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+10 votes

$P$ 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
asked in Mathematical Logic by Veteran (68.8k points)
edited by | 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.

4 Answers

+12 votes
Best answer

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


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. 

answered by Loyal (3.2k points)
selected by
+14 votes
(B) Only I, II and III. Draw truth table to check, evaluating individual expression will consume lot of time with no guaranteed answer.
answered by Boss (6.3k points)
+10 votes

answer B

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

answered by Veteran (10.9k points)
edited by

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)    

Thanks a lot :)
You are welcome :)
+2 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

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

write 4th similarly is $p+q$ i.e. not equivalent to 1st, 2nd and 3rd.
answered by Loyal (2.9k points)

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