Question

Prove that a language A is decidable if and only if it is finite or there is a computable function f : N → {0,1)' such that range(f) = A and each f(n) comes strictly before f(n + 1) in the standard enumeration of 10, 1*.

Answer 1

If and only if A language is computably enumerable in increasing order then this language will be decidable. That is, a language A is decidable if and only if it is finite or there is a total computable function f: N → {0,1}* for all numbers n,

Such that range f=(A) and each f(n) comes strictly before f(n+1) in the standard enumeration of {0,1}* i.e.

f(n) < f(n + 1)

If A is finite, then it will be decidable.

Let A is not finite, and a function f: N → {0,1}* and

for all n, f(n) < f(n + 1).

So, we can present formulation of a Turing Machine that decides A, given below:-

STEP-1: TAKE input X

STEP-2: Initialize I to 0.

STEP-3: while f(I) < X do

STEP-4: Increment the value of I by 1

STEP-5: END while

STEP-6: if f(I) == X then

STEP-7: Take the value of X and HALT

STEP-8: else refuse the value of X and HALT

Vice-Versa, Let A is decidable. If A is finite, then it is computably enumerable by definition mentioned above. If A is infinite, then there is a computable function f: N → {0,1}* that forms enumeration of A in the increasing order.

So there is a computable function f that enumerates A ,

it means A is computably enumerable and by all observation given above we can say that range f=(A) and each f(n) comes strictly before f(n+1) in the standard enumeration of {0,1}* i.e.

f(n) < f(n + 1)