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GATE | GATE CS 2019 | Question 21

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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 XOR (⊕) 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. XOR (⊕) 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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Last Updated : 20 Sep, 2021
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