I don’t think so, at least if you assume that each concrete solution can be expressed in finite length in a formal language with a finite alphabet, and can be mechanically checked (which is generally the case for mathematical proofs). Because then you could just enumerate all strings of that language until you find one that describes the solution to the given problem, which by your second item would be guaranteed to exist, and thus the procedure be guaranteed to terminate, contradicting your first item.

Thanks for the answer, it helps piece together the puzzle. I believe there's a problem with this reasoning:

> each concrete solution can be expressed in finite length in a formal language with a finite alphabet

Suppose that's the case, the problem is that the resulting language of all the finite proofs can still be infinite and we again cannot enumerate and check all the solutions (since the final set is infinite). Therefore we're short of a method to decide yes/no for each question.

This appears to be exactly the case in the example provided by @ykonstant

> Consider the sequence of yes/no problems P_K = {Is there a solution to Q=0 in [-K,K]^n?} parametrized by a positive integer K.

For each yes/no question, the workload is finite, but for the union of all yes/no questions, the workload is not finite.

> Suppose that's the case, the problem is that the resulting language of all the finite proofs can still be infinite and we again cannot enumerate and check all the solutions (since the final set is infinite).

The set of all strings in the language is only countably infinite (= same size as tne natural numbers), which means you will reach the solution string after a finite time (just count 1, 2, 3, ... until you reach the corresponding natural number).

What @ykonstant saying is something different: They want to solve all infinitely many single problems. That may not be possible to do in finite time. But each individual one may be solvable in finite time, which is the question you were asking.

You are right, I was wrong. If there's a way to evaluate each single yes/no question in a finite number of steps, then the set of all problems is decidable, since any question can be resolved in a finite number of steps.

That would seem to imply that each problem (seen as a set of sub-problems) contains some undecidable sub-problems.. so is there something like the most primitive undecidable problem..

The way you are phrasing your question is confusing; without the parenthesis, your two statements are identical. The only meaningful way to interpret "The whole thing is undecidable" is whether you can or cannot decide all (infinitely many) statements at once.

If that is what you mean, then yes: take any undecidable Diophantine equation Q=0 of, say, n variables. Consider the sequence of yes/no problems P_K = {Is there a solution to Q=0 in [-K,K]^n?} parametrized by a positive integer K. Each of those problems is decidable in isolation, but the totality cannot be, since that would decide if Q=0 has a solution or not.

Thank you for suggesting this example.

For the undecidable Diophantine equation Q=0 of n variables - I'm assuming that the fact that someone could hit a solution randomly (by randomly generating the n variables and getting lucky) does not contradict with its undecidability.

Still I'm not clear what happens if such a solution would be randomly hit for an equation that's mapped to another hard question such as ZFC consistency. It would imply that a question such as ZFC consistency could be solved by randomly generating an answer, which seemingly doesn't make sense?