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45 votes

Let $\oplus$ denote the exclusive OR (XOR) operation. Let '$1$' and '$0$' denote the binary constants. Consider the following Boolean expression for $F$ over two variables $P$ and $Q$:

$$F(P,Q)=\left( \left(1 \oplus P \right) \oplus \left( P \oplus Q \right )\right ) \oplus \left(\left(P \oplus Q\right) \oplus \left(Q \oplus 0\right)\right)$$

The equivalent expression for $F$ is

  1. $P+Q$
  2. $\overline{P+Q}$
  3. $P \oplus Q$
  4. $\overline {P \oplus Q}$
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7 Answers

3 votes
3 votes

We need to simplify the above expression. As the given operation is XOR, we shall see property of XOR. Let A and B be boolean variable. In A XOR B, the result is 1 if both the bits/inputs are different, else 0. Now,

( ( 1 X P) X (P X Q) ) X ( (P X Q) X (Q X 0) )

( P' X P X Q ) X ( P X Q X Q ) ( as 1 X P = P' and Q X 0 = Q )

(1 X Q) X ( P X 0) ( as P' X P = 1 , and Q X Q = 0 )

Q' X P ( as 1 X Q = Q' and P X 0 = P )

PQ + P'Q' ( XOR Expansion, A X B = AB' + A'B )

This is the final simplified expression.

Now we need to check for the options.

If we simplify option D expression.

( P X Q )' = ( PQ' + P'Q )' ( XOR Expansion, A X B = AB' + A'B )

((PQ')'.(P'Q)') ( De Morgan's law )

( P'+ Q).(P + Q') ( De Morgan's law )

P'P + PQ + P'Q' + QQ'

PQ + P'Q' ( as PP' = 0 and QQ' = 0 ) 

Hence both the equations are same. Therefore Option D. 
1 votes
1 votes
observe the common term p ex or q in both

consider a case where p and q are equal

then p ex or q results in 0

1)in first p is ex ored  with 1

2)in second q is ex ored with 0

so if p, q are same then either of one oresults in 1 and another to 0

1 ex or  0

it is exnor
0 votes
0 votes

Substitute p=1 and q=1 in the given equation and the options-

((1⊕1)⊕(1⊕1))⊕((1⊕1)⊕(1⊕0))= 1


Therefore we can eliminate options B and C.


now to eliminate one of A and D we can substitute P=0 and Q=1

((1⊕0)⊕(0⊕1))⊕((0⊕1)⊕(1⊕0))=0


Therefore D is the correct option

Answer:

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