UGC NET Computer Science December 2019 | Question 109

Question 109

---

Consider the following statements:
S1: ∀x P(x) v ∀x Q(x) and ∀x (P(x) v Q(x)) are not logically equivalent.
S2: ∃x P(x) ∧ ∃x Q(x) and ∃x (P(x) ∧ Q(x)) are not logically equivalent
Which of the following statements is/are correct?

1. 1. Only S1
2. 2. Only S2
3. 3. Both S1 and S2
4. 4. Neither S1 nor S2

Answer : 3. Both S1 and S2

Explanation Question 109

Here we are considering the example to prove that given statements are true:

Let P(x) be the statement that x is an even number and
Let Q(x) be the statement that x is an odd number,
where x denotes the domain of all positive integers.

∴ D = N = natural numbers and
P(x) = "x is even",
Q(x) = "x is odd".

First prove that statement, S1 is true → ∀x P(x) v ∀x Q(x) and ∀x (P(x) v Q(x)) are not logically equivalent.
Because here, All positive numbers are not even. So, (∀x P(x)) is false
and all positive numbers are not odd. So, (∀x Q(x)) is false
So, ∀x P(x) v ∀x Q(x) is false
But ∀x (P(x) ∨ Q(x)) will always be true, because x will always be either even or odd.
Hence, ∀x P(x) v ∀x Q(x) and ∀x (P(x) v Q(x)) are not logically equivalent.

Now prove that statement, S2 is true → ∃x P(x) ∧ ∃x Q(x) and ∃x (P(x) ∧ Q(x)) are not logically equivalent
Because here, there exist some positive number that is even. So, ∃x P(x) is true
and there exist some positive number that is odd. So, ∃x Q(x) is true
So, ∃x P(x) ∧ ∃x Q(x) is true
But ∃x (P(x) ∧ Q(x)) will always be false, because x will always be either even or odd.
So, ∃x P(x) ∧ ∃x Q(x) and ∃x (P(x) ∧ Q(x)) are not logically equivalent

Reference 1: Show that ∀xP (x) ∨ ∀xQ(x) and ∀x(P (x) ∨ Q(x)) are not logically equivalent.

Reference 2: Equivalence using quantifier - section 1.4 #45 s

So, option 3 is correct answer

Previous	Next
UGC NET CS December 2019 - Question 108	UGC NET CS December 2019 - Question 110