasons why the Goldbach conjecture could turn out to be unprovable: either it is false or it is independent.

If it is false, then there is some counterexample somewhere that we may or may not find someday. This isn't particularly interesting.

If it is independent, then it is not provable that a counterexample exists (this is what you were saying in the question, but for this you need the negation of the Goldbach conjecture to be unprovable). What does this mean? Well, if the conjecture was false, then there would be some counterexample, some large integer not expressible as a sum of two primes. But then we could (in principle) write a proof that this integer is a counterexample, going through all of the pairs of primes below it and adding them. There is a finite number of cases to check, so such a proof would exist. But we were assuming that it is not provable that a counterexample exists. Therefore the conjecture must in fact be true!

Now, this argument is very dependent on this particular context. First of all, we have a standard model of PA, where we can (by default) interpret things as true or false. By the completeness theorem, if the Goldbach conjecture were independent, it would be true in some model of PA, but that model might not be the standard model.

Secondly, it is important that the Goldbach conjecture is a very simple statement. It has a single quantifier (it is what is called a $\Pi^0_1$ statement). As soon as we would have to deal with more quantifiers (for example, checking that "for every foo some bar exists such that baz") we wouldn't be able to convert the existence of an arbitrary counterexample into a proof as we did here.

Answer 2

Yes.

When there is a standard interpretation of a theory, for example the natural numbers have a standard model. Then if something is independent of the theory, but true in the standard model then it is often regarded as true.

Good examples are the Paris-Harrington theorem and the Goodstein theorem, which are true in $\Bbb N$ but the first-order theory of $\sf PA$ cannot prove them.

It should be remarked that not all mathematical theories have standard models, or even "intended interpretation". There is no such model of $\sf ZFC$, so in a very deep sense the continuum hypothesis is independent from $\sf ZFC$. (In a fit of nomenclature confusion, there is a concept of standard models of $\sf ZFC$, but this is not the same things as the standard model of $\sf PA$.)