Created by cameronfdowner
about 9 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.