GATE CSE 2008 | Question: 31

Question

$P$ and $Q$ are two propositions. Which of the following logical expressions are equivalent?

1. $P ∨ \neg Q$
2. $\neg(\neg P ∧ Q)$
3. $(P ∧ Q) ∨ (P ∧ \neg Q) ∨ (\neg P ∧ \neg Q)$
4. $(P ∧ Q) ∨ (P ∧ \neg Q) ∨ (\neg P ∧ Q)$

• A. Only I and II
• B. Only I, II and III
• C. Only I, II and IV
• D. All of I, II, III and IV

Comments

Answer could be obtained via checking minimized SOP form.

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option 1 and 2 can be easily seen that they are equivalent.

Option III- PQ+P(~Q)+(~P)(~Q)=P(Q+(~Q))+(~P)(~Q)=P+(~P)(~Q)

→   P+(~P)(~Q)=(P+(~P))+(P+~Q)=P+(~Q).

Option IV-Proceed same as in Option III.

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To solve faster prefer solving using the case method.

Answer 1

I and II are present in all options so need to check.

For III and IV

$(P∧Q)∨(P∧\neg Q) \equiv P∧(Q \vee\neg Q)$  (By distributive law)
$\quad \quad \equiv P∧T$
$\quad \quad \equiv P$

For III.

$P∨( \neg P∧\neg Q) \equiv P\vee \neg Q$ (By Absorption Law)

For IV.

$P∨(¬P∧Q) \equiv P∨Q$ (By Absorption Law)

So Option B is correct.

I, II, III are logically equivalent. 

Comments

First two are easy. For last two, use K map to simplify them without any error.

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Also for option C, you can use intuition. For example take

P         Q

0         0

0         1

1         0

1         1

are possible right and now if you observe the given equation (P∧Q)∨(P∧¬Q)∨(¬P∧¬Q) you can see all the terms except (¬P∧Q) is present. So you can write the whole equation as ~(¬P∧Q) which is same as option A and B. Similarly you can verify option D.

Answer 2

(B) Only I, II and III. Draw truth table to check, evaluating individual expression will consume lot of time with no guaranteed answer.

Answer 3

answer B

I. P ∨ ~Q
II. ~ ( ~P ∧ Q) apply De Morgan's law. we get  P ∨ ~Q
III. (P ∧ Q) ∨ (P ∧~ Q) ∨ ( ~P ∧ ~Q)

=[(P ∧ Q) ∨ (P ∧ ~Q)] ∨ ( ~P ∧ ~Q)

=[P(Q V ~Q)] ∨ ( ~P ∧ ~Q)

=P ∨ ( ~P ∧ ~Q)

= P ∨ ~Q
 

Comments

Hi @Anu , P + (^~P^~Q) , how you are deriving it to be P+^~Q. Can you please explain ?

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P+(~P~Q) = (P+~P)(P+~Q)  (OR is distributive over AND)

               =1(P+~Q)  (P+~P=1)    

               =(P+~Q)

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Thanks a lot :)

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You are welcome :)

Answer 4

$$Option\ B\ is\ correct$$

writing in digital like boolean algebra is much faster

1st $p+q'$

2nd $(p'.q)'=p+q'$

3rd is (P∧Q)∨(P∧¬Q)∨(¬P∧¬Q) which is also
$p.q+p.q'+p'q'$

$=(p.q+p.q')+(p.q'+p'q')$        //A+A=A
=$p+q'$

write 4th similarly is $p+q$ i.e. not equivalent to 1st, 2nd and 3rd.

Comments

@bhuv why you add p.q' twice in second last statement.How can you interpret where we add this and where we dont

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@yash See I have written in comments (//......) it called Idempotent law you can write A+A+A....=A any number of times similarly for A.A.A.......=A for any number of times. We use such law to simplify our expression. You'll learn "when to use and when not" by practicing more these type of questions.

 

you can find more such Laws here https://www.electronics-tutorials.ws/boolean/bool_6.html