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Logical equivalence - class-XII

Description: logical equivalence
Number of Questions: 42
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$\left( {p \Rightarrow q} \right) \to \left[ {\left( {r \vee p} \right) \Rightarrow \left( {r \vee q} \right)} \right]$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a contradiction

  2. a tautology

  3. a tautology and a contradiction

  4. neither a tautology nor a contradiction

Show answer
Correct Option: A

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

logical equivalence mathematical logic discrete mathematics business maths maths

  1. tautology

  2. contradiction

  3. dualoty

  4. double implication

Show answer
Correct Option: A

Which of the following is not correct ?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p \vee \sim p $ is a tautology.

  2. $\sim (\sim p) \leftrightarrow p$ is a tautology.

  3. $p \wedge \sim p $ is a contradiction.

  4. $([(p \wedge p ) \rightarrow q] \rightarrow p)$ is a tautology.

Show answer
Correct Option: B,D
Explanation:

$\because [(p \wedge p ) \rightarrow q ] \rightarrow p \equiv ( p

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

when $p$ is false and $q$ is true (or false) then

$(p \rightarrow q) $ is true i.e. $(p \rightarrow q ) ]\rightarrow p$ is false

Hence $[(p \wedge p ) \rightarrow q] \rightarrow p$ is not a tautology.

Which of the following statement is a tautology?

logical equivalence mathematical logic discrete mathematics business maths maths

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

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

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

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

Show answer
Correct Option: A
Explanation:

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

$\equiv (\sim p \vee p ) \vee (p \vee \sim q )$    (by distributive law)

$ \equiv t \vee \sim q \equiv t $                      t is a tautology

Hence $(\sim p  \vee \sim, q ) \vee (p  \vee \sim q )$ is a tautology.

$ p\Rightarrow p \vee q$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.

Show answer
Correct Option: A
Explanation:
$p$       $q$      $p\vee q$       $p\rightarrow (p\vee q)$
T T     T            T
T F    T            T
F T    T              T
F F     F            T

Since, all the entries in the last column has true value. So, the given statement is a tautology

$p\Rightarrow \sim p$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.

Show answer
Correct Option: D
Explanation:
 $p$  $\sim p$  $p\Rightarrow \sim p$
 T  F  F
 F  T  T

So the result of the Truth table show that $p\Rightarrow \sim p$ is neither a tautology not a contradiction

$ p\wedge  (\sim p)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.

Show answer
Correct Option: B
Explanation:

Truth table is,

$p$         $\sim p$ $p\wedge (\sim p)$
T                      F                      F
F T F

Hence given statement is contradiction.

Which of the following is a contradiction?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p\vee q$

  2. $p\wedge q$

  3. $p\vee (\sim p)$

  4. $p\wedge (\sim p)$

Show answer
Correct Option: D
Explanation:
$p$  $q$  $p\vee q$  $p\wedge q$
$ T$ $ T$ $ T$ $ T$
$ T$ $ F$ $ T$ $ F$
$ F$ $T$ $T$ $F$
$ F$ $F$ $ F$ $ F$
So $p\vee q$ and $p\wedge q$ are not contradiction 

Now check other options:

| $p$ |  $\sim p$ |  $p\vee (\sim p)$ |  $p\wedge (\sim p)$ | | --- | --- | --- | --- | |  $T$ | $ F$ | $ T$ | $ F$ | | $ F$ | $T$ | $T$ | $F$      |
Hence $p\wedge (\sim p)$ is contradiction 

Note: If a compound statement is always False , then it is called contradiction 

The proposition $(p\rightarrow \sim p)\wedge (\sim p\rightarrow q)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. a tautology and a contradiction

Show answer
Correct Option: C
Explanation:
$p$      $q$      $\sim p$      $p\rightarrow \sim p$     $\sim p \rightarrow q$     $(p\rightarrow \sim p)\wedge(\sim p\rightarrow q)$
T T    F      F       T         F
T F    F      F       T         F
F T    T      T       T         T
F F    T      T       F         F

The statement $(p-q)\rightarrow [(\sim p \rightarrow q)\rightarrow q]$ is 

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology

  2. equivalent to $\sim p \rightarrow q$

  3. equivalent to $p\rightarrow \sim q$

  4. a fallacy

Show answer
Correct Option: A

Which of the following proposition is a contradiction?

logical equivalence mathematical logic discrete mathematics business maths maths

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

  2. $(p\rightarrow q)\vee (p\wedge \sim q)$

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

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

Show answer
Correct Option: C
Explanation:

