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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
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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:
pq
TT
TF
FT
FF

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$.
pq$ \sim q$
TTF
TFT
FTF
FFT

Now, let us add $ \sim q \wedge p$ in the table:-
pq$ \sim q$$ \sim q \wedge p$
TTFF
TFTT
FTFF
FFTF

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:-
pq$ \sim q$$ \sim q \wedge p$$\left( { \sim q \wedge p} \right) \wedge q$
TTFFF
TFTTT
FTFFF
FFTFF

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:
pq$ \sim q$$ \sim q \wedge p$
TTFF
TFTT
FTFF
FFTF

Now, we will just have to find the value of $p \wedge \sim p$ :
pq$ \sim p$$p \wedge \sim p$
TTFF
TFFF
FTTF
FFTF

Now let us combine these two tables to get the following table:-
pq$ \sim q \wedge p$$p \wedge \sim p$$\left( { \sim q \wedge p} \right) \wedge \left( {p \wedge \sim p} \right)$
TTFFF
TFTFF
FTFFF
FFFFF

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:
pq$ \sim q$$ \sim q \wedge p$
TTFF
TFTT
FTFF
FFTF

Now, we will just have to find the value of $p \vee \sim p$ :
pq$ \sim p$$p \vee \sim p$
TTFT
TFFT
FTTT
FFTT

Now let us combine these two tables to get the following table:-
pq$ \sim q \wedge p$$p \vee \sim p$$\left( { \sim q \wedge p} \right) \vee \left( {p \vee \sim p} \right)$
TTFTT
TFTTT
FTFTT
FFFTT

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:-
pq$p \wedge q$
TTT
TFF
FTF
FFF

So, for this we will then get:-
pq$p \wedge q$$\left( {p \wedge q} \right) \wedge \left( {p \wedge q} \right)$
TTTT
TFFF
FTFF
FFFF

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”.