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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