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Apr 6, 2013 at 20:44 comment added user6976 The point of my previous request is: if you are looking for an undecidable mass problem (a set of inputs and a set of admissible inputs, the problem is when an input is admissible), then you need to look for an algorithm (in your case - an algorithm to find an algorithm). If your problem is not a mass problem, you are looking for a non-provable statement which is a completely different ball game (independence from ZFC and such things). Please clarify.
Apr 6, 2013 at 20:17 comment added user6976 @Loïc: It may help if you formulate your question in a most standard way, as a "mass problem". What exactly is the problem which you want undecidable, i.e. what is the input set and when an input should be recognized?
Apr 6, 2013 at 20:12 comment added user6976 I see. The new formulation is much less vague. I (or somebody else) need to think more about it.
Apr 6, 2013 at 14:19 comment added Loïc Teyssier Thanks for your answer! Yet I don't feel it relates fully to my admitedly vague quetion, although its nature pleases me better than other answers do. Maybe my edit clarifies what I meant.
Apr 5, 2013 at 18:43 history edited user6976 CC BY-SA 3.0
fixed an error; added 154 characters in body
Apr 5, 2013 at 18:39 comment added user6976 Yes, I was not careful enough. I will change the answer.
Apr 5, 2013 at 18:17 comment added Benjamin Steinberg @Mark, I believe Slobodoskii's theorem doesn't say exactly what you want. It says you cannot decide give a finite presentation <X|R> whether a word in X is 1 in every X-generated finite group satisfying the relations R. Note <X|R> can present an infinite group. If you know a finite presentation presents a finite group, you can solve the yes part by enumerating all consequences of R and the no part enumerating finite groups.
Apr 5, 2013 at 14:52 history answered user6976 CC BY-SA 3.0