Question
Download Solution PDFWhich one of the following is NOT a valid identity?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDF1. (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)
|
x |
y |
z |
(x ⊕ y) ⊕ z |
x ⊕ (y ⊕ z) |
|
0 |
0 |
0 |
0 |
0 |
|
0 |
O |
1 |
1 |
1 |
|
0 |
1 |
0 |
0 |
1 |
|
0 |
1 |
1 |
0 |
0 |
|
1 |
0 |
0 |
1 |
1 |
|
1 |
0 |
1 |
0 |
0 |
|
1 |
1 |
0 |
0 |
0 |
|
1 |
1 |
1 |
1 |
1 |
Therefore exclusive OR is associative and hence (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)
2.
|
x |
y |
z |
(x + y) ⊕ z |
x ⊕ (y + z) |
|
0 |
0 |
0 |
0 |
0 |
|
0 |
O |
1 |
1 |
1 |
|
0 |
1 |
0 |
0 |
1 |
|
0 |
1 |
1 |
0 |
1 |
|
1 |
0 |
0 |
1 |
1 |
|
1 |
0 |
1 |
0 |
0 |
|
1 |
1 |
0 |
1 |
0 |
|
1 |
1 |
1 |
0 |
0 |
(x + y) ⊕ z ≠ x ⊕ (y + z) ∴ is it not a valid identity
3.
|
xy = 0 |
Checking validity |
||
|
X |
y |
x + y |
x ⊕ y |
|
0 |
0 |
0 |
0 |
|
0 |
1 |
1 |
1 |
|
1 |
0 |
1 |
1 |
x + y = x ⊕ y / if xy = 0
4.
(xy + x'y')'
= (x’ + y’).(x+y) / Demorgan’s Law
= x’y +xy’
= x ⊕ y
Last updated on Jan 8, 2025
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