1
GATE CSE 2004
MCQ (Single Correct Answer)
+2
-0.6
The following propositional statement is $$$\left( {P \to \left( {Q \vee R} \right)} \right) \to \left( {\left( {P \wedge Q} \right) \to R} \right)$$$
A
Satisfiable but not valid
B
Valid
C
A contradiction
D
None of the above
2
GATE CSE 2004
MCQ (Single Correct Answer)
+2
-0.6
Let $$p, q, r$$ and $$s$$ be four primitive statements. Consider the following arguments:

$$P:\left[ {\left( {\neg p \vee q} \right) \wedge \left( {r \to s} \right) \wedge \left( {p \vee r} \right)} \right] \to \left( {\neg s \to q} \right)$$
$$Q:\left[ {\left( {\neg p \wedge q} \right) \wedge \left[ {q \to \left( {p \to r} \right)} \right]} \right] \to \neg r$$
$$R:\left[ {\left[ {\left( {q \wedge r} \right) \to p} \right] \wedge \left( {\neg q \vee p} \right)} \right] \to r$$
$$S:\left[ {p \wedge \left( {p \to r} \right) \wedge \left( {q \vee \neg r} \right)} \right] \to q$$

Which of the above arguments are valid?

A
$$P$$ and $$Q$$ only
B
$$P$$ and $$R$$ only
C
$$P$$ and $$S$$ only
D
$$P, Q, R$$ and $$S$$
3
GATE CSE 2003
MCQ (Single Correct Answer)
+2
-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?

A
$$\left( {\left( {P \vee R} \right) \wedge \left( {Q \vee \neg R} \right)} \right) \Rightarrow \left( {P \vee Q} \right)$$ is logically valid
B
$$\left( {P \vee Q} \right) \Rightarrow \left( {\left( {P \vee R} \right) \wedge \left( {Q \vee \neg R} \right)} \right)$$ is logically valid
C
$$\left( {P \vee Q} \right)$$ is satisfiable if and only if $${\left( {P \vee R} \right) \wedge \left( {Q \vee \neg R} \right)}$$ is satisfiable
D
$$\left( {P \vee Q} \right) \Rightarrow $$ FALSE if and only if both $$P$$ and $$Q$$ are unsatisfiable
4
GATE CSE 2000
MCQ (Single Correct Answer)
+2
-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
A
True
B
False
C
Same as truth value of $$b$$
D
Same as truth value of $$d$$
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