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

1. Only I and II
2. Only I, II and III
3. Only I, II and IV
4. All of I, II, III and IV
edited | 882 views

I. P ~Q
II. ~ ( ~PQ)
III. (PQ)(P~ Q)( ~P ~Q)
IV. (PQ)(P~ Q)( ~PQ)

what is the fastest way to solve these type of questions?by truth table or to solve the expression individually or any other if the expression has more than 2 variables.
Answer could be obtained via checking minimized SOP form.

I and II are present in each option so need to check.

For III and IV

(P∧Q)∨(P∧~ Q) = P∧(QV~Q)  (By distributive law)

= P∧T

=P

For III.

P∨( ~P∧~Q) = P∨~Q (By Absorption Law)

For IV.

P∨(¬P∧Q) = P∨Q (By Absorption Law)

So Option B is correct.

I, II, III are logically equivalent.

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

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

=[(∧ Q) ∨ (∧ ~Q)] ∨ ( ~∧ ~Q)

=[P(Q V ~Q)] ∨ ( ~∧ ~Q)

=P ∨ ( ~∧ ~Q)

∨ ~Q

edited

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

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

=1(P+~Q)  (P+~P=1)

=(P+~Q)
Thanks a lot :)
You are welcome :)
$$Option\ B\ is\ correct$$

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.