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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?

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? Correct Answer S2 and S3

∀x

Let, X ≡ (∀z z|x ⇒ ((z =x) ∨ (z=1)))

Y ≡ ∃w (w > x)

Z ≡ (∀z z|w ⇒ ((w = z) ∨ (z=1))

∀x

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.

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