Question

Question: Let $\mathbf{\text{S=}}\left\{\left\{M\right\}|MisaTMandL\left(M\right)=\left\{\{M\text{'}\right\}\right\}\mathbf{.}$Show that S nor S^' neither is Turing recognizable.

Short answer

Answer:

The language $S=\left\{\left\{M\right\}|MisaTMandL\left(M\right)=\left\{\{M\text{'}\right\}\right\}.$for S nor S ^'is turing recognizable is proved.

Step-by-step solution

Step 1: Turing machine.

A Turing Machine (TM) is a mathematical model which consists of an infinite length tape divided into cells on which input is given. It consists of a head which reads the input tape. A state register stores the state of the Turing machine.

Step 2:

For this problem, let’s assume that a Turing machine can recognize its own code one . Where the language should be recognizable by Turing machine for this instance it is must to show that showing the variables contains in the language which is evalualated by the grammar for Turing machine that is $A_{TM}{⩽}_{M}Sand{A}_{TM}{⩽}_M{S}^{\text{'}}$ , which also imply $A_{TM}{⩽}_{M}Sand{A}_{TM}{⩽}_M{S}^{\text{'}}$, respectively.

Decidable languages are closed under complementation. To design a machine M for the complement of a language, and simulate the machine for language on an input.

If it accepts then accept and vice versa.Decidable languages are closed under inverse homeomorphisms.

For the first step give the reduction from $A_{TM}toS$.

And given an instance $\left\{M,w\right\}$of $A_{TM}$, we construct a machine which given an input x, rejects if$\left\{x\ne M\text{'}\right\}$ and simulates M on w if $x=\left(M\right)i.$

Thus, if$L\left(M\text{'}\right)=\left\{\left\{M\text{'}\right\}\right\}$ accepts w and∅otherwise. Similarly, for the reduction from ATM to S, we make M0 accept if x = hM0 i and simulate on x otherwise.

In this case, it gives$L\left(M\text{'}\right)=\Sigma *$ if M accepts w and .

Hence, here the language $S=\left\{\left\{M\right\}|MisaTMandL\left(M\right)=\left\{\{M\text{'}\right\}\right\}.$for is turing recognizable is proved.