GATE CSE 2006 | Mathematical Logic Question 38 | Discrete Mathematics

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 37 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

Consider the following first order logic formula in which $$R$$ is a binary relation symbol.
$$\forall x\forall y\left( {R\left( {x,\,y} \right) \Rightarrow R\left( {y,x} \right)} \right).$$

The formula is

Satisfiable and valid

Satisfiable and so is its negation

Unsatisfiable but its negation is valid

Satisfiable but its negation is unsatisfiable

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 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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