Let X 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) = ( ( 1 X P) X (P X Q) ) X ( (P X Q) X (Q X 0) ) 

The equivalent expression for F is
(A) P + Q
(B) (P + Q)\’
(C) P X Q
(D) (P X Q)\’


Answer: (D)

Explanation:
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. 

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