GATE CSE 2008 | Mathematical Logic Question 38 | Discrete Mathematics

GATE CSE 2008

MCQ (Single Correct Answer)

-0.6

Which of the following first order formulae is logically valid? Here $$\alpha \left( x \right)$$ is a first order formulae with $$x$$ as a free variable, and $$\beta $$ is a first order formula with no free variable.

$$\left[ {\beta \to \left( {\exists x,\alpha \left( x \right)} \right)} \right] \to \left[ {\forall x,\beta \to \alpha \left( x \right)} \right]$$

$$\left[ {\exists x,\beta \to \alpha \left( x \right)} \right] \to \left[ {\beta \to \left( {\forall x,\alpha \left( x \right)} \right)} \right]$$

$$\left[ {\left( {\exists x,\alpha \left( x \right)} \right) \to \beta } \right] \to \left[ {\forall x,\alpha \left( x \right) \to \beta } \right]$$

$$\left[ {\left( {\forall x,\alpha \left( x \right)} \right) \to \beta } \right] \to \left[ {\forall x,\alpha \left( x \right) \to \beta } \right]$$

GATE CSE 2008

MCQ (Single Correct Answer)

-0.6

If $$P$$, $$Q$$, $$R$$ are Boolean variables, then $$(P + \bar{Q}) (P.\bar{Q} + P.R) (\bar{P}.\bar{R} + \bar{Q})$$ simplifies to

$$P.\overline Q $$

$$P.\overline R $$

$$P.\overline Q + R$$

$$P.\overline R + Q$$

GATE CSE 2007

MCQ (Single Correct Answer)

-0.6

Which one of these first-order logic formulae is valid?

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

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

$$\exists x\left( {P\left( x \right) \wedge Q\left( x \right)} \right) \Leftrightarrow \left( {\left( {\exists xP\left( x \right)} \right) \wedge \left( {\exists xQ\left( x \right)} \right)} \right)$$

$$\forall x\exists yP\left( {x,y} \right) \Rightarrow \exists y\forall xP\left( {x,y} \right)$$

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

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