Question

Which of the following is not true for any two statements $p$ and $q$?

• A. $~\left[p\vee \left(~q\right)\right]\equiv \left(~p\right)\wedge q$
• B. $~(p\vee q)=(~p)\vee (~q)$
• C. $q\wedge ~q$ is contradiction
• D. $~[p\wedge (~p\left)\right]$ is a tautology

Answer 1

The correct option is B

$~(p\vee q)=(~p)\vee (~q)$

Explanation for correct option

A tautology is a compound statement that is always true.

Signs of tautology:

Symbol	Meaning
$~$	Not
$\wedge$	And
$\vee$	Or
$\to$	Implies or If-then

Make truth table to check the relation$~\left(p\vee q\right)=$$\left(~p\right)\vee \left(~q\right)$

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

From the table it clear that $~\left(p\vee q\right)\ne$$\left(~p\right)\vee \left(~q\right)$is is

Explanation for incorrect option

Option (A)

Make truth table to check the relation $~\left[p\vee \left(~q\right)\right]\equiv \left(~p\right)\wedge q$

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

Thus the relation$~\left[p\vee \left(~q\right)\right]\equiv \left(~p\right)\wedge q$ is true.

Option (C)

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

Since the results are false. So it is a contradiction.

For option (D)

$p$	$~p$	$\left[p\wedge \left(~p\right)\right]$	$~\left[p\wedge \left(~p\right)\right]$
$T$	$F$	$F$	$T$
$F$	$T$	$F$	$T$

Since the results are true. So $~\left[p\wedge \left(~p\right)\right]$ is a tautology

Hence option (B) is the correct option.