Contradiction is a preposition which is always false(F).
Here $\sim \equiv negation$ and $\wedge \equiv AND$
Let $ x=(\sim p \wedge q) \wedge (\sim q)$
If $p=F$ and $q=F$ then $x=F$
If $p=F$ and $q=T$ then $x=F$
If $p=T$ and $q=F$ then $x=F$
If $p=T$ and $q=T$ then $x=F$
Hence option (c) is correct

Which of the following is not true (where $p, q$ and $r$ take truth values and $t$ is a tautology, $c$ is a contradiction)

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p\wedge p\equiv p$

  2. $p\vee t=t$

  3. $p\wedge c= p$

  4. $p\vee (q\wedge r)=(p\vee q)\vee (p\vee r)$

Show answer
Correct Option: A

$p,q,r$ are $3$ statement such that $(p\rightarrow q)\wedge (q\rightarrow r)\Rightarrow (p\rightarrow r)$ is 

logical equivalence mathematical logic discrete mathematics business maths maths

  1. Tautology

  2. Contradiction

  3. $P\wedge q$

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

Show answer
Correct Option: A

The statement $[p \wedge (p \rightarrow q)]\rightarrow q$,is :

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a fallacy

  2. a tautology

  3. neither a fallacy nor a tautology

  4. not a compound statement

Show answer
Correct Option: B
Explanation:

The statement

$[p \wedge (p \rightarrow q)] \rightarrow q$
a tautology 

$p,q,r$ are $3$ statement such that $(p \rightarrow q)\wedge (q \rightarrow r)\Rightarrow (P \rightarrow r)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. Tautology

  2. Contradiction

  3. $P \wedge q$

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

Show answer
Correct Option: A

Which of the following is a tautology?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p\wedge (\sim p)$

  2. $p\wedge c$

  3. $p\vee t$

  4. $p\wedge p$

Show answer
Correct Option: A

The proposition $p\vee (\sim p\vee q)$ is a 

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tatutology

  2. a contradiction

  3. Logically equivalent to $p$ & $q$

  4. both $1$ & $2$

Show answer
Correct Option: A

The only statement among the followings that is a tautology is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $A\vee(A\wedge B)$

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

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

  4. $A\wedge (A\vee B)$

Show answer
Correct Option: A

The simplifed form of $(p \vee q)\vee (\sim p \wedge q)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $T$

  2. $p \wedge q$

  3. $F$

  4. $p \vee q$

Show answer
Correct Option: A

If p, q two propositions then $(p \vee \sim q) \wedge ( \sim p \wedge q)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. both a tautology and a contradiction

Show answer
Correct Option: A

The only statement among the following taht is a tautology is -

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $A\wedge (A\vee B)$

  2. $A\vee (A\wedge B)$

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

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

Show answer
Correct Option: A

The contrapositive of the statement "if  $2 ^ { 2 } = 5 ,$  then  $1$  get first class" is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. If I do not get a first class, then $2 ^ { 2 } = 5$

  2. If I do not get a first class, then $2 ^ { 2 } \neq 5$

  3. If I get a first class, then $2 ^ { 2 } = 5$

  4. If I get a first class, then $2 ^ { 3 } = 5$

Show answer
Correct Option: B
Explanation:

$P:{ 2 }^{ 2 }=5$

$q:I$ get first class
the contrapositive of $p\rightarrow q$ is $\sim q\rightarrow \sim p$. Hence the answer is if $I$ do not get a first class, then ${ 2 }^{ 2 }\neq 5$
Correct Answer : Option B.

The proposition $( P \Longrightarrow \sim p) ^ (\sim p \Longrightarrow P)$ is 

logical equivalence mathematical logic discrete mathematics business maths maths

  1. Contingency

  2. Neither Tautology nor contradiction

  3. contradiction

  4. Tautology

Show answer
Correct Option: A

Which of  the following is logically equivalent to : $\sim \left[\sim p\rightarrow q\right]$

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p\vee\sim q$

  2. $\sim p\wedge q$

  3. $\sim p\vee q$

  4. $\sim p\wedge \sim q$

Show answer
Correct Option: A

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

logical equivalence mathematical logic discrete mathematics business maths maths

  1. is a tautology

  2. is equivalent to $( p \wedge q ) \vee ( - q )$

  3. is equivalent to $p \vee q$

  4. is a contradiction

Show answer
Correct Option: A

The simplicity $ \sim(p \rightarrow q) \longleftrightarrow(\sim p \vee \sim q) $ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. tautology

