@Arjun Sir, Is **L' = {<M>| L(M) = ϕ} **correct. I mean complement of everything as nothing!

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+6 votes

$L= \{\langle M \rangle\mid L(M) = \Sigma^*\} $

A. $L$ is RE but $L'$ is not RE

B. Both $L$ and $L'$ are RE

C. $L$ is not RE but $L'$ is RE

D. Both $L$ and $L'$ are not RE

A. $L$ is RE but $L'$ is not RE

B. Both $L$ and $L'$ are RE

C. $L$ is not RE but $L'$ is RE

D. Both $L$ and $L'$ are not RE

+14 votes

Best answer

Yes. Both L and L' are not RE. We can have the same reduction as done for L(M) is infinite.

Lets assume L is RE. So, we have a TM N which will say "yes" if given an encoding of a TM whose language is $\Sigma ^*$. Now using N we try to solve non-halting problem as follows:

A non halting problem is given as follows: Given an encoding of TM <H> and a word w, whether H does not halt on w. That is, we have to decide if H does not halt on w. This problem is proved to be not RE and so no TM can not even say "yes" for "yes" case of the problem.

Now, given a halting problem (<H>, w), we proceed as follows: We make a new TM say A, which on input x, just simulates H on w for |x| steps. If H halts on w, A goes to reject state. Otherwise it accepts. So, $L(A) =\Sigma^*$ if H does not halt on w and L(A) = a finite set if H halts on w. (If H halts in |x| steps for w, any string with length greater than |w|, would certainly be not in L, making L a finite set and hence can never equal $\Sigma^*$).

Now, we just give the encoding of A to our assumed TM N. If N says "accept", we have that L(A) is $\Sigma^*$ => H does not halt on w => we solved "yes" case of not halting problem, which is not possible. Hence, our initial assumption of L is RE is false. L is not RE.

Proving L' is not RE is easy.

L' = {<M> | L(M) is not equal to $\Sigma^*$}

L(M) is not equal to $\Sigma^*$ is a non-monotonic property of TM. Because we can take TM $T_{yes}$ with $L(T_{yes})=\phi$ and $T_{no}$ with $L(T_{no})=Σ^*$. Here, $T_{yes}$ satisfies the property (of being not being equal to $\Sigma^*$) and $T_{no}$ does not satisfy that property and $L(T_{yes})⊂L(T_{no})$, making this a non-monotonic property of TM. And hence this becomes not RE as per Rice's theorem second part.

So, both L and L' are not RE making (D) the correct answer.

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