Question

Is the statement $\exists x\forall y\left[x+y=y\right]$a member of Th$\left(N,+\right)$? Why or why not? What about the statement $\exists x\forall y\left[x+y=x\right]$?

Short answer

The statement is true.

Step-by-step solution

Turing Machine

A Turing Machine is computational model concept that runs on the unrestricted grammar of Type-zero. It accepts recursive enumerable language. It comprises of an infinite tape length where read and write operation can be perform accordingly.

The given statement is true.

The statement $\text{'}\exists x\forall y\left[x+y=y\right]amemberofTh\left(N,+\right)\text{'}$ is TRUE.

This is because of the fact that for any set of standard value in universal set N over operator $\left(+\right)$.

Because if we use value N=$\left\{0,1,2,.......\right\}$then we will get x as 0 according to above statement.

Now, statement$\text{'}\exists x\forall y\left[x+y=x\right]\text{'}$ is not a member of Th$\left(N,+\right)$ .

This is because this statement will fail in the case if we set y=1 , then for any value of x, the statement ,$\exists x\forall y\left[x+y=x\right]\text{'}$will not hold true.