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Prove:
Majority = ((~A && B && C) || (A && ~B && C) ||
(A && B && ~C) || (A
&& B && C))
|
|
|
|
| (A && ~B && C) |
| (A && B && C) |
|
|---|---|---|---|---|---|---|---|
|
|
|
|
| 0 |
| 0 |
|
|
|
|
|
| 0 |
| 0 |
|
|
|
|
|
| 0 |
| 0 |
|
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 |
| 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 |
|
|
|
|
| 0 |
| 1 |
|
Look at the equation:
((~A && B && C) || (A && ~B && C) || (A && B && ~C) || (A && B && C))
The first three groups contains two "normal" variables and the NOT
of the other -- i.e., each corresponds to a case of having two true values.
The fourth term corresponds to the case where all three are true.
What is the significance of the NOT terms?
CS1104 Main Page
Last Updated 02/14/2000
© L.Heath, 2000, updated by J.A.N. Lee, 2000/02/10