$$\eqalign{ & S\left( {p,q,r} \right) = \left( { \sim p} \right) \vee \left[ { \sim \left( {q \wedge r} \right)} \right] \cr & S\left( { \sim p, \sim q, \sim r} \right) = \sim \left( { \sim p} \right) \vee \left[ { \sim \left( { \sim q \, \wedge \sim r} \right)} \right] \cr & = p \vee \left[ { \sim \left( { \sim q} \right) \vee \sim \left( { \sim r} \right)} \right] = p \vee \left( {q \vee r} \right) \cr} $$

$$p$$ : $$x$$ is an irrational number
$$q$$ : $$y$$ is a transcendental number
$$r$$ : $$x$$ is a rational number if $$f y$$ is a transcendental number.
clearly $$r: \sim p \leftrightarrow q$$
Let us use truth table to check the equivalence of $$‘r’$$ and $$‘q$$ or $$p’,'r'$$  and $$ \sim \left( {p \leftrightarrow \sim q} \right)$$

				1	2		3
$$p$$	$$q$$	$$ \sim p$$	$$ \sim q$$	$$ \sim p \leftrightarrow q$$	$$q{\text{ or }}p$$	$$p \leftrightarrow \, \sim q$$	$$ \sim \left( {p \leftrightarrow \, \sim q} \right)$$
T	T	F	F	F	T	F	T
T	F	F	T	T	T	T	F
F	T	T	F	T	T	T	F
F	F	T	T	F	F	F	T

From columns (1), (2) and (3), we observe, none of the these statements are equivalent to each other.
∴ Statement las well as statement 2 both are false.
∴ None of the options is correct.

Statement - 2 : $$\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to \sim p} \right) \equiv \left( {p \to q} \right) \leftrightarrow \left( {p \to q} \right)$$
which is always true.
So statement - 2 is true
Statement - 1 : $${\left( {{p^ \wedge } \sim q} \right)^ \wedge }\left( { \sim {p^ \wedge }q} \right)$$
$$\eqalign{ & = {p^ \wedge } \sim {q^ \wedge } \sim {p^ \wedge }q \cr & = {p^ \wedge } \sim {p^ \wedge } \sim {q^ \wedge }q \cr & = {f^ \wedge }f \cr & = f \cr} $$
So statement - 1 is true

It is correct.
$$\because \sqrt 3 $$  is not rational but it is real.

$$\eqalign{ & \sim \left[ { \sim s \vee \left( { \sim r \wedge s} \right)} \right] \cr & = s \wedge \sim \left( { \sim r \wedge s} \right) \cr & = s \wedge \left( {r \vee \sim s} \right) \cr & = \left( {s \wedge r} \right) \vee \left( {s \wedge \sim s} \right) \cr & = \left( {s \wedge r} \right) \vee 0 \cr & = s \wedge r \cr} $$

Contrapositive of $$p \Rightarrow q{\text{ is }} \sim q \Rightarrow \, \sim p$$
$$\therefore $$ contrapositive of $$\left( {p \vee q} \right) \Rightarrow r{\text{ is}}$$
$$ \sim r \Rightarrow \, \sim \left( {p \vee q} \right){\text{ i}}{\text{.e}}{\text{.,}} \sim r \Rightarrow \left( { \sim p \wedge \sim q} \right)$$

Statement given in option $$\left( C \right)$$ is correct.
$$ \sim \left[ {p \vee \left( { \sim q} \right)} \right] = \left( { \sim p} \right) \wedge \sim \left( { \sim q} \right) = \left( { \sim p} \right) \wedge q$$

$$p$$	$$q$$	$$p \wedge q$$	$$p \vee q$$	$$ \sim \left( {p \vee q} \right)$$	$$\left( {p \wedge q} \right) \wedge \sim \left( {p \vee q} \right)$$
T	T	T	T	F	F
T	F	F	T	F	F
F	T	F	T	F	F
F	F	F	F	T	F

$$\therefore \left( {p \wedge q} \right) \wedge \left( { \sim \left( {p \vee q} \right)} \right)$$     is a contradiction.

Given : $$P :$$ He is intelligent.
$$Q =$$  He is strong.
Symbolic form of
“It is not true that he is either intelligent or strong” is $$ \sim \left( {P \vee Q} \right)$$

$$ \sim \left\{ {\left( {p \vee \left( { \sim q} \right)} \right) \wedge q} \right\} = \left( { \sim \left( {p \vee \left( { \sim q} \right)} \right)} \right) \vee \left( { \sim q} \right)$$
By De Morgan's Law,
$$ \sim \left( {A \wedge B} \right) = \left( { \sim A} \right) \vee \left( { \sim B} \right)$$
$$ = \left( {\left( { \sim p} \right) \wedge \left( { \sim \left( { \sim q} \right)} \right)} \right) \vee \left( { \sim q} \right)$$       [Using De Morgan’s law again]
$$ = \left( { \sim p \wedge q} \right) \vee \left( { \sim q} \right)\left[ {\because \,\, \sim \left( { \sim q} \right) = q} \right]$$