The statement $(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))$ is equivalent to _________.
The negation of the statement $((A \wedge(B \vee C)) \Rightarrow(A \vee B)) \Rightarrow A$ is
Among the two statements
$(\mathrm{S} 1):(p \Rightarrow q) \wedge(p \wedge(\sim q))$ is a contradiction and
$(\mathrm{S} 2):(p \wedge q) \vee((\sim p) \wedge q) \vee(p \wedge(\sim q)) \vee((\sim p) \wedge(\sim q))$ is a tautology
The converse of $((\sim p) \wedge q) \Rightarrow r$ is
The statement $\sim[p \vee(\sim(p \wedge q))]$ is equivalent to :
The negation of the statement $(p \vee q) \wedge (q \vee ( \sim r))$ is :
The negation of $(p \wedge(\sim q)) \vee(\sim p)$ is equivalent to :
Negation of $(p \Rightarrow q) \Rightarrow(q \Rightarrow p)$ is :
Among the statements
(S1) : $(p \Rightarrow q) \vee((\sim p) \wedge q)$ is a tautology
(S2) : $(q \Rightarrow p) \Rightarrow((\sim p) \wedge q)$ is a contradiction
Statement $\mathrm{(P \Rightarrow Q) \wedge(R \Rightarrow Q)}$ is logically equivalent to :
Which of the following statements is a tautology?
The negation of the expression $q \vee \left( {( \sim \,q) \wedge p} \right)$ 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
Consider the following statements:
P : I have fever
Q: I will not take medicine
$\mathrm{R}$ : I will take rest.
The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to :
Among the statements :
$(\mathrm{S} 1)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow(\mathrm{p} \Rightarrow \mathrm{r})$
$(\mathrm{S} 2)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow((\mathrm{p} \Rightarrow \mathrm{r}) \vee(\mathrm{q} \Rightarrow \mathrm{r}))$
If $p,q$ and $r$ are three propositions, then which of the following combination of truth values of $p,q$ and $r$ makes the logical expression $\left\{ {(p \vee q) \wedge \left( {( \sim p) \vee r} \right)} \right\} \to \left( {( \sim q) \vee r} \right)$ false?
Let $\Delta ,\nabla \in \{ \wedge , \vee \} $ be such that $\mathrm{(p \to q)\Delta (p\nabla q)}$ is a tautology. Then
The statement $\left( {p \wedge \left( { \sim q} \right)} \right) \Rightarrow \left( {p \Rightarrow \left( { \sim q} \right)} \right)$ is
Let p and q be two statements. Then $ \sim \left( {p \wedge (p \Rightarrow \, \sim q)} \right)$ is equivalent to
The compound statement $\left( { \sim (P \wedge Q)} \right) \vee \left( {( \sim P) \wedge Q} \right) \Rightarrow \left( {( \sim P) \wedge ( \sim Q)} \right)$ is equivalent to
The statement $(p \Rightarrow q) \vee(p \Rightarrow r)$ is NOT equivalent to
The statement $(p \wedge q) \Rightarrow(p \wedge r)$ is equivalent to :
Let
$\mathrm{p}$ : Ramesh listens to music.
$\mathrm{q}$ : Ramesh is out of his village.
$\mathrm{r}$ : It is Sunday.
$\mathrm{s}$ : It is Saturday.
Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as
Let the operations $*, \odot \in\{\wedge, \vee\}$. If $(\mathrm{p} * \mathrm{q}) \odot(\mathrm{p}\, \odot \sim \mathrm{q})$ is a tautology, then the ordered pair $(*, \odot)$ is :
If the truth value of the statement $(P \wedge(\sim R)) \rightarrow((\sim R) \wedge Q)$ is F, then the truth value of which of the following is $\mathrm{F}$ ?
$(p \wedge r) \Leftrightarrow(p \wedge(\sim q))$ is equivalent to $(\sim p)$ when $r$ is
Negation of the Boolean expression $p \Leftrightarrow(q \Rightarrow p)$ is
The statement $(\sim(\mathrm{p} \Leftrightarrow \,\sim \mathrm{q})) \wedge \mathrm{q}$ is :
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:
Which of the following statements is a tautology ?
The conditional statement
$((p \wedge q) \to (( \sim p) \vee r)) \vee ((( \sim p) \vee r) \to (p \wedge q))$ is :
Negation of the Boolean statement (p $\vee$ q) $\Rightarrow$ (($\sim$ r) $\vee$ p) is equivalent to :
Let $\Delta$ $\in$ {$\wedge$, $\vee$, $\Rightarrow$, $\Leftrightarrow$} be such that (p $\wedge$ q) $\Delta$ ((p $\vee$ q) $\Rightarrow$ q) is a tautology. Then $\Delta$ is equal to :
Let p, q, r be three logical statements. Consider the compound statements
${S_1}:(( \sim p) \vee q) \vee (( \sim p) \vee r)$ and
${S_2}:p \to (q \vee r)$
Then, which of the following is NOT true?
Which of the following statement is a tautology?
The boolean expression $( \sim (p \wedge q)) \vee q$ is equivalent to :
Let r $\in$ {p, q, $\sim$p, $\sim$q} be such that the logical statement
r $\vee$ ($\sim$p) $\Rightarrow$ (p $\wedge$ q) $\vee$ r
is a tautology. Then r is equal to :
Let $\Delta$, $\nabla $ $\in$ {$\wedge$, $\vee$} be such that p $\nabla$ q $\Rightarrow$ ((p $\Delta$ q) $\nabla$ r) is a tautology. Then (p $\nabla$ q) $\Delta$ r is logically equivalent to :
The negation of the Boolean expression (($\sim$ q) $\wedge$ p) $\Rightarrow$ (($\sim$ p) $\vee$ q) is logically equivalent to :
Consider the following two propositions:
$P1: \sim (p \to \sim q)$
$P2:(p \wedge \sim q) \wedge (( \sim p) \vee q)$
If the proposition $p \to (( \sim p) \vee q)$ is evaluated as FALSE, then :
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 number of choices for $\Delta \in \{ \wedge , \vee , \Rightarrow , \Leftrightarrow \} $, such that
$(p\Delta q) \Rightarrow ((p\Delta \sim q) \vee (( \sim p)\Delta q))$ is a tautology, is :
(S1) : (p $\to$ q) $ \vee $ ($ \sim $ q $\to$ p) is a tautology .
(S2) : (p $ \wedge $ $ \sim $ q) $ \wedge $ ($\sim$ p $\wedge$ q) is a fallacy.
Then :