I. *P*∨ ~*Q*

II. ~ ( ~*P*∧*Q*)

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

IV. (*P*∧*Q*)∨(*P*∧~ *Q*)∨( ~*P*∧*Q*)

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+17 votes

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

- $P ∨ \neg Q$
- $\neg(\neg P ∧ Q)$
- $(P ∧ Q) ∨ (P ∧ \neg Q) ∨ (\neg P ∧ \neg Q)$
- $(P ∧ Q) ∨ (P ∧ \neg Q) ∨ (\neg P ∧ Q)$

- Only I and II
- Only I, II and III
- Only I, II and IV
- All of I, II, III and IV

+19 votes

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.

+16 votes

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

+11 votes

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*

+6 votes

$$Option\ B\ is\ correct$$

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.

0

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

0

@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

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