GATE | GATE CS 2019 | Question 21

Which one of the following is NOT a valid identity?

(A) (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)
(B) (x + y) ⊕ z = x ⊕ (y + z)
(C) x ⊕ y = x + y, if xy = 0
(D) x ⊕ y = (xy + x′y′)′


Answer: (B)

Explanation: According to Exor (⊕) operation,

x y x⊕y
0 0 0
0 1 1
1 0 1
1 1 0

Therefore, option (D),

x⊕y 
= (x'y + xy′)
= (x'+y').(x+y)
= (x⊙y)'
= (xy + x′y′)′ 

It also clearly shows that, if atleast one of x and y is 0, then it works as (x+y).

x⊕y = x + y,  if xy = 0

You can notice that it works as (x+y) except last row in given Truth table, because only last row does not satisfy (x.y)=0. So, option (C) is also correct.

Exor (⊕) operation also satisfies associative law, i.e.,

(x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)

So, option (A) is also correct.

But, option (B) is not correct because,

(x+y)⊕z 
= (x+y)'.z + (x+y).z'
= (x'y').z + xz' + yz'

And,
x⊕(y+z)
= x'.(y+z) + x.(y+z)'
= x'y + x'z + x.y'z'

Therefore,
(x+y)⊕z ≠ x⊕(y+z)


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