
Which of the following statements is a tautology?
A. $\left( { \sim q \wedge p} \right) \wedge q$
B. $\left( { \sim q \wedge p} \right) \wedge \left( {p \wedge \sim p} \right)$
C. $\left( { \sim q \wedge p} \right) \vee \left( {p \vee \sim p} \right)$
D. $\left( {p \wedge q} \right) \wedge \left( {p \wedge q} \right)$
Answer
501.9k+ views
Hint: We will first create all the possible values and everything in a table. After that, we will find the possible values of all the inside values inside the parenthesis.
Complete step-by-step solution:
Let us first consider the possibilities. We have two statements here, p and q. Both can be true and false both. So, we can 2 into 2 = 4 possibilities.
So, let us create a table for this:
Here, we now have the required table.
Option A: Now the first option given to us is: $\left( { \sim q \wedge p} \right) \wedge q$
Now, here we have two statements p and q.
Now, we will find the possibilities of $ \sim q$.
Now, let us add $ \sim q \wedge p$ in the table:-
Now, we will just get the truth table for the first option that is $\left( { \sim q \wedge p} \right) \wedge q$ using this data in above table:-
Option B: Now we have the option $\left( { \sim q \wedge p} \right) \wedge \left( {p \wedge \sim p} \right)$.
For this we will use the table:
Now, we will just have to find the value of $p \wedge \sim p$ :
Now let us combine these two tables to get the following table:-
Option C: Now we have the option $\left( { \sim q \wedge p} \right) \vee \left( {p \vee \sim p} \right)$
For this we will use the table:
Now, we will just have to find the value of $p \vee \sim p$ :
Now let us combine these two tables to get the following table:-
Option D: Now we have the option $\left( {p \wedge q} \right) \wedge \left( {p \wedge q} \right)$.
For this we will need to use the following table:-
So, for this we will then get:-
Now, tautology is a statement which is always true.
Hence, the answer is option C) $\left( { \sim q \wedge p} \right) \vee \left( {p \vee \sim p} \right)$.
Note: The students must know that Tautology is a statement which is always true.
Here, we can clearly see that since in option C we have $p \vee \sim p$ which is no matter what is always going to be true always. Hence, we have the option C as a tautology.
The word “tautology” comes from the work “the same”.
Complete step-by-step solution:
Let us first consider the possibilities. We have two statements here, p and q. Both can be true and false both. So, we can 2 into 2 = 4 possibilities.
So, let us create a table for this:
p | q |
T | T |
T | F |
F | T |
F | F |
Here, we now have the required table.
Option A: Now the first option given to us is: $\left( { \sim q \wedge p} \right) \wedge q$
Now, here we have two statements p and q.
Now, we will find the possibilities of $ \sim q$.
p | q | $ \sim q$ |
T | T | F |
T | F | T |
F | T | F |
F | F | T |
Now, let us add $ \sim q \wedge p$ in the table:-
p | q | $ \sim q$ | $ \sim q \wedge p$ |
T | T | F | F |
T | F | T | T |
F | T | F | F |
F | F | T | F |
Now, we will just get the truth table for the first option that is $\left( { \sim q \wedge p} \right) \wedge q$ using this data in above table:-
p | q | $ \sim q$ | $ \sim q \wedge p$ | $\left( { \sim q \wedge p} \right) \wedge q$ |
T | T | F | F | F |
T | F | T | T | T |
F | T | F | F | F |
F | F | T | F | F |
Option B: Now we have the option $\left( { \sim q \wedge p} \right) \wedge \left( {p \wedge \sim p} \right)$.
For this we will use the table:
p | q | $ \sim q$ | $ \sim q \wedge p$ |
T | T | F | F |
T | F | T | T |
F | T | F | F |
F | F | T | F |
Now, we will just have to find the value of $p \wedge \sim p$ :
p | q | $ \sim p$ | $p \wedge \sim p$ |
T | T | F | F |
T | F | F | F |
F | T | T | F |
F | F | T | F |
Now let us combine these two tables to get the following table:-
p | q | $ \sim q \wedge p$ | $p \wedge \sim p$ | $\left( { \sim q \wedge p} \right) \wedge \left( {p \wedge \sim p} \right)$ |
T | T | F | F | F |
T | F | T | F | F |
F | T | F | F | F |
F | F | F | F | F |
Option C: Now we have the option $\left( { \sim q \wedge p} \right) \vee \left( {p \vee \sim p} \right)$
For this we will use the table:
p | q | $ \sim q$ | $ \sim q \wedge p$ |
T | T | F | F |
T | F | T | T |
F | T | F | F |
F | F | T | F |
Now, we will just have to find the value of $p \vee \sim p$ :
p | q | $ \sim p$ | $p \vee \sim p$ |
T | T | F | T |
T | F | F | T |
F | T | T | T |
F | F | T | T |
Now let us combine these two tables to get the following table:-
p | q | $ \sim q \wedge p$ | $p \vee \sim p$ | $\left( { \sim q \wedge p} \right) \vee \left( {p \vee \sim p} \right)$ |
T | T | F | T | T |
T | F | T | T | T |
F | T | F | T | T |
F | F | F | T | T |
Option D: Now we have the option $\left( {p \wedge q} \right) \wedge \left( {p \wedge q} \right)$.
For this we will need to use the following table:-
p | q | $p \wedge q$ |
T | T | T |
T | F | F |
F | T | F |
F | F | F |
So, for this we will then get:-
p | q | $p \wedge q$ | $\left( {p \wedge q} \right) \wedge \left( {p \wedge q} \right)$ |
T | T | T | T |
T | F | F | F |
F | T | F | F |
F | F | F | F |
Now, tautology is a statement which is always true.
Hence, the answer is option C) $\left( { \sim q \wedge p} \right) \vee \left( {p \vee \sim p} \right)$.
Note: The students must know that Tautology is a statement which is always true.
Here, we can clearly see that since in option C we have $p \vee \sim p$ which is no matter what is always going to be true always. Hence, we have the option C as a tautology.
The word “tautology” comes from the work “the same”.
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