$\sim (p\vee (q\wedge r\left)\right)=\sim p\wedge \sim (q\wedge r)=\sim p\wedge (\sim q\vee \sim r)=(\sim p\wedge \sim q)\vee (\sim p\wedge \sim r)$

Given that $(p\wedge q)\vee (\sim r)$ is false.
$\therefore i\left)\right(p\wedge q\left)isfalseii\right)\sim risfalse(p\wedge q)isfalse$
Hence, p and q cannot be true simulaneously.
Also, ~r is false. Hence, r must be true.
Only option D satisfies this condition

P: He is intelligent
Q: He is strong
P $\vee$ Q: He is intelligent or strong
~(P $\vee$ Q): It is not true that he is intelligent or strong

We know that $\sim (p\wedge q)=\sim p\vee \sim q$

$\therefore \sim (\sim p\wedge q)=\sim (\sim p)\vee \sim q=p\vee \sim q$

A logical statement is a statement which has a definite truth value which is invariant. Here, 'Paris is the capital of India' is a statement because we can definitely say that it is false.

p: 7 is not greater than 4
q: Paris is in France
~p: 7 is greater than 4
~q: Paris is not in France
$Now,\sim (p\vee q)=\sim p\wedge \sim q$

This is the statement '7 is greater than 4 and Paris is not in France'.

p: 4 is an even prime number - False
q: 6 is a divisor of 12 - True
r: the HCF of 4 and 6 is 12  - True
It can be seen that the statement $\sim p\vee (q\wedge r)$ is true.

$\sim (p\Rightarrow q)=p\wedge \sim q\sim (\sim p\Rightarrow q)=\sim p\wedge \sim q$

p: It rains
q: I shall go to school
$p\Rightarrow q:$ If it rains, I shall go to school
$\sim (p\Rightarrow q)=p\wedge \sim qp\wedge \sim q:\text{It rains and I shall not go to school}$

The negation of 'there exists a rational number x such that its square is 2' is 'there does not exist a rational number x such that its square is 2'.