edited by
1,839 views
6 votes
6 votes

Match List-I with List-II:

$$\begin{array}{|c|c|c|c|} \hline {} & \text{List-I} & {} & \text{List-II} \\ \hline  (a) & p \rightarrow q & (i) & \rceil ( q \rightarrow \rceil p) \\ \hline  (b) & p \vee q & (ii) & p \wedge \rceil q \\ \hline  (c) & p \wedge q & (iii) & \rceil p \rightarrow q \\ \hline (d) &  \rceil ( p \rightarrow q) & (iv) & \rceil p \vee q \\ \hline \end{array}$$

Choose the correct option from those given below:

  1. (a) – (ii); (b) – (iii); (c) – (i); (d) – (iv)
  2. (a) – (ii); (b) – (i); (c) – (iii); (d) – (iv)
  3. (a) – (iv); (b) – (i); (c) – (iii); (d) – (ii)
  4. (a) – (iv); (b) – (iii); (c) – (i); (d) – (ii)
edited by

3 Answers

2 votes
2 votes
  1. $p\rightarrow q$ = $\neg p \vee q$
  2. $p\vee q$
  3. $p\wedge q$
  4. $\neg(p\rightarrow q)$ = $\neg (\neg p \vee q) = p\ \wedge \neg q$

  1.  $\neg(q\rightarrow \neg p)$ = $\neg (\neg q\ \vee \neg p) = q\ \wedge p$
  2. $p\ \wedge \neg q$
  3. $\neg p\rightarrow q$ = $ p \vee q $
  4. $ \neg p \vee q $

a - iv , b-iii, c-i, d-ii

$\therefore$ Option $4.$ is the correct answer.

edited by
0 votes
0 votes
(a) p $\rightarrow$ q, we can write  ˥p ˅ q, so the option is (iv)

So the answer could be 3 or 4, now check for 4 and by solving, we will get:

(b) p ˅ q = ˥(˥p)˅q = ˥p$\rightarrow$ q,

so the option is (iii), we can say by checking the given options that the answer is 4, but for the sake of proof:

(c) p ˄ q = q ˄ p = ˥(˥q ˅  ˥p) = ˥(q $\rightarrow$ ˥p) , so the option is (i)

(d) p ˄ ˥q = ˥(˥p ˅ q ) = ˥(p$\rightarrow$q) , so the option is (ii)

 

So, the correct answer will be  (4)
edited by
0 votes
0 votes
ANS :D

p ->q  = ~p∨q

p ∨ q = ~p->q

p∧q = ~(q->~p)

~(p->q)=p∧~q
Answer:

Related questions

6 votes
6 votes
2 answers
1
Arjun asked Jul 2, 2019
7,608 views
Which of the following is principal conjunctive normal form for $[(p\vee q)\wedge\ \neg p \rightarrow \neg q ]$ ?$p\ \vee \neg q$$p \vee q $$\neg p \vee q$$\neg p\ \vee ...
2 votes
2 votes
2 answers
3
Arjun asked Jul 2, 2019
7,063 views
How many ways are there to place $8$ indistinguishable balls into four distinguishable bins?$70$$165$$^8C_4$$^8P_4$
3 votes
3 votes
2 answers
4
Arjun asked Jul 2, 2019
5,752 views
How many bit strings of length ten either start with a $1$ bit or end with two bits $00$ ?$320$$480$$640$$768$