Question

Which of the following is a tautology from the given options?

• A. $\left(p\to q\right)\wedge \left(q\to p\right)$
• B. $\left(~p\right)\wedge \left(p\vee q\right)\to q$
• C. $\left(q\to p\right)\vee ~\left(p\to q\right)$
• D. $\left(~q\right)\vee \left(p\wedge q\right)\to q$

Answer 1

The correct option is B

$\left(~p\right)\wedge \left(p\vee q\right)\to q$

The explanation for the correct option:

The truth table of $p\vee q$:

$p$	$q$	$\left(p\vee q\right)$
T	T	T
T	F	T
F	T	T
F	F	F

The truth table of $p\wedge q$:

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

Truth table of $~p$

$p$	$~p$
T	F
F	T

A tautology is a compound statement in Maths that always results in a Truth value.

It doesn’t matter what the individual part consists of, the result in tautology is always true.

The possible truth table of $\left(~p\right)\wedge \left(p\vee q\right)\to q$:

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

All outputs of $\left(~p\right)\wedge \left(p\vee q\right)\to q$ are true.

So, it is a tautology.

Hence, the correct option is $B)\left(~p\right)\wedge \left(p\vee q\right)\to q$.

The explanation for the incorrect options:

Truth table for $p\to q$

$p$	$q$	$\left(p\to q\right)$
T	T	T
T	F	F
F	T	T
F	F	T

The truth table of $\left(p\to q\right)\wedge \left(q\to p\right)$:

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

All outputs of $\left(p\to q\right)\wedge \left(q\to p\right)$ are not true.

So, it is not a tautology.

Hence, the option $A)$ is incorrect.

The truth table of $\left(q\to p\right)\vee ~\left(p\to q\right)$:

$p$	$q$	$\left(p\to q\right)$	$\left(q\to p\right)$	$~\left(p\to q\right)$	$\left(q\to p\right)\vee ~\left(p\to q\right)$
T	T	T	T	F	T
T	F	F	T	T	T
F	T	T	F	F	F
F	F	T	T	F	T

All outputs of $\left(q\to p\right)\vee ~\left(p\to q\right)$ are not true.

So, it is not a tautology.

Hence, the option $C)$ is incorrect.

The truth table of $\left(~q\right)\vee \left(p\wedge q\right)\to q$:

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

All outputs of $\left(~q\right)\vee \left(p\wedge q\right)\to q$ are not true.

So, it is not a tautology.

Hence, the option $D)$ is incorrect.