in Mathematical Logic edited by
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20 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
in Mathematical Logic edited by
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3 Comments

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

0
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.
0
Answer could be obtained via checking minimized SOP form.
1

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6 Answers

24 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. 

edited by

2 Comments

First two are easy. For last two, use K map to simplify them without any error.
2
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.
1
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. ∨ ~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 by
by

4 Comments

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

1
P+(~P~Q) = (P+~P)(P+~Q)  (OR is distributive over AND)

               =1(P+~Q)  (P+~P=1)    

               =(P+~Q)
4
Thanks a lot :)
1
You are welcome :)
0
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.
by

2 Comments

@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
0
0 votes
Just Use digital Logic Concept.

convert V → + and ^ → .

So whose answer is P + Q’.

They are matched :)
0 votes
The first two options are straightforward.

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.
Answer:

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