GATE CSE 2006 | Mathematical Logic Question 43 | Discrete Mathematics

GATE CSE 2006

MCQ (Single Correct Answer)

-0.6

Let $$P, Q$$, and $$R$$ be sets. Let $$\Delta $$ denote the symmetric difference operator defined as $$P\Delta Q = \left( {P \cup Q} \right) - \left( {P \cap Q} \right)$$. Using venn diagrams, determine which of the following is/are TRUE.

($${\rm I}$$) $$P\Delta \left( {Q \cap R} \right) = \left( {P\Delta Q} \right) \cap \left( {P\Delta R} \right)$$
($${\rm I}{\rm I}$$) $$P \cap \left( {Q\Delta R} \right) = \left( {P \cap Q} \right)\Delta \left( {P \cap R} \right)$$

$${\rm I}$$ only

$${\rm I}$$$${\rm I}$$ only

Neither $${\rm I}$$ nor $${\rm I}$$$${\rm I}$$

Both $${\rm I}$$ and $${\rm I}$$$${\rm I}$$

GATE CSE 2006

MCQ (Single Correct Answer)

-0.6

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 2006

MCQ (Single Correct Answer)

-0.6

Which one of the first order predicate calculus statements given below correctly expresses the following English statement?

Tigers and lion attack if they are hungry of threatened.

$$\forall x[(tiger(x) \wedge lion(x)) \to $$$$\{ (hungry(x) \vee threatened(x)) \to attacks(x)\} ]$$

$$\forall x[(tiger(x) \vee lion(x)) \to $$$$\{ (hungry(x) \wedge threatened(x)) \to attacks(x)\} ]$$

$$\forall x[(tiger(x) \vee lion(x)) \to $$$$\{ attacks(x) \to (hungry(x)) \vee threatened(x))\} ]$$

$$\forall x[(tiger(x) \vee lion(x)) \to $$$$\{ (hungry(x) \vee threatened(x)) \to attacks(x)\} ]$$

GATE CSE 2005

MCQ (Single Correct Answer)

-0.6

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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