Question

Statement 1: $\sim \left( {p \leftrightarrow \sim q} \right)$ is equivalent to $p \leftrightarrow q$
Statement 2: $\sim \left( {p \leftrightarrow \sim q} \right)$ is a tautology
$\left( a \right)$ Statement – 1 is true, statement – 2 is true; statement – 2 is a correct explanation for statement – 1.
$\left( b \right)$ Statement – 1 is true, statement – 2 is true; statement – 2 is not a correct explanation for statement – 1.
$\left( c \right)$ Statement – 1 is true, statement – 2 is false.
$\left( d \right)$ Statement – 1 is false, statement – 2 is true.

Answer 1

In this particular question use the concept that $\sim$ is the symbol of negation i.e. opposite of something i.e. if a is true then $\sim a$ is false, and use the concept that $\leftrightarrow$ is the symbol of bi-conditional i.e. if p and q have the same value then it is true otherwise it is false so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Statement 1: $\sim \left( {p \leftrightarrow \sim q} \right)$ is equivalent to $p \leftrightarrow q$
Now as we know that there are four possible cases which are given below in the table.

S.No	p	q
1	T	T
2	T	F
3	F	T
4	F	F

Now as we know that $\sim$ is the symbol of negation i.e. opposite of something i.e. if a is true then $\sim a$ is false.

S.No	p	q	$\sim p$	$\sim q$
1	T	T	F	F
2	T	F	F	T
3	F	T	T	F
4	F	F	T	T

Now as we know that $\leftrightarrow$ is the symbol of bi-conditional i.e. if p and q have the same value then it is true otherwise it is false

S.No	p	q	$\sim p$	$\sim q$	$p \leftrightarrow \sim q$	$\sim \left( {p \leftrightarrow \sim q} \right)$	$p \leftrightarrow q$
1	T	T	F	F	F	T	T
2	T	F	F	T	T	F	F
3	F	T	T	F	T	F	F
4	F	F	T	T	F	T	T

So as we see that $\sim \left( {p \leftrightarrow \sim q} \right)$ is equivalent to $p \leftrightarrow q$
Hence statement 1 is absolutely true.
Statement 2: $\sim \left( {p \leftrightarrow \sim q} \right)$ is a tautology
Now as we know that tautology is a condition in which all of the cases are true.
But in the seventh column of above table ${2^{nd}}{\text{ and }}{{\text{3}}^{rd}}$ cases are false.
So statement 2 is false.
Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall which symbol refers to what otherwise we cannot solve these types of questions, and always recall that tautology is a condition in which all of the cases are true, if any case is false then it is not a tautology.