[layer8]

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.

[najdan33]

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.

[layer8]

> 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 I'm saying is that if each individual problem has a solution in the form of an individual finite proof (the premise of your second point), then the above gives you a general algorithm for finding the proof for any given of those individual problems (contradicting the premise of your first point).

What this doesn't give you is a way to prove that a proof exists for all individual prohlems, because that indeed would take infinite time.

Ykonstant is correct in that your use of "undecidable" was confused; I just ignored that.

[najdan33]

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..