Which one of the following is NOT logically equivalent to eg ∃ x (∀ y ( ) ∀ z ( ))? A) ∀ x (∃ z ( eg ) → ∀ y ( )) B) ∀ x (∀ z ( ) → ∃ y ( eg )) C) ∀ x (∀ y ( ) → ∃ z ( eg )) D) ∀ x (∃ y ( eg ) → ∃ z ( eg ))

Official answer: See verified solution. Explanation: Let us simplify the given statement: P = eg ∃ x (∀ y( ) ∀ z( )) Using the quantifier negation law $ eg ∃ x ∀ x eg$: P ∀ x eg (∀ y( ) ∀ z( )) Applying De Morgan's Law $ eg(A B) eg A eg B$: P ∀ x ( eg ∀ y( ) eg ∀ z( )) ∀ x (∃ y( eg ) ∃ z( eg )) Now, let us test the options using $p → q eg p q$: - Option (B): $∀ x (∀ z( ) → ∃ y( eg )) ∀ x ( eg ∀ z( ) ∃ y( eg )) ∀ x (∃ z( eg ) ∃ y( eg ))$ (Equivalent) - Option (C): $∀ x (∀ y( ) → ∃ z( eg )) ∀ x ( eg ∀ y( ) ∃ z( eg )) ∀ x (∃ y( eg ) ∃ z( eg ))$ (Equivalent) - Option (A): $∀ x (∃ z( eg ) → ∀ y( )) ∀ x ( eg ∃ z( eg ) ∀ y( )) ∀ x (∀ z( ) ∀ y( ))$ (Not Equivalent) - Option (D): $∀ x (∃ y( eg ) → ∃ z( eg )) ∀ x ( eg ∃ y( eg ) ∃ z( eg )) ∀ x (∀ y( ) ∃ z( eg ))$ (Not Equivalent) Since both (A) and (D) are NOT logically equivalent to the given expression, there are two correct options. Hence, the official answer key awarded full marks to all candidates (Marks to All).

A) ∀ x (∃ z ( eg ) → ∀ y ( ))

B) ∀ x (∀ z ( ) → ∃ y ( eg ))

C) ∀ x (∀ y ( ) → ∃ z ( eg ))

D) ∀ x (∃ y ( eg ) → ∃ z ( eg ))

Propositional and First-Order Logic

GATE CSE 2013 · Q47

MCQ2 marksMedium

Time00:00

Which one of the following is NOT logically equivalent to $eg \exists x (\forall y (\alpha) \wedge \forall z (\beta))$ ?

Select one answerSingle correct

Pick an option to check

Solution

Step-by-step verified solution

Official answer: See verified solution

Propositional and First-Order Logic

Let us simplify the given statement: $P = eg \exists x (\forall y(\alpha) \wedge \forall z(\beta))$

Using the quantifier negation law $eg \exists x \equiv \forall x eg$ : $P \equiv \forall x eg (\forall y(\alpha) \wedge \forall z(\beta))$

Applying De Morgan's Law $eg(A \wedge B) \equiv eg A \vee eg B$ : $P \equiv \forall x ( eg \forall y(\alpha) \vee eg \forall z(\beta)) \equiv \forall x (\exists y( eg \alpha) \vee \exists z( eg \beta))$

Now, let us test the options using $p \to q \equiv eg p \vee q$ :

Option (B): $\forall x (\forall z(\beta) \to \exists y(

eg \alpha)) \equiv \forall x ( eg \forall z(\beta) \vee \exists y( eg \alpha)) \equiv \forall x (\exists z( eg \beta) \vee \exists y( eg \alpha))$ (Equivalent)

Option (C): $\forall x (\forall y(\alpha) \to \exists z(

eg \beta)) \equiv \forall x ( eg \forall y(\alpha) \vee \exists z( eg \beta)) \equiv \forall x (\exists y( eg \alpha) \vee \exists z( eg \beta))$ (Equivalent)

Option (A): $\forall x (\exists z(

eg \beta) \to \forall y(\alpha)) \equiv \forall x ( eg \exists z( eg \beta) \vee \forall y(\alpha)) \equiv \forall x (\forall z(\beta) \vee \forall y(\alpha))$ (Not Equivalent)

Option (D): $\forall x (\exists y(

eg \alpha) \to \exists z( eg \beta)) \equiv \forall x ( eg \exists y( eg \alpha) \vee \exists z( eg \beta)) \equiv \forall x (\forall y(\alpha) \vee \exists z( eg \beta))$ (Not Equivalent)

Since both (A) and (D) are NOT logically equivalent to the given expression, there are two correct options. Hence, the official answer key awarded full marks to all candidates (Marks to All).

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This Propositional and First-Order Logic previous year question is Q47 of the GATE CSE 2013 paper. Find more GATE CSE PYQs on Propositional and First-Order Logic in Engineering Mathematics, or browse all GATE CSE PYQs.

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Propositional and First-Order Logic in GATE CSE

Propositional and First-Order Logic has 18 GATE CSE PYQs across 16 papers from 2011 to 2026. 7 of them are 2-mark MCQs like this one, and it is the only one in GATE CSE 2013.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.