$\left( {p \wedge \sim q \wedge \sim r} \right) \vee \left( { \sim p \wedge q \wedge \sim r} \right) \vee \left( { \sim p \wedge \sim q \wedge r} \right)$ is equivalent to-
$ \sim \left( {\left( {p \wedge q} \right) \vee \left( {q \wedge r} \right) \vee \left( {r \wedge p} \right)} \right)$
$p \vee q \vee r$
$ \left( {\left( {p \wedge q} \right) \vee \left( {q \wedge r} \right) \vee \left( {r \wedge p} \right)\left( {p \vee q \vee r} \right)} \right)$
$\left( { \sim \left( {(p \wedge q} \right) \vee \left( {q \wedge r} \right) \vee \left( {r \wedge p)} \right) \wedge \left( {p \vee q \vee r} \right)} \right)$
Negation of the Boolean statement $( p \vee q ) \Rightarrow((\sim r ) \vee p )$ is equivalent to
If the Boolean expression $( p \wedge q ) \circledast( p \otimes q )$ is a tautology, then $\circledast$ and $\otimes$ are respectively given by
Consider the two statements :
$(\mathrm{S} 1):(\mathrm{p} \rightarrow \mathrm{q}) \vee(\sim \mathrm{q} \rightarrow \mathrm{p})$ is a tautology
$(S2): (\mathrm{p} \wedge \sim \mathrm{q}) \wedge(\sim \mathrm{p} \vee \mathrm{q})$ is a fallacy.
Then :
The statement $B \Rightarrow((\sim A ) \vee B )$ is equivalent to
For the statements $p$ and $q$, consider the following compound statements :
$(a)$ $(\sim q \wedge( p \rightarrow q )) \rightarrow \sim p$
$(b)$ $((p \vee q) \wedge \sim p) \rightarrow q$
Then which of the following statements is correct?