1
GATE CSE 2015 Set 2
MCQ (Single Correct Answer)
+2
-0.6
Which one of the following well formed formulae is a tautology?
A
$$\forall x\,\exists y\,R\left( {x,y} \right) \leftrightarrow \exists y\forall x\,R\left( {x,y} \right)$$
B
$$\left( {\forall x\left[ {\exists y\,R\left( {x,y} \right) \to S\left( {x,y} \right)} \right]} \right) \to \forall x\exists y\,S\left( {x,y} \right)$$
C
$$\left[ {\forall x\,\exists y\,\left( {P\left( {x,y} \right)} \right. \to R\left( {x,y} \right)} \right] \leftrightarrow \left[ {\forall x\,\exists y\,\left( {\neg P\left( {x,y} \right)V\,R\left( {x,y} \right)} \right.} \right]$$
D
$$\forall x\,\forall y\,P\left( {x,y} \right) \to \forall x\forall y\,P\left( {y,x} \right)$$
2
GATE CSE 2014 Set 3
MCQ (Single Correct Answer)
+2
-0.6
The CORRECT formula for the sentence, "not all rainy days are cold" is
A
$$\forall d\left( {Rainy\left( d \right) \wedge \sim Cold\left( d \right)} \right)$$
B
$$\forall d\left( { \sim Rainy\left( d \right) \to Cold\left( d \right)} \right)$$
C
$$\exists d\left( { \sim Rainy\left( d \right) \to Cold\left( d \right)} \right)$$
D
$$\exists d\left( {Rainy\left( d \right) \wedge \sim Cold\left( d \right)} \right)$$
3
GATE CSE 2014 Set 2
MCQ (Single Correct Answer)
+2
-0.6
Which one of the following Boolean expressions is NOT A tautology?
A
$$\left( {\left( {a \to b} \right) \wedge \left( {b \to c} \right)} \right) \to \left( {a \to c} \right)$$
B
$$\left( {a \leftrightarrow c} \right) \to \left( { \sim b \to \left( {a \wedge c} \right)} \right)$$
C
$$\left( {a \wedge b \wedge c} \right) \to \left( {c \vee a} \right)$$
D
$$A \to \left( {b \to a} \right)$$
4
GATE CSE 2014 Set 1
MCQ (Single Correct Answer)
+2
-0.6
Which one of the following propositional logic formulas is TRUE when exactly two of $$p, q,$$ and $$r$$ are TRUE?
A
$$\left( {\left( {p \leftrightarrow q} \right) \wedge r} \right) \vee \left( {p \wedge q \wedge \sim r} \right)$$
B
$$\left( { \sim \left( {p \leftrightarrow q} \right) \wedge r} \right) \vee \left( {p \wedge q \wedge \sim r} \right)$$
C
$$\left( {\left( {p \to q} \right) \wedge r} \right) \vee \left( {p \wedge q \wedge \sim r} \right)$$
D
$$\left( { \sim \left( {p \leftrightarrow q} \right) \wedge r} \right) \wedge \left( {p \wedge q \wedge \sim r} \right)$$
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