Chapter 0: Q9P (page 1)

Use the recursion theorem to give an alternative proof of Rice’s theorem in Problem 5.28.

Short Answer

Expert verified

Answer:

Rice’s Theorem is proved.

Step by step solution

Rice Theorem

Let’s first recall the Rice’s Theorem which was in Problem 5.28.

Let be a language consisting of Turing machine descriptions, then P must fulfil the following conditions:

P is nontrivial—it contains some, but not all, TM descriptions.

P is a property of the TM’s language—whenever we have

$< M 1 > \in P i f f < M_{2} > \in P .$ Here, $M_{1} a n d M_{2} a r e a n y T M s .$

Proving rice Theorem

Now, let us assume as Turing Machine that is decider of P and P satisfy both the conditions of Rice’s Theorem which is mentioned above. The second condition says that there are two Turing Machine M₁ and M₂ , where M₁ $\in$ P and M₂ $€$ P.

Now we will use M₁ and M₂ to construct a Turing Machine R as follows:

R=on input(w)

Obtain own description using recursion theorem.
Run X on <R>
If X accepts <R> , run M₂ on w

Else

If rejects , run M₁ on w

<R> $\in$ P, then X will accepts =<R> and L(R) = L( $M_{2}$ )

But since, <B> $€$ P so this makes as contradict statement. This is due to the fact that P will agree on Turing Machines that have same language.

Also, <R> $€$ P can also be conclude from above. This shows our assumption was wrong.

So every property which satisfies Rice’s Theorem are undecidable.

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Use the recursion theorem to give an alternative proof of Rice’s theorem in Problem 5.28.

Short Answer

Step by step solution

Rice Theorem

Proving rice Theorem

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