True

False

Tautology

Contradiction

Contingent

Not statement

$a _{r}=a _{n-r}$

$a _{2r}=a _{n-r}$

$a _{r}=a _{2n-r}$

$None\ of\ these$

The sky if blue or milk is white

The sky is blue and milk is white

The sky is white and milk is blue

If the sky is blue then milk is white

$\sim q\rightarrow \sim p$

$\sim p\rightarrow \sim q$

$q\rightarrow p$

$p\leftrightarrow q$

$\sim { p }{ \wedge }\sim q$

$\sim \left( p\vee q \right) \vee \left( p\vee \left( \sim p \right) \right) $

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

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

$p \vee q$

$p \wedge q$

$p \rightarrow  q$

none of these

$q \vee r$

$q \wedge r$

$q \rightarrow r$

none of these

$\sim (p\Rightarrow q)$

$\sim p$

$\sim p\Rightarrow q$

$\sim q$

If he is hard working, then he is not intelligent.

If he is not hard working, then he is intelligent.

If he is not intelligent, then he is not had working.

If he is not intelligent, then he is hard working.

$ p\Rightarrow q$

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

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

$ (\sim q)\Rightarrow p$

$p$

$\sim p$

$\sim q$

$q$

All slow learners attend this school.

All slow learners do not attend this school.

Some slow learners attend this school.

Some slow learners do not attend this school.

No slow learners do not attend this school.

$p\vee (\sim  q\wedge r)$

$p\vee q\wedge r$

$p\vee (\sim  q\wedge \sim r)$

$\sim p\vee (\sim q\wedge r)$

tautology.

contradiction.

neither a tautology nor a contradiction.

tautology and contradiction.

$\dfrac {1}{2}$ is rational or $\sqrt {3}$ is irrational

$\dfrac {1}{2}$ is not rational or $\sqrt {3}$ is not irrational

$\dfrac {1}{2}$ is not rational or $\sqrt {3}$ is irrational

$\dfrac {1}{2}$ is rational and $\sqrt {3}$ is irrational

$\sim q\Rightarrow p$

$p\Rightarrow q$

$\sim p\vee q$

$p\Leftrightarrow q$

$p \leftrightarrow q$

$p \rightarrow q$

$p \leftarrow q$

$p \vee q$

$p \vee q$

$p \rightarrow q$

$p \leftarrow q$

None of these

F, T, F

F, F, F

T, T, T

T, F, F

$(p\wedge ∼q)\wedge ∼p$

$(p\wedge ∼q)\vee  ∼p$

$(p\wedge ∼q)\vee ∼p$

none of these

If I do not have a Siberian Husky, then I do not have a dog.

If I have a dog, then I have a Siberian Husky.

If I do not have a dog, then I do not have a Siberian Husky.

If I do not have a Siberian Husky, then I have a dog.

$(p\vee q)$

$[p\wedge(q)]$

$p\wedge(q)$

$p\vee(q)$

a tautology

a contradiction

neither a tautology nor a contradiction

cannot come to any conclusion

you want of success and you find a way

you want of success and you do not find a way

if you do not want to succeed then you will find a way

if you want of success then you cannot find a way

$\left( { \sim p \vee q} \right) - \left( {p \vee \sim q} \right)$

$\left( { \sim p \vee \sim q} \right) \to p \vee q$

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

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

Open the door

What an intelligent student!

Are you going to Delhi

All prime numbers are odd numbers

neither tautology nor contradiction

A tautology

A contradiction

Cannot be determined

$p \to q$

$p \to (q \vee p)$

$p \to (q \to p)$

$p \to (q \wedge p)$

$(~p \vee ~q) \equiv (p \wedge q)$

$(p \rightarrow q) \equiv (~q \rightarrow ~p)$

$~(p \rightarrow ~q) \equiv (p \wedge ~q)$

$~(p \leftrightarrow q) \equiv (p \rightarrow q) \wedge (q \rightarrow p)$

$\sim p \wedge q$

$\sim p \vee q$

$p \wedge q$

$p \vee q$

$7$

$6$

$2$

$5$

$\sim p \vee q$

$\sim p \wedge q$

$p\vee \sim q$

$p\wedge \sim q$

$p \vee \sim p$

$T \rightarrow F$

$T \leftrightarrow F$

$p \wedge \sim p$

If $x$ is bass then $x$ is bad.

Boss implies bad,

Bad is necessary condition for bass.

$x$ is boss iff $x$ is bad.

$p \vee q$

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

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

none

$I$ & $II$

$I,\ II$ & $IV$

$III$

$All$

$S \rightarrow \sim Q$

$\sim Q \rightarrow S$

$\sim S \rightarrow \sim Q$

$Q \rightarrow \sim S$

Equivalent to $p\wedge q$

Tautology

Fallacy

Neither tautology nor fallacy

$\displaystyle k\left ( k-1-l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.

