Question

Evaluate $\left(p\wedge ~q\right)\wedge \left(~p\wedge q\right)$

• A. A tautology
• B. A contradiction
• C. Both a tautology and a contradiction
• D. Neither a tautology nor a contradiction

Answer 1

The correct option is B

A contradiction

Explanation for the correct option:

Finding the value :

We use the truth table method to find whether it is tautology or a contradiction.

Using associative laws and commutative laws,

$\left(p\wedge ~q\right)\wedge \left(~p\wedge q\right)=\left(p\wedge ~p\right)\wedge \left(~q\wedge q\right)$

where $~p$ is inverse of $p$

Similarly $~q$ is inverse of $q$

If $p$is $T$and $~p$ is $F$ then $\left(p\wedge ~p\right)$ is $F$

similarly if $q$is $T$ and $~q$is $F$ then $\left(~q\wedge q\right)$ is $F$.

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

Hence, $(p\wedge ~q)\wedge \left(~p\wedge q\right)$ is a contradiction as it has all false $F$.

Hence, option (B) is the correct answer.