GATE CSE 2014 Set 1 | Mathematical Logic Question 20 | Discrete Mathematics

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

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

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

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

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

GATE CSE 2013

MCQ (Single Correct Answer)

-0.6

What is the logical translation of the following statement?
"None of my friends are perfect."

$$\exists x\left( {F\left( x \right) \wedge \neg P\left( x \right)} \right)$$

$$\exists x\left( {\neg F\left( x \right) \wedge P\left( x \right)} \right)$$

$$\exists x\left( {\neg F\left( x \right) \wedge \neg P\left( x \right)} \right)$$

$$\neg \exists x\left( {F\left( x \right) \wedge P\left( x \right)} \right)$$

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