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# GATE | GATE-CS-2014-(Set-3) | Question 65

• Last Updated : 20 Sep, 2021

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)’

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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