  2. contradiction

  3. neither t nor e

  4. None of these.

Show answer
Correct Option: A

Consider :
Statement - I :$(p\wedge \sim q)\wedge (\sim p\wedge q)$ is a fallacy.
Statement - II :$(p\rightarrow q)\leftrightarrow (\sim q\rightarrow \sim p)$ is a tautology.

logical equivalence mathematical logic discrete mathematics business maths maths

  1. Statement - I is true: Statement - II is true: Statement - II is a correct explanation for Statement - I.

  2. Statement - I is true: Statement - II is true: Statement - II is not a correct explanation for Statement - I.

  3. Statement - I is true; Statement - II is false.

  4. Statement - I is false; Statement - II is true.

Show answer
Correct Option: A

The statement (p ^ q) ^ (-pv - q) is _______________.

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology

  2. a contradiction

  3. a contingency

  4. neither a tautology nor a contradiction

Show answer
Correct Option: A

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

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. neither a tautology nor a contradiction.

  4. none of these.

Show answer
Correct Option: A

The statement $\sim (p \rightarrow q) \leftrightarrow  (\sim p \vee \sim q)$ is 

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. None of these

Show answer
Correct Option: C
Explanation:

When $p$ and $q$ both are true then 

$(p \rightarrow  q) and (\sim p \vee \sim q)$ both are false

i.e. $\sim (p \rightarrow  q) \leftrightarrow  (\sim p \vee  \sim q)$ is true when $p$ and $q$ both are false then  

$\sim (p \rightarrow  q)$ is false and $(\sim p \vee \sim q)$ is true

i.e. $\sim (p \rightarrow  q) \leftrightarrow  (\sim p \vee  \sim q)$ is false

Hence $\sim (p \rightarrow  q) \leftrightarrow  (\sim p \vee  \sim q)$ is neither tautology nor contradiction

Which of the following statement is a contradiction ?

logical equivalence mathematical logic discrete mathematics business maths maths

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

  2. $(p \rightarrow q) \vee (p \wedge \sim q)$

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

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

Show answer
Correct Option: C

If $p$ is any statement, $t$ is a tautology and $c$ is a contradiction, then which for the following is NOT correct?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p \wedge (\sim c) \equiv p$

  2. $p \vee (\sim t) \equiv p$

  3. $t \vee c \equiv p \vee t$

  4. $(p\wedge t) \vee (p \vee c) \equiv (t \wedge c)$

Show answer
Correct Option: D
Explanation:

A.   $p \wedge (\sim c )\equiv p \wedge t  \equiv p$

B.   $p \vee (\sim t )\equiv p \vee c \equiv p$

C.   $ t \vee c \equiv t \equiv p \vee t$

Clearly option 'A', 'B', 'C' are correct.

Hence option 'D' is the correct choice.

If $p$ is any statement, $t$ and $c$ are a tautology and a contradiction respectively, then which of the following is INCORRECT?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p \wedge t \equiv p $

  2. $ p \wedge c \equiv c$

  3. $p \vee t \equiv p $

  4. $ p \vee c \equiv p$

Show answer
Correct Option: C
Explanation:

Truth table,

$p$             $p\wedge t$        $p\wedge c$          $p\vee  t$      $p\vee c$
T T            F               T          T
F F            F            T         F

Hence option 'C' is the correct choice.

The statement $(p \rightarrow ~p) \wedge (~ p \rightarrow p)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. neither a tautology nor a contradiction.

  4. None of these.

Show answer
Correct Option: A
Explanation:

If $p$ is a true statement, then $p \rightarrow p$ is true.

Also, if $p$ is a false statement, then $p \rightarrow p$ is true.
Then, $(p \rightarrow p) \wedge (p \rightarrow p)$ is always true.
Hence, the given statement is a tautology.

Which of the following statement is a contradiction?

logical equivalence mathematical logic discrete mathematics business maths maths

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

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

  3. $(p \rightarrow q) \rightarrow p$

  4. $\sim p \vee \sim q$

Show answer
Correct Option: A
Explanation:

We check for contradiction for all the given options.

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

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

All F so this is a contradiction.


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

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

So not a contradiction.


C. $\left( p\longrightarrow q \right) \rightarrow p$

 $p$  $q$  $\left( p\longrightarrow q \right) $  $\left( p\longrightarrow q \right) \rightarrow p$
 T  T  T  T
 T  F  F  T
 F  T  T  F
 F  F  T  F

So it is also not a contradiction.


D. $\sim p\vee \sim q$

 $p$  $q$ $\sim p$ $\sim q$ $\sim p\vee \sim q$
 T  T  F  F  F
 T  F  F  T  T
 F  T  T  T  T
 F  F  T  T  T

So it is also not a contradiction.