$\displaystyle k\left ( k+1-l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.

$\displaystyle k\left ( k-1+l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired zero.

$\displaystyle k\left ( k+1-l \right )= m\left ( n-k+1 \right )$ is equivalent to the desired one.

$\sim p \vee [\sim q \vee (p \vee q) \vee \sim r]$

$ p \vee [q \vee (\sim p \wedge \sim q) \vee r]$

$ \sim p \vee [\sim q \vee (p\wedge q) \vee \sim r]$

$ \sim p \vee [\sim q \wedge (p\wedge q) \wedge \sim r]$

$p \rightarrow \sim q $

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

$ \sim ( p \rightarrow \sim q)$

None of these

$p \rightarrow \sim q$

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

$\sim (p \rightarrow \sim q)$

None of these.

$(p \wedge q) \vee (p \vee q)$

$(p \rightarrow q) \vee (q \rightarrow p)$

$(\sim p \vee q) \vee (p \vee \sim q)$

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

$(~p \vee ~q) \equiv (p \wedge q)$

$(p \rightarrow q) \equiv (~q \rightarrow ~p)$

$~(p \rightarrow ~q) \equiv (p \wedge ~q)$

none of these

$ \sim (p \vee \sim q)$ and $ (\sim p \wedge q )$

$\sim (p \rightarrow q )$ and $(p \wedge \sim q )$

$(p \rightarrow q) $ and $(\sim q \rightarrow \sim p) $

$(p \rightarrow q )$ and $(\sim p \wedge q)$

$(p \wedge q) \vee (p \vee q)$

$(p \rightarrow q) \vee (q \rightarrow p)$

$(\sim p \vee q) \vee (p \vee \sim q)$

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

$(\sim p\vee q)$

$(p\vee \sim q)$

$(\sim p\wedge q)$

$(p\wedge \sim q)$

$\sim(p\rightarrow q) \equiv \sim p \wedge q$

$\sim(p\vee q) \equiv \sim p \vee \sim q$

$\sim (p \implies q ) \equiv (p \land \sim q )$

$\sim(p \wedge q) \equiv \sim p \wedge \sim q$

$p\rightarrow q\equiv\sim p\rightarrow\sim q$

$\sim(p \rightarrow\sim q)\equiv\sim p\wedge q$

$\sim(\sim p\rightarrow\sim q)\equiv\sim p\wedge q$

$\sim (p\leftrightarrow q) \equiv(\sim(p\rightarrow q))\wedge\sim(q\rightarrow p)$

$\displaystyle p\wedge q$

$\displaystyle p\wedge \sim q$

$\displaystyle \sim p\wedge q$

$\displaystyle \sim p\wedge \sim q$

Reena is not beaufiful and Meena is not healthy.

Reena is not beautiful or Meena is not healthy.

Reena is not healthy or Meena is not beautiful.

None of these.

$2^2 = 5$ and I do not get first class

$2^2 = 5$ or I do not get first class

$2^2 \neq 5$ or I get first class

None of these.

$2^2 = 5$ and I donot get first class

$2^2 = 5$ or I do not get first class

$2^2 \neq 5$ or I get first class

None of these

$(p \rightarrow q)\wedge (q \rightarrow p)$

$(p \wedge q)\vee (q \rightarrow p)$

$(p \wedge q)\rightarrow (q \vee p)$

none of these

You watch television if and only if your mind is free.

You watch television and your mind is free.

You watch television or your mind is free.

