GATE CSE 2006 | Mathematical Logic Question 45 | Discrete Mathematics

GATE CSE 2006

MCQ (Single Correct Answer)

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A logical binary relation $$ \odot $$, is defined as follows: GATE CSE 2006 Discrete Mathematics - Mathematical Logic Question 42 English

Let ~ be the unary negation (NOT) operator, with higher precedence then $$ \odot $$. Which one of the following is equivalent to $$A \wedge B?$$

$$\left( { \sim A \odot B} \right)$$

$$\left( { \sim A \odot \sim B} \right)$$

$$ \sim \left( { \sim A \odot \sim B} \right)$$

$$ \sim \left( { \sim A \odot B} \right)$$

GATE CSE 2005

MCQ (Single Correct Answer)

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Let $$P, Q$$ and $$R$$ be three atomic prepositional assertions. Let $$X$$ denotes $$\left( {P \vee Q} \right) \to R$$ and $$Y$$ denote $$\left( {P \to R} \right) \vee \left( {Q \to R} \right)$$.

Which one of the following is a tautology?

$$X \equiv Y$$

$$X \to Y$$

$$Y \to X$$

$$\neg Y \to X$$

GATE CSE 2005

MCQ (Single Correct Answer)

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What is the first order predicate calculus statement equivalent to the following?
Every teacher is liked by some student

$$\forall \left( x \right)\left[ {teacher\left( x \right) \to \exists \left( y \right)\left[ {student\left( y \right) \to likes\left( {y,\,x} \right)} \right]} \right]$$

$$\forall \left( x \right)\left[ {teacher\left( x \right) \to \exists \left( y \right)\left[ {student\left( y \right) \wedge likes\left( {y,\,x} \right)} \right]} \right]$$

$$\exists \left( y \right)\forall \left( x \right)\left[ {teacher\left( x \right) \to \left[ {student\left( y \right) \wedge likes\left( {y,x} \right)} \right]} \right]$$

$$\forall \left( x \right)\left[ {teacher\left( x \right) \wedge \exists \left( y \right)\left[ {student\left( y \right) \to likes\left( {y,\,x} \right)} \right]} \right]$$

GATE CSE 2005

MCQ (Single Correct Answer)

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Let $$P(x)$$ and $$Q(x)$$ be arbitrary predicates. Which of the following statement is always TRUE?

$$\left( {\forall x\left( {P\left( x \right) \vee Q\left( x \right)} \right)} \right) \Rightarrow \left( {\left( {\forall xP\left( x \right)} \right) \vee \left( {\forall xQ\left( x \right)} \right)} \right)$$

$$\left( {\forall x\left( {P\left( x \right) \Rightarrow Q\left( x \right)} \right)} \right) \Rightarrow \left( {\left( {\forall xP\left( x \right)} \right) \Rightarrow \left( {\forall xQ\left( x \right)} \right)} \right)$$

$$\left( {\left( {\forall x\left( {P\left( x \right)} \right) \Rightarrow \left( {\forall xQ\left( x \right)} \right)} \right) \Rightarrow \left( {\forall x\left( {P\left( x \right) \Rightarrow Q\left( x \right)} \right)} \right)} \right)$$

$$\left( {\left( {\forall x\left( {P\left( x \right)} \right)} \right)} \right) \Leftrightarrow \left( {\forall x\left( {Q\left( x \right)} \right)} \right) \Rightarrow \left( {\forall x\left( {P\left( x \right) \Leftrightarrow Q\left( x \right)} \right)} \right)$$

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