The statement $\sim (p \rightarrow q )\leftrightarrow (\sim p \vee \sim q)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. neither a tautology nor a contradiction.

  4. None of these.

Show answer
Correct Option: C
Explanation:

When $p$ and $q$ both are true then

$\sim (p \rightarrow q ) $ and $(\sim p \vee \sim q) $ both are false

i.e. $\sim (p \rightarrow q ) \leftrightarrow (\sim p \vee \sim q ) $ is true when $p$ and $q$ both are false then $\sim (p \rightarrow q ) $ is false and $(\sim p \vee \sim q)$ is true 

i.e. $\sim (p \rightarrow q ) \leftrightarrow (\sim p \vee \sim q)$ is false 

Hence $ \sim (p \rightarrow q ) \leftrightarrow (\sim p \vee \sim q) $ is neither tautology nor contradiction.

Which of the following is a tautology?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p\implies p\wedge q$

  2. $p\implies p\vee q$

  3. $(p\vee q)\implies(p\wedge q)$

  4. None of these

Show answer
Correct Option: B
Explanation:

A tautology is a statement that is always true.
The statement $ p\Longrightarrow p\vee q $ is read as " if p is true, then either p or q is true. " as the symbol $ \vee  $ denotes OR.
Hence, the given statement is a tautology.

Which of the following statements is/are true?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p\wedge (\sim p)$ is a contradiction.

  2. $(p\rightarrow q)\Leftrightarrow (\sim q \rightarrow \sim p)$ is a contradiction.

  3. $\sim(\sim p) \Leftrightarrow p$ is a tautology.

  4. $p\vee (\sim p)$ is a tautology.

Show answer
Correct Option: A,C,D
Explanation:

A,C,D obvious and for B
p $\rightarrow$ q is same as $\sim$ q $\rightarrow$ $\sim$p
$\therefore$ it is tautology not contradiction.

If $p$ is any statement $t$ and $c$ are tautology and contradiction respectively, then which of the following is(are) correct?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p\wedge t \equiv p$

  2. $p \wedge c \equiv c$

  3. $p\vee t \equiv c$

  4. $p \vee c \equiv p$

Show answer
Correct Option: A,B,D
Explanation:

Tautology is the preposition which is always true and contradiction is a preposition which is always false.
Here $\wedge \equiv AND, \vee \equiv OR$, $T$=true and $F$=false
Given statement=$p$, tautology=$t$ and contradiction=$c$
a)$p\wedge t\equiv p$ irrespective of value of $p$ as $t=T$ always
And if $p=T$ then $T\wedge T=T$ and if $p=F$ then $F\wedge T=F$
Thus option (a) is correct.
b)$p\wedge c\equiv c$ irrespective of value of $p$ as$ c=F$ always
And if $p=T$ then $T\wedge F=F$ and if $p=F$ then $F\wedge F=F$
Thus option (b) is correct
c)$p\vee t\equiv c$
If $p=T$ then $T\vee T\equiv T\neq c$ and if $p=F$ then $F\vee T\equiv T \neq c$
Thus option (c) is not correct
d)$p\vee c\equiv p$
If $p=T$ then $T\vee F\equiv T$ and if $p=F$ then $F\vee F\equiv F$
Thus it depends on the value of $p$
Hence option (d) is correct

Which one of the following statements is a tautology?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $\left( p\vee q \right) \rightarrow q$

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

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

  4. $p\rightarrow (p\rightarrow q)$

Show answer
Correct Option: B
Explanation:
$ p$  $q$  $p\to q$  $q\to p$ $pvq$  $p\to (p\to q)$  $(pvq)\to q$  $pv(p\to q)$   $pv(q\to q$
 T  T  T  T  T  T  T  T  T
 T  F  F  T  T  F  F  T  T
 F  T  T  F  T  T  T  T  F
 F  F  T  T  F  T  T  T  T

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

If $p$ and $q$ are two statement, then $(p  \wedge \sim q) \wedge   (\sim p  \wedge   q)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a fallacy

  2. a tautology

  3. neither tautology nor a fallacy

  4. none of these

Show answer
Correct Option: A
Explanation:
$p$         $q$          $p \land \sim q$       $q \land \sim p $          $(p \land \sim q ) \land(q \land \sim p )$
T T     F       F          F
T F     T       F          F
F T      F       T          F
F F     F       F          F


$\therefore\ (p \land \sim q ) \land(q \land \sim p )$ is a fallacy

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