I. *P*∨ ~*Q*

II. ~ ( ~*P*∧*Q*)

III. (*P*∧*Q*)∨(*P*∧~ *Q*)∨( ~*P*∧ ~*Q*)

IV. (*P*∧*Q*)∨(*P*∧~ *Q*)∨( ~*P*∧*Q*)

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## 6 Answers

Best answer

**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.

### 2 Comments

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.

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.

$$Option\ B\ is\ correct$$

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.

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.

### 2 Comments

@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

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

ie (q--->p)

now, the easiest way is check the condition of implication in (iii)option.

ie if Q is true , p has to be true . If Q is false, P Can be anything.

that’s exactly what is (iii) option.

Now, in 4th option.If Q is true p can be either true or false. Hence, 4th option is not equivalent to the remaining.

Answer: (i) , (ii), (iii) are equivalent.