Question

Consider the following two statements :

Statement-$1$ : $~\left(p\leftrightarrow ~q\right)$ is equivalent to $p\leftrightarrow q$.

Statement-$2$ : $~\left(p\leftrightarrow ~q\right)$ is a tautology.

Then which one of the following choices is correct $?$

• A. Both statement-$1$ and statement-$2$ are true and statement-$2$ is a correct explanation for statement-$1$
• B. Both statement-$1$ and statement-$2$ are true and statement-$2$is not a correct explanation for statement-$1$.
• C. Statement-$1$ is true but statement-$2$ is false.
• D. Both statement-$1$ and statement-$2$ are true.

Answer 1

The correct option is C

Statement-$1$ is true but statement-$2$ is false.

Explanation for the correct option :

Step-1 : (Construct the truth table of $p\to q$)

Here, we shall use the truth values of the compound statement $p\to q$again and again. So, first we will construct the truth table of $p\to q$, where $p$and $q$are any two mathematical statements.

Suppose, $p$and $q$are any two mathematical statements. Then the truth values of $p\to q$can be found from the following truth table ($T$ stands for True and $F$ stands for False) :

$p$	$q$	$p\to q$
$T$	$T$	$T$
$T$	$F$	$F$
$F$	$T$	$T$
$F$	$F$	$T$

Note that the truth value of $p\to q$ is $T$ whenever the truth value of $p$is $F$ and in that case it does not depend on the truth values of $q$. Also $T\to T$ is always true and $T\to F$ is always false and hence the above table is constructed.

Step-2 : (Construct the truth table of $q\to p$)

Similarly the truth table of $q\to p$ can be given as follows :

$p$	$q$	$q\to p$
$T$	$T$	$T$
$T$	$F$	$T$
$F$	$T$	$F$
$F$	$F$	$T$

Step-3 : (Construct the truth table of $p\leftrightarrow q$)

We know that $p\leftrightarrow q$ is true whenever $p\to q$ and $q\to p$ both are true. So, the truth table of $p\leftrightarrow q$ can be given as follows :

$p$	$q$	$p\to q$	$q\to p$	$p\leftrightarrow q$
$T$	$T$	$T$	$T$	$T$
$T$	$F$	$F$	$T$	$F$
$F$	$T$	$T$	$F$	$F$
$F$	$F$	$T$	$T$	$T$

Step-4 : Construct the truth table of $~(p\leftrightarrow ~q)$

We know that the truth value of $~p$ is $F$ if and only if the truth value of $p$ is $T$. So, we can construct the truth table of $~(p\leftrightarrow ~q)$ as follows :

$p$	$q$	$~q$	$p\to ~q$	$~q\to p$	$p\leftrightarrow ~q$	$~(p\leftrightarrow ~q)$
$T$	$T$	$F$	$F$	$T$	$F$	$T$
$T$	$F$	$T$	$T$	$T$	$T$	$F$
$F$	$T$	$F$	$T$	$T$	$T$	$F$
$F$	$F$	$T$	$T$	$F$	$F$	$T$

Step-5 : Check whether $~(p\leftrightarrow ~q)$ is equivalent to $p\leftrightarrow q$ and whether $~(p\leftrightarrow ~q)$ is a tautology.

We know that two statements $p$ and $q$ are equivalent if and only if they have the same set of truth values. Now, from Step-3 and Step-4, we get :

$p$	$q$	$~q$	$p\leftrightarrow q$	$~(p\leftrightarrow ~q)$
$T$	$T$	$F$	$T$	$T$
$T$	$F$	$T$	$F$	$F$
$F$	$T$	$F$	$F$	$F$
$F$	$F$	$T$	$T$	$T$

From the above table, we can see that $~(p\leftrightarrow ~q)$ and $p\leftrightarrow q$ has same set of truth values. Hence, $~(p\leftrightarrow ~q)$ is equivalent to $p\leftrightarrow q$. So, Statement-$1$ is true.

Also we can see that $~(p\leftrightarrow ~q)$ has truth values $T$ as well as $F$ i.e. all the truth values of $~(p\leftrightarrow ~q)$ are not $T$. Hence $~(p\leftrightarrow ~q)$ is not a tautology (A formula that is always true for every value of its propositional variables). So, Statement-2 is false.

Hence, option (C) is the correct answer.