Question

Consider the first order predicate formula ϕ:

∀x [(∀z z|x ⇒ ((z = x) ∨ (z = 1))) ⇒ ∃w (w > x) ∧ (∀z z|w ⇒ ((w = z) ∨ (z = 1)))]

Here ‘a|b’ denotes that ‘a divides b’, where a and b are integers. Consider the following

sets:

S1. {1,2,3, ... , 100}

S2. Set of all positive integers

S3. Set of all integers

Which of the above sets satisfy φ?

1. S1 and S2
2. S1 and S3
3. S2 and S3
4. S1, S2 and S3

Answer 1

Correct Answer - Option 3 : S2 and S3

∀x [(∀z z|x ⇒ ((z=x) ∨ (z=1))) ⇒ ∃w (w > x) ∧ (∀z z|w ⇒((w=z) V (z=1)))]

Let, X ≡ (∀z z|x ⇒ ((z =x) ∨ (z=1)))

Y ≡ ∃w (w > x)

Z ≡ (∀z z|w ⇒ ((w = z) ∨ (z=1))

∀x [X ⇒ Y ⇒ Z]

X is true: If x is prime number

Y is true: If there exist a w which is greater than x

Z is true. If w is prime

S1 = {1,2,3, ... , 100}

Let x = 97

z = 97

∴ z|x ⇒ (z = 1) ≡ T ⇒ T ≡ T

prime number greater than 97 is 101

But S1 is not included in set. Hence there doesn’t exist w which is greater than x

T ⇒ F.Z ≡ T ⇒ F ≡ F

Therefore, S1 does not satisfy φ

S2. Set of all positive integers satisfy φ

S3. Set of all integers

If x is negative, X ≡ T

F ⇒ Y.Z ≡ T

Also, set of all positive integers satisfy φ

Therefore, option 3 is correct.