GATE CSE 2012 | Mathematical Logic Question 19 | Discrete Mathematics

PREVIOUS NEXT

GATE CSE 2012

MCQ (Single Correct Answer)

-0.3

The truth table GATE CSE 2012 Discrete Mathematics - Mathematical Logic Question 34 English

Represents the Boolean function

$$X$$

$$X+Y$$

$$X \oplus Y$$

$$Y$$

GATE CSE 2008

MCQ (Single Correct Answer)

-0.3

A set of Boolean connectives is functionally complete if all Boolean function can be synthesized using those, Which of the following sets of connectives is NOT functionally complete?

EX-NOR

implication, negation

OR, negation

NAND

GATE CSE 2004

MCQ (Single Correct Answer)

-0.3

Identify the correct translation into logical notation of the following assertion.

$$Some\,boys\,in\,the\,class\,are\,taller\,than\,all\,the\,girls$$
Note: taller$$\left( {x,\,y} \right)$$ is true if $$x$$ is taller than $$y$$.

$$\left( {\exists x} \right)\left( {boy\left( x \right) \to \left( {\forall y} \right)\left( {girl\left( y \right) \wedge taller\left( {x,y} \right)} \right)} \right)$$

$$\left( {\exists x} \right)\left( {boy\left( x \right) \wedge \left( {\forall y} \right)\left( {girl\left( y \right) \wedge taller\left( {x,y} \right)} \right)} \right)$$

$$\left( {\exists x} \right)\left( {boy\left( x \right) \to \left( {\forall y} \right)\left( {girl\left( y \right) \to taller\left( {x,y} \right)} \right)} \right)$$

$$\left( {\exists x} \right)\left( {boy\left( x \right) \wedge \left( {\forall y} \right)\left( {girl\left( y \right) \to taller\left( {x,y} \right)} \right)} \right)$$

GATE CSE 2004

MCQ (Single Correct Answer)

-0.3

Let $$a(x,y)$$, $$b(x,y)$$ and $$c(x,y)$$ be three statements with variables $$x$$ and $$y$$ chosen from some universe. Consider the following statement: $$$\left( {\exists x} \right)\left( {\forall y} \right)\left[ {\left( {a\left( {x,\,y} \right) \wedge b\left( {x,\,y} \right)} \right) \wedge \neg c\left( {x,\,y} \right)} \right]$$$

Which one of the following is its equivalent?

$$\left( {\forall x} \right)\left( {\exists y} \right)\left[ {\left( {a\left( {x,\,y} \right) \vee b\left( {x,\,y} \right)} \right) \to c\left( {x,\,y} \right)} \right]$$

$$\left( {\exists x} \right)\left( {\forall y} \right)\left[ {\left( {a\left( {x,\,y} \right) \vee b\left( {x,\,y} \right)} \right) \wedge \neg c\left( {x,\,y} \right)} \right]$$

$$ - \left[ {\left( {\forall x} \right)\left( {\exists y} \right)\left[ {\left( {a\left( {x,\,y} \right) \wedge b\left( {x,\,y} \right)} \right) \to c\left( {x,\,y} \right)} \right]} \right]$$

$$ - \left[ {\left( {\forall x} \right)\left( {\exists y} \right)\left[ {\left( {a\left( {x,\,y} \right) \vee b\left( {x,\,y} \right)} \right) \to c\left( {x,\,y} \right)} \right]} \right]$$

PREVIOUS NEXT