GATE | GATE CS 2013 | Question 47

**(A)** A

**(B)** B

**(C)** C

**(D)** D

**Answer:** **(A)** **(D)** **Explanation:**

Given statement is :

**¬ ∃ x ( ∀y(α) ∧ ∀z(β) ) **

where ¬ is a negation operator, ∃ is Existential Quantifier with the meaning of"there Exists", and ∀ is a Universal Quantifier with the meaning" for all ", and α, β can be treated as predicates. here we can apply some of the standard results of Propositional and 1st order logic on the given statement, which are as follows : [Result 1: ¬(∀x P(x)) <=> ∃ x¬P(x), i.e. negation of "for all" gives "there exists" and negation also gets applied to scope of quantifier, which is P(x) here. And also negation of "there exists" gives "for all", and negation also gets applied to scope of quantifier ] [Result 2: ¬ ( A ∧ B ) = ( ¬A ∨ ¬B ) ] [Result 3: ¬P ∨ Q <=> P -> Q ] [Result 4: If P ->Q, then by Result of Contrapositive, ¬Q -> ¬P ]

Now we need to use these results as shown below:

¬ ∃ x ( ∀y(α) ∧ ∀z(β) ) [ Given ] => ∀ x (¬∀y(α) ∨ ¬∀z(β) ) [ after applyingResult 1&Result 2] => ∀ x ( ∀y(α) -> ¬∀z(β) ) [after applyingResult 3] =>∀ x ( ∀y(α) -> ∃z(¬β) )[after applyingResult 1] which is same as the statement C. Hence the Given Statement is logically Equivalent to the statement C. Now, we can also prove that given statement is logically equivalent to the statement in option B. Let's see how ! The above derived statement is : ∀ x ( ∀y(α) -> ∃z(¬β) ) Now this statement can be written as (or equivalent to) :=> ∀ x ( ∀z(β) -> ∃y(¬α) )[after applyingResult 4] And this statement is same as statement B. Hence the Given statement is also logically equivalent to the statement B. So, we can conclude that the Given statement isNOTlogically equivalent to the statementsAandD.

Hence, the correct answer is Option A and Option D. But in GATE 2013, marks

were given to all for this question.

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