GATE CSE 2004 | Mathematical Logic Question 45 | Discrete Mathematics

GATE CSE 2004

MCQ (Single Correct Answer)

-0.6

The following propositional statement is $$$\left( {P \to \left( {Q \vee R} \right)} \right) \to \left( {\left( {P \wedge Q} \right) \to R} \right)$$$

Satisfiable but not valid

Valid

A contradiction

None of the above

GATE CSE 2003

MCQ (Single Correct Answer)

-0.6

The following resolution rule is used in logic programming. Derive clause $$\left( {P \vee Q} \right)$$ from clauses $$\left( {P \vee R} \right)$$, $$\left( {Q \vee \neg R} \right)$$.

Which of the following statements related to this rule is FALSE?

$$\left( {\left( {P \vee R} \right) \wedge \left( {Q \vee \neg R} \right)} \right) \Rightarrow \left( {P \vee Q} \right)$$ is logically valid

$$\left( {P \vee Q} \right) \Rightarrow \left( {\left( {P \vee R} \right) \wedge \left( {Q \vee \neg R} \right)} \right)$$ is logically valid

$$\left( {P \vee Q} \right)$$ is satisfiable if and only if $${\left( {P \vee R} \right) \wedge \left( {Q \vee \neg R} \right)}$$ is satisfiable

$$\left( {P \vee Q} \right) \Rightarrow $$ FALSE if and only if both $$P$$ and $$Q$$ are unsatisfiable

GATE CSE 2000

MCQ (Single Correct Answer)

-0.6

Let $$a, b, c, d$$ be propositions. Assume that the equivalences $$a \leftrightarrow \left( {b \vee \neg b} \right)$$ and $$b \leftrightarrow c$$ hold. Then the truth value of the formulae $$\left( {a\, \wedge \,b} \right) \to \left( {\left( {a \wedge c} \right) \vee d} \right)$$ is always

True

False

Same as truth value of $$b$$

Same as truth value of $$d$$

GATE CSE 1996

MCQ (Single Correct Answer)

-0.6

Which one of the following is false? Read $$ \wedge $$ as AND, $$ \vee $$ as OR, $$ \sim $$ as NOT, $$ \to $$ as one way implication and $$ \leftrightarrow $$ two way implication.

$$\left( {\left( {x \to y} \right) \wedge x} \right) \to y$$

$$\left( {\left( { \sim x \to y} \right) \wedge \left( { \sim x \to \sim y} \right)} \right) \to x$$

$$\left( {x \to \left( {x \vee y} \right)} \right)$$

$$\left( {\left( {x \vee y} \right) \leftrightarrow \left( { \sim x \to \sim y} \right)} \right)$$

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