Question

If $\left( p\wedge \sim q \right)\wedge \left( p\wedge r \right)\to \sim p\vee q$ is false , then the truth values of p, q and r are respectively:
 $\begin{align}
  & \left( A \right)F,T,F \\
 & \left( B \right)T,F,T \\
 & \left( C \right)F,F,F \\
 & \left( D \right)T,T,T \\
\end{align}$

Answer 1

Hint: For this problem, first we have to make a truth table for$\left( p\wedge \sim q \right)$ after that for $\left( p\wedge r \right)$
And at last for $\to \sim p\vee q$, then we have to observe from the table the truth value of $p$ and after that q and r.

Complete step-by-step solution:
We have to find the truth value of p, q, r when the $\left( p\wedge \sim q \right)\wedge \left( p\wedge r \right)\to \sim p\vee q$ is false.
Now we will have to make the truth table with the help of its operation

$a$	$b$	$a\to b$	$a\vee b$	$a\wedge b$	$\sim a$	$\sim b$
T	T	T	T	T	F	F
T	F	F	T	F	F	T
F	T	T	T	F	T	F
F	F	T	F	F	T	T

With the using of the above table we can make $\left( p\wedge \sim q \right)\wedge \left( p\wedge r \right)\to \sim p\vee q$ truth table:

$p$	$q$	$r$	$\sim p$	$\sim q$	$\left( p\wedge \sim q \right)$	$p\wedge r$	$\left( p\wedge \sim q \right)\wedge \left( p\wedge r \right)$	$\sim p\vee q$	$\left( p\wedge \sim q \right)\wedge \left( p\wedge r \right)\to \sim p\vee q$
T	T	T	F	F	F	T	F	T	T
T	T	F	F	F	F	F	F	T	T
T	F	T	F	T	T	T	T	F	F
T	F	F	F	T	T	F	F	F	T
F	T	T	T	F	F	F	F	T	T
F	T	F	T	F	F	F	F	T	T
F	F	T	T	T	F	F	F	T	T
F	F	F	T	T	F	F	F	T	T

It is given that $\left( p\wedge \sim q \right)\wedge \left( p\wedge r \right)\to \sim p\vee q$ is false. From the above table we can say that it's possible when $p$ is true , q is false and r is also true.

Option B is the correct option.

Note: The symbol $\wedge $ is used for and. Symbol $\vee $ is used for or. Symbol$\sim $ is used for not.
Sometimes students are confused between $a\wedge b\,\,and\,\,a\vee b$ and they make mistakes. Both the symbols have different meanings, if we use $\wedge \,$ symbol instead of $\vee $ then our truth table is wrong.