GATE CSE 1996 | Mathematical Logic Question 55 | Discrete Mathematics

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.

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

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

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

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

GATE CSE 1995

MCQ (Single Correct Answer)

-0.6

If the proposition $$\neg p \Rightarrow q$$ is true, then the truth value of the proposition $$\neg p \vee \left( {p \Rightarrow q} \right)$$ where $$'\neg '$$ is negation, $$' \vee '$$ is inclusive or and $$' \Rightarrow '$$ is implication, is

true

multiple-valued

false

cannot be determined

GATE CSE 1994

True or False

-0

Let $$p$$ and $$q$$ be propositions. Using only the truth table decide whether $$p \Leftrightarrow q$$ does not imply $$p \to \sim q$$ is true or false.

TRUE

FALSE

GATE CSE 1990

MCQ (Single Correct Answer)

-0.6

Indicate which of the following well-formed formula are valid:

$$\left( {\left( {{\rm P} \Rightarrow Q} \right) \wedge \left( {Q \Rightarrow R} \right)} \right) \Rightarrow \left( {{\rm P} \Rightarrow R} \right).$$

$$\left( {{\rm P} \Rightarrow Q} \right) \Rightarrow \left( { \sim P \Rightarrow \sim Q} \right)$$

$$\left( {{\rm P}\, \wedge \,\left( { \sim {\rm P}\,\,V \sim Q} \right)} \right) \Rightarrow Q\left( { \sim {\rm P} \Rightarrow \sim Q} \right)$$

$$\left( {\left( {{\rm P} \Rightarrow R} \right) \vee \left( {Q \Rightarrow R} \right)} \right) \Rightarrow \left( {\left( {\left( {{\rm P} \vee Q} \right) \Rightarrow R} \right)} \right)$$