Chapter 0: Q5E (page 1)

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

Short Answer

Expert verified

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 $' \exists x \forall y [x + y = y] a m e m b e r o f T h (N, +)'$ is TRUE.

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

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

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

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

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

Short Answer

Step by step solution

Turing Machine

The given statement is true.

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Is the statement ∃x∀yx+y=ya member of ThN,+? Why or why not? What about the statement ∃x∀yx+y=x?

Short Answer

Step by step solution

Turing Machine

The given statement is true.

One App. One Place for Learning.

Most popular questions from this chapter

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