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State Distributive law and verify it using truth table.
Answer:
There are two Distributive laws:
(i) A.(B+C) =(A.B) + (A.C)
(ii) A+(B.C) = (A+B).(A+C)
Proof:
(i)
|
A |
B |
C |
B+C |
A.(B+C) |
A.B |
A.C |
(A.B)+(A.C) |
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
|
0 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
|
0 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
|
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
1 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
|
1 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
|
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
(ii)
|
A |
B |
C |
B.C |
A+(B.C) |
A+B |
A+C |
(A+B).(A+C) |
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
|
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
|
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
|
1 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
|
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
|
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
|
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
