1

I have been asked to prove if the following set is decidible, semi-decidible or not semi-decidible:

In other words, it is the set of inputs such that exists a Turing Machine encoded with the natural y with input p that returns its input.

Consider the set K as the set of naturals such that the Turing machine encoded with x and input x stops. This is demonstrated to be a non-decidible set.

I think that what I need is to find a reduction of K to L, but I don't know how to prove that L is decidible, semi-decidible or not semi-decidible.

Zoe
  • 27,060
  • 21
  • 118
  • 148
Mick
  • 41
  • 3
  • Is the following restatement of your desired set accurate? "The set of all natural numbers {n1, n2, …} encoding Turing machines {m1, m2, …} which eventually halt (or halt-accept only and not halt-reject) and which, after halting, have the same tape contents they were given as input." So the TM that immediately halt-accepts is an example of the kind of TMs your numbers encode, but the TM that changes the first input symbol and then halt-accepts is not? – Patrick87 Apr 29 '19 at 15:51

1 Answers1

1

L may not look decidable at first glance, because there is this nasty unbounded quantifier included, which seems to make necessary a possibly infinite search when you look for a y satisfying the condition for a specific p.

However, the answer is much simpler: There is a turing machine M which always returns its input, i.e. M(p) = p holds for all p in the considered language. Let y be a code of M. Then you can use this same y for all p, showing that L contains all words of the language. Hence L is of course decidable.

In fact, this is an example to demonstrate the principle of extensionality (if two sets have the same elements and one is decidable, then the other is decidable too, even if it doesn't look so).

Bilbo
  • 63
  • 1
  • 7