Mathematical Reasoning

The compound statement $(P \vee Q) \wedge ( \sim P) \Rightarrow Q$ is equivalent to :

$(\mathrm{S} 1)~(p \Rightarrow q) \vee(p \wedge(\sim q))$ is a tautology $(\mathrm{S} 2)~((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)$ is a contradiction. Then

Which of the following Boolean expression is a tautology?

Among the statements : $(\mathrm{S} 1)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow(\mathrm{p} \Rightarrow \mathrm{r})$ $(\mathrm{S} 2)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow((\mathrm{p} \Rightarrow \mathrm{r}) \vee(\mathrm{q} \Rightarrow \mathrm{r}))$

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

Which of the following statements is a tautology?

The statement A $\to$ (B $\to$ A) is equivalent to :

The Boolean expression $\sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right)$ is equvalent to :

If (p $\wedge$ $\sim$ q) $\wedge$ (p $\wedge$ r) $\to$ $\sim$ p $\vee$ q is false, then the truth values of $p, q$ and $r$ are, respectively :

The Boolean expression ~(p $\Rightarrow$ (~q)) is equivalent to :

Let F₁(A, B, C) = (A $\wedge$ $\sim$ B) $\vee$ [$\sim$C $\wedge$ (A $\vee$ B)] $\vee$ $\sim$ A and F₂(A, B) = (A $\vee$ B) $\vee$ (B $\to$ $\sim$A) be two logical expressions. Then :

If p $\to$ ($\sim$ p$\vee$ $\sim$ q) is false, then the truth values of p and q are respectively :

The negation of $\sim s \vee \left( { \sim r \wedge s} \right)$ is equivalent to :

The Boolean expression $\left( {p \wedge \sim q} \right) \vee q \vee \left( { \sim p \wedge q} \right)$ is equivalent to :

The negation of the Boolean expression x $\leftrightarrow$ ~ y is equivalent to :

The statement $p \to \left( {q \to p} \right)$ is equivalent to

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.

Consider : Statement − I : $\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)$ is a fallacy. Statement − II :$\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to \sim p} \right)$ is a tautology.

The statement $\sim \left( {p \leftrightarrow \sim q} \right)$ is :