None of these

$\sim[p\vee (\sim q)]=(\sim p)\wedge q$

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

$q\wedge \sim q$ is a contradiction

$\sim (p\wedge (\sim p))$ is a tautology

Tautology

Contradiction

Neither tautology nor contradiction

None of these

Equivalent to $\sim p \leftrightarrow q$

A tautology

A fallacy

Equivalent to $p \leftrightarrow q$

neither tautology nor contradiction

A tautology

A contradiction

Cannot be determined

$A\wedge \left( A\vee B \right) $

$A\vee \left( A\wedge B \right) $

$[A\wedge (A\rightarrow B)]\rightarrow B$

$B\rightarrow [A\wedge (A\vee B)]$

$145^{o}$

$150^{o}$

$155^{o}$

$160^{o}$

$6\alpha$

$12\alpha$

$20\alpha$

$24\alpha$

$(i)$True and $(ii)$False

$(i)$False and $(ii)$True

$(i)$True and $(ii)$True

$(i)$False and $(ii)$False

$(i)$True $(ii)$False

$(i)$True $(ii)$True

$(i)$False $(ii)$False

$(i)$False $(ii)$True

(p$\wedge$ q) $\vee$ T

(p$\wedge$ q) $\vee$ F

(p$\vee$ q) $\vee$ F

(p$\vee$ q) $\vee$ T

(i) True (ii) False

(i) False (ii) True

(i) True (ii) True

(i) False (ii) False

Idempotent Law

Commutative Law

Associative Law

Conditional Law

$\sim(p\rightarrow\sim q)$

$(p\rightarrow\sim q)$

$(\sim p\rightarrow\sim q)$

$(\sim p\rightarrow q)$

$\left( { \sim q \wedge r} \right) \to \sim p$

$\left( {q \wedge \sim r} \right) \to \sim p$

$p \to \left( { \sim r \vee q} \right)$

$p \wedge \left( {q \vee r} \right)$

$p \wedge \sim q$

$p \wedge q$

$\sim p \wedge \sim q$

$\sim p \vee \sim q$

Statement I is true and Statement II is the correct explanation for statement.

Statement I is true and Statement II is true. Statement II is not the correct explanation for the Statement I.

Statement I is true but Statement II is false.

Statement I is false but Statement II is true

$ \sim p$

p

q

$ \sim q$

$10000\ Rs$

$100000\ Rs$

$1000000\ Rs$

$2000000\ Rs$

$\left( {p \to q} \right) \cong \left( { \sim q \to \sim p} \right)$

$ \sim \left( {p \vee q} \right) \cong \left( { \sim p \vee \sim q} \right)$

$ \sim \left( {p \to q} \right) \cong \left( {p \vee \sim q} \right)$

$ \sim \left( {p \wedge q} \right) \cong \left( { \sim p \wedge \sim q} \right)$

$p$

$q$

$ \sim q $

$ \sim p $

True

False

$q$

$p\vee q$

$p$

$p\vee \sim q$

$p \vee \sim q$

$p \wedge \sim q$

$ \sim p \vee q$

$ \sim p \wedge q$

True

False

$\left(p \rightarrow q\right)\leftrightarrow \left( \sim q \rightarrow \sim p \right) $

$\left[ \left( p\rightarrow q \right) \wedge \left( q\rightarrow r \right) \right]\rightarrow \left( p\rightarrow r \right) $

$\left( \sim p\vee q \right) \leftrightarrow \left( p\wedge \sim q \right) $

$p \rightarrow \sim p$

$p \vee ( q \wedge r )$

$( p \wedge q ) \wedge r$

$( p \vee q ) \vee r$

$p \rightarrow ( q \wedge r )$

$p \text { is true and } q \text{ is true}$

$p \text { is false and } q  \text{ is true}$

$p \text { is true and } q \text{ is false}$

both $p$ and $q$ are false

Mathematics is interesting implies that Mathematics is difficult

Mathematics is interesting implies and is implied by Mathematics is difficult

Mathematics is interesting and Mathematics is difficult

Mathematics is interesting or Mathematics is difficult

$p\vee \left( \sim q \right) \vee \sim p$

$\left( p\vee \sim q \right) \vee \sim p$

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

none of these

$p \rightarrow \sim q$

$q \rightarrow \sim p$

$q \rightarrow p$

$\sim p \rightarrow \sim q$

$\sim q \rightarrow \sim p$

$(a)False$ and $(b)$ False

$(a)True$ and $(b)$ False

$(a)False$ and $(b)$ True

$(a)True$ and $(b)$ True

p$\rightarrow$q and $F$

q$\rightarrow$p and $T$

q$\rightarrow$p and $F$

p$\rightarrow$q and $T$

DeMorgan's Law

Conditional Law

Involution Law

Complement Law

$\sim P\vee Q$

$\sim P\wedge Q$

$\sim Q\wedge P$

$\sim Q\vee P$

Complement Law

Identity Law

Involution Law

Absorption Law

True

False

Associative law

Commutative Law

Involution Law

Conditional Law

Commutative Law

DeMorgan's Law

Complement Law

Conditional Law

$\sim p \vee q$

$ p \wedge \sim q$

$\sim p \wedge q$

$ p \vee q$

$\sim p\vee \sim q$

$ p\vee \sim q$

$\sim p\vee q$

$\sim p\wedge q$

$p \leftrightarrow q$

$p \rightarrow q$

$p \leftrightarrow \sim q$

$\sim p \rightarrow q$

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

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

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

None of these

Conditional Law

Demorgan's Law

Absorption Law

Identity Law

Idempotent Law

Conditional Law

Involution Law

Commutative Law

F, F

T, T

F, T

T, F

True

False

$\sim q$

$q$

$\sim p$

$p$

True

False

$~ q \rightarrow p$

$ ~ p v q$

$p \leftrightarrow q$

$p \rightarrow q$