4140215
Description
Flashcards by cameronfdowner, updated more than 1 year ago
|
Created by cameronfdowner
over 8 years ago
|
|
| Question | Answer |
| A \(\cdot\)1 = | A |
| A \(\cdot\)0 = | 0 |
| A + 1 = | 1 |
| A + 0 = | A |
| Zero and Unit Rules. | A + 0 = A, A + 1 = 1, A \(\cdot\)0 = 0, A \(\cdot\)1 = A |
| A \(\cdot\) \(\bar A\) = | 0 |
| A + \(\bar A\) = | 1 |
| \(\overline {(\bar A)} \) = | A |
| Complement Relations | \(\overline {(\bar A)} \) = A, A + \(\bar A\) = 1, A \(\cdot\) \(\bar A\) = 0 |
| A \(\cdot\) A = | A |
| A + A = | A |
| Idempotence | A \(\cdot\) A = A, A + A = A |
| A \(\cdot\) B = | B \(\cdot\) A |
| A + B = | B + A |
| Commutative Laws | A \(\cdot\) B = B \(\cdot\) A, A + B = B + A |
| A + (A \(\cdot\) B) = | A |
| A \(\cdot\) (A + B) = | A |
| A + (\(\bar A\) \(\cdot\) B) = | A + B |
| Absorption Laws | A + (\(\bar A\) \(\cdot\) B) = A + B, A \(\cdot\) (A + B) = A, A + (A \(\cdot\) B) = A |
| A \(\cdot\) (B + C) = | (A \(\cdot\) B) + (A \(\cdot\) C) |
| A + (B \(\cdot\) C) = | (A + B) \(\cdot\) (A + C) |
| Distributive Laws | A \(\cdot\) (B + C) = (A \(\cdot\) B) + (A \(\cdot\) C), A + (B \(\cdot\) C) = (A + B) \(\cdot\) (A + C) |
| A + B + C = | A + (B + C) = (A + B) + C |
| A \(\cdot\) B \(\cdot\) C = | A \(\cdot\) (B \(\cdot\) C) = (A \(\cdot\) B) \(\cdot\) C |
| Associative Laws | A \(\cdot\) B \(\cdot\) C = A \(\cdot\) (B \(\cdot\) C), A + B + C = A + (B + C) |
| \(\overline {A + B + C} \) = | \(\bar A\) \(\cdot\) \(\bar B\) \(\cdot\) \(\bar C\) |
| \(\overline {A \cdot\ B \cdot\ C} \) = | \(\bar A\) + \(\bar B\) + \(\bar C\) |
| De Morgan's Theorem | \(\overline {A + B + C} \) = \(\bar A\) \(\cdot\) \(\bar B\) \(\cdot\) \(\bar C\), \(\overline {A \cdot\ B \cdot\ C} \) = \(\bar A\) + \(\bar B\) + \(\bar C\) |
Want to create your own Flashcards for free with GoConqr? Learn more.