Tag: different forms of theoretical statements

Questions Related to different forms of theoretical statements

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

  1. $\displaystyle p\wedge q$

  2. $\displaystyle p\wedge \sim q$

  3. $\displaystyle \sim p\wedge q$

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


Correct Option: D
Explanation:
$\sim p$  $\sim q$  $\sim p \rightarrow q$  $\sim (\sim p \rightarrow q)$  $p \wedge q$  $p \wedge \sim q$   $\sim p \wedge q$   $\sim p \wedge \sim q$  
T
F
F

The values in column 6 and column 10 are same.

Hence, option D is correct.

The dual of the following statement "Reena is healthy and Meena is beautiful" is

  1. Reena is not beaufiful and Meena is not healthy.

  2. Reena is not beautiful or Meena is not healthy.

  3. Reena is not healthy or Meena is not beautiful.

  4. None of these.


Correct Option: C
Explanation:

Let $p$ denote the statement "Reena is healthy" 


and $q$ denote the statement "Meena is beautiful"

Now the given statement is $p\wedge q$

Now the Dual of this statement will be obtained by replacing $\vee$ by 

$\wedge$ and $\wedge$ by $\vee$ and inversing the true value of the statement.

So the Dual of $p\wedge q$ will be $\sim p\vee \sim q$

The statement $\sim p$ will be "Reena is not healthy"

The statement $\sim q$ will be "Meena is not beautiful"

So the dual statement will be $\sim p\vee \sim q$ or "Reena is not healthy or Meena is not beautiful."

The statement "If $2^2 = 5$ then I get first class" is logically equivalent to

  1. $2^2 = 5$ and I do not get first class

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

  3. $2^2 \neq 5$ or I get first class

  4. None of these.


Correct Option: C
Explanation:

There can be two cases
$2^{2}=5$ $\rightarrow$ first class.
$2^{2}\neq 5$\rightarrow not a first class.
Hence logically equivalent statement will be 
$2^{2}=5$ or $2^{2}\neq 5$ but $2^{2}=5$ statement is equivalent to getting first class.
Hence
First class or $2^{2}\neq 5$.

The statement "If $2^2 = 5$ then I get first class" is logically equivalent to

  1. $2^2 = 5$ and I donot get first class

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

  3. $2^2 \neq 5$ or I get first class

  4. None of these


Correct Option: C
Explanation:

Obviously, ${ 2 }^{ 2 }\neq 5$, then the statement will be ${ 2 }^{ 2 }\neq 5$ or $I$ get first class.

Logically equivalent statement to $p \leftrightarrow  q$ is

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

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

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

  4. none of these


Correct Option: A
Explanation:
 $p$  $q$  $p\leftrightarrow q$
 T  T  T
 T  F  F
 F  T  F
 F  F  T
 $p$  $q$  $p\rightarrow q$  $q\rightarrow p$ $\left( p\longrightarrow q \right) \wedge \left( q\longrightarrow p \right) $ $p\wedge q$  $\left( p\wedge q \right) \vee \left( q\longrightarrow p \right) $ $q\vee p$  $\left( p\wedge q \right) \longrightarrow \left( q\vee p \right) $ 
 T  T  T  T  T  T  T  T  T
 F  F  T  F  F  T  T  T
 F  T  T  F  F  F  F  T  T
 F  T  T  T  F  T  F  T

Which one of the statement gives the same meaning of statement
If you watch television, then your mind is free and if your mind is free then you watch television

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

  2. You watch television and your mind is free.

  3. You watch television or your mind is free.

  4. None of these


Correct Option: B
Explanation:
"You watch television and your mind is free".
The above statement gives or suits for the same meaning of the structure given because it is logically correct.

Which of the following is NOT true for any two statements $p$ and $q$?

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

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

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

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


Correct Option: B
Explanation:
$p$ and $q$ are two statements.
$A) LHS = \sim [pv (\sim q)]$
By De morgon's laws
$\sim(pr (\sim q))= \sim pnq$
$\therefore (A) $ is true .

$B) \sim(p v q) = (\sim p) \vee (\sim q)$
According to demorgon's laws, this is false.
$\because \sim (p \vee q) = (\sim p)\wedge (\sim q)$. 
$\therefore (B)$ is false.

$C) q \wedge \sim  q$ is a contradiction because $'q'$ and $\sim q$ are opposite statements i.e, cannot be there at the same time.

$D) \sim (p \wedge (\sim p))$
$p \wedge (\sim p)$ is a contradiction, which is evident from option $(C)$. $\therefore $ opposite of a contradiction is a tautology .
$\therefore [B]$ is wrong.

If p and q are two statements, then statement $p\Rightarrow q\wedge \sim q$.

  1. Tautology

  2. Contradiction

  3. Neither tautology nor contradiction

  4. None of these


Correct Option: A

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

  1. Equivalent to $\sim p \leftrightarrow q$

  2. A tautology

  3. A fallacy

  4. Equivalent to $p \leftrightarrow q$


Correct Option: C

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

  1. neither tautology nor contradiction

  2. A tautology

  3. A contradiction

  4. Cannot be determined


Correct Option: C