GATE CSE 2014 Set 2 | Mathematical Logic Question 19 | Discrete Mathematics

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GATE CSE 2014 Set 2

MCQ (Single Correct Answer)

-0.6

Which one of the following Boolean expressions is NOT A tautology?

$$\left( {\left( {a \to b} \right) \wedge \left( {b \to c} \right)} \right) \to \left( {a \to c} \right)$$

$$\left( {a \leftrightarrow c} \right) \to \left( { \sim b \to \left( {a \wedge c} \right)} \right)$$

$$\left( {a \wedge b \wedge c} \right) \to \left( {c \vee a} \right)$$

$$A \to \left( {b \to a} \right)$$

GATE CSE 2014 Set 1

MCQ (Single Correct Answer)

-0.6

Which one of the following propositional logic formulas is TRUE when exactly two of $$p, q,$$ and $$r$$ are TRUE?

$$\left( {\left( {p \leftrightarrow q} \right) \wedge r} \right) \vee \left( {p \wedge q \wedge \sim r} \right)$$

$$\left( { \sim \left( {p \leftrightarrow q} \right) \wedge r} \right) \vee \left( {p \wedge q \wedge \sim r} \right)$$

$$\left( {\left( {p \to q} \right) \wedge r} \right) \vee \left( {p \wedge q \wedge \sim r} \right)$$

$$\left( { \sim \left( {p \leftrightarrow q} \right) \wedge r} \right) \wedge \left( {p \wedge q \wedge \sim r} \right)$$

GATE CSE 2014 Set 3

MCQ (Single Correct Answer)

-0.6

The CORRECT formula for the sentence, "not all rainy days are cold" is

$$\forall d\left( {Rainy\left( d \right) \wedge \sim Cold\left( d \right)} \right)$$

$$\forall d\left( { \sim Rainy\left( d \right) \to Cold\left( d \right)} \right)$$

$$\exists d\left( { \sim Rainy\left( d \right) \to Cold\left( d \right)} \right)$$

$$\exists d\left( {Rainy\left( d \right) \wedge \sim Cold\left( d \right)} \right)$$

GATE CSE 2013

MCQ (More than One Correct Answer)

-0

Which one of the following is NOT logically equivalent to $$\neg \exists x\left( {\forall y\left( \alpha \right) \wedge \left( {\forall z\left( \beta \right)} \right)} \right)?$$

$$\forall x\left( {\exists z\left( {\neg \beta } \right) \to \forall y\left( \alpha \right)} \right)$$

$$\forall x\left( {\forall z\left( \beta \right) \to \exists y\left( {\neg \alpha } \right)} \right)$$

$$\forall x\left( {\forall y\left( \alpha \right) \to \exists z\left( {\neg \beta } \right)} \right)$$

$$\forall x\left( {\exists y\left( {\neg \alpha } \right) \to \exists z\left( {\neg \beta } \right)} \right)$$

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