GATE CSE 2009 | Mathematical Logic Question 33 | Discrete Mathematics

GATE CSE 2009

MCQ (Single Correct Answer)

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The binary operation ◻ is defined as follows: GATE CSE 2009 Discrete Mathematics - Mathematical Logic Question 19 English

Which one of the following is equivalence to $$P \vee Q$$?

$$\neg \,Q$$ ◻ $$\neg \,P$$

$$P$$ ◻ $$\neg \,Q$$

$$\neg \,P$$ ◻ $$Q$$

$$\neg \,P$$ ◻ $$\neg \,Q$$

GATE CSE 2008

MCQ (Single Correct Answer)

-0.6

Which of the following is the negation of $$$\left[ {\forall x,\alpha \to \left( {\exists y,\beta \to \left( {\forall u,\exists v,\gamma } \right)} \right)} \right]?$$$

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

$$\left[ {\exists x,\alpha \to \left( {\forall y,\beta \to \left( {\exists u,\forall v,\neg \gamma } \right)} \right)} \right]$$

$$\left[ {\forall x,\neg \alpha \to \left( {\exists y,\neg \beta \to \left( {\forall u,\exists v,\neg \gamma } \right)} \right)} \right]$$

$$\left[ {\exists x,\alpha \wedge \left( {\forall y,\beta \wedge \left( {\exists u,\forall v,\neg \gamma } \right)} \right)} \right]$$

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

$$P.\overline Q $$

$$P.\overline R $$

$$P.\overline Q + R$$

$$P.\overline R + Q$$