Mathematical reasoning practice · Mixed difficulty

Consider

Statement-1: $(p \wedge \sim q) \wedge(\sim p \wedge q)$ is a fallacy.

Statement-2: $(p \rightarrow q) \leftrightarrow(\sim q \rightarrow \sim p)$ is a tautology.

If $P$ and $Q$ are two statements, then which of the following compound statement is a tautology?

If $p$ and $q$ are two statements. Then negation of compound statement $(\sim p \vee q)$ is

Negation of the conditional, "If it rains, I shall go to school" is

Let $\mathrm{S}$ be a non-empty subset of $\mathrm{R}$. Consider the following statement :
$\mathrm{P}$ : There is a rational number $x \in \mathrm{S}$ such that $x>0$.
Which of the following statements is the negation of the statement P ?

Let $p$ be the proposition Mathematics is interesting and let $q$ be the proposition that Mathematics is difficult, then the symbol $p \wedge q$ means

The contrapositive of the statement “If you are born in India, then you are a citizen of India”, is

The negation of $\sim s \vee ( \sim r \wedge s)$ is equivalent to

Consider the following statements :
P : Ramu is intelligent.
Q : Ramu is rich.
R : Ramu is not honest.
The negation of the statement “Ramu is intelligent and honest if and only if Ramu is not rich” can be expressed as :

Consider the following two statements :

P: If 7 is an odd number, then 7 is divisible by 2 .

$Q$ : If 7 is a prime number, then 7 is an odd number.

If $V_1$ is the truth value of the contrapositive of $\mathrm{P}$ and $\mathrm{V}_2$ is the truth value of contrapositive of $\mathrm{Q}$, then the ordered pair $\left(\mathrm{V}_1, \mathrm{~V}_2\right)$ equals:

Which of the following is logically equivalent to $\sim \left( { \sim p \to q} \right)$ ?

The inverse of the statement $(p \wedge \sim q) \rightarrow r$ is

Let $p:$ I am well.

$q:$ I will not take rest

$r$ : I will not sleep properly, then

"If I am not well then I will not take rest and I will not sleep properly" is logically equivalent to

The contrapositive of $2x + 3 = 9 \Rightarrow x \ne 4$ is

If $S\left( {p,\;q,\;r} \right) = \left( { \sim P} \right) \vee \left[ { \sim \left( {q \wedge r} \right)} \right]$ is a compound statement, then $S\left( { \sim p, \sim q,\; \sim r} \right)$ is

Negation of the Boolean expression $p \Leftrightarrow(q \Rightarrow p)$ is

The negative of the statement $\sim \,p\, \wedge \,(p\, \vee q)\,$ is

Negation of the statement If a number is prime then it is odd' is.

Let $\mathrm{p}, \mathrm{q}, \mathrm{r}$ denote arbitrary statements. Then the logically equivalent of the statement $p \Rightarrow(q \vee r)$ is:

Consider the following statements:
A : Rishi is a judge.
B : Rishi is honest.
C : Rishi is not arrogant.
The negation of the statement “if Rishi is a judge and he is not arrogant, then he is honest” is

The statement $(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))$ is equivalent to

If $p:$ Ashok works hard

$q$ : Ashok gets good grade

The verbal form for $(\sim p \rightarrow q)$ is

The negation of the following statement : $P$ : Neha lives in Ludhiana or she lives in Gurudaspur.

The logical statement $(p \wedge \sim q) \rightarrow(p \rightarrow \sim q)$ is a:

If $p, q, r$ are substatements with truth values. $T, T, F$ then the Statement $r \rightarrow(p \wedge \sim q) \vee(\sim q \wedge \sim r)$ will be

If $p$ and $q$ are two statements. Then negation of compound statement $(\sim p \vee q)$ is

The statement $p \Leftrightarrow q$ is equal to -

The negative of the proposition : “If a number is divisible by 15, then it is divisible by 5 or 3”

Converse of the statement:

If a number $n$ is even, then $n^2$ is even, is

The proposition $p \to \sim \left( {p \wedge \sim q} \right)$ is