Question

$\left( p\wedge q\leftrightarrow r \right)$ is true and q is false then which one of the following is a tautology?
A. $\left( q\vee r \right)\to \left( p\wedge r \right)$
B. $p\vee r$
C. $p\wedge r$
D. $\left( p\vee r \right)\to \left( p\wedge r \right)$

Answer 1

Hint: We first find the conditions of $q$ and $r$ using the truth table of the given conditions. We find the tautology using the first statement being false. The individual possibilities of the terms give the final solution.

Complete step by step solution:
It is given that $\left( p\wedge q\leftrightarrow r \right)$ is true and q is false. The biconditional of $p\wedge q$ with r is true. The tautology has to be the statement which is always true.
We try to find the truth table of $p\wedge q$ given the condition that q is false. The value of p will be used twice as true and false.

p	q	$p\wedge q$
T	F	F
F	F	F

Therefore, irrespective of the value of p, the value of $p\wedge q$ will always be false.
The biconditional statement $\left( p\wedge q\leftrightarrow r \right)$ can be true when both of them are similar. The cases will be when both of them are false or both of them are true.

r	$p\wedge q$	$\left( p\wedge q\leftrightarrow r \right)$
T	T	T
F	T	F
T	F	F
F	F	T

Now we already have that $p\wedge q$ will always be false which gives us that r has to be false for $\left( p\wedge q\leftrightarrow r \right)$ to be true. The final conditions for $q$ and $r$ are false. $p$ can be both.
The table will be

p	q	r
T	F	F
F	F	F

The only option satisfying is A as the starting statement in the implication of $\left( q\vee r \right)\to \left( p\wedge r \right)$ is always false.

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

Note: We do not need to check the full tautology if the first statement is false. If it is true only then we have to check if the second statement is true or not. If that is true then the tautology is true again, otherwise not.