Pembuktian Hukum Aljabar Boolean
T1. Hukum Komutatif
| A | B | A + B | B + A |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 |
| A | B | AB | BA |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 |
(a) (A + B) + C = A + (B + C)
| A | B | C | A + B | B + C | (A+B)+C | A+(B+C) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
(b) (A B) C = A (B C)
| A | B | C | AB | BC | (AB)C | A(BC) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
(a) A (B + C) = A B + A C
| A | B | C | B +C | AB | AC | A(B+C) | (AB)+(AC) |
| 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 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| A | B | C | BC | A+B | A+C | A+(BC) | (A+B)(A+C) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 | 0 | 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 |
T4. Hukum Identity
(a) A + A = A
| A | A | A+A |
| 0 1 | 0 1 | 0 1 |
(b) A A = A
| A | A | A A |
| 0 1 | 0 1 | 0 1 |
T5.
(a) AB + AB’ = A
| A | B | B’ | AB | AB’ | AB+AB’ |
| 0 0 1 1 | 0 1 0 1 | 1 0 1 0 | 0 0 0 1 | 0 0 1 0 | 0 0 1 1 |
| A | B | B’ | A+B | A+B’ | (A+B)(A+B’) |
| 0 0 1 1 | 0 1 0 1 | 1 0 1 0 | 0 1 1 1 | 1 0 1 1 | 0 0 1 1 |
T6. Hukum Redudansi
(a)A + A B = A
| A | B | AB | A+AB |
| 0 0 1 1 | 0 1 0 1 | 0 0 0 1 | 0 0 1 1 |
(b) A (A + B) = A
| A | B | A+B | A(A+B) |
| 0 0 1 1 | 0 1 0 1 | 0 1 1 1 | 0 0 1 1 |
T7
(a) 0 + A = A
| 0 | A | 0+A |
| 0 0 | 0 1 | 0 1 |
(b) 0 A = 0
| 0 | A | 0 A |
| 0 0 | 0 1 | 0 0 |
T8
(a) 1 + A = 1
| 1 | A | 1+A |
| 1 1 | 0 1 | 1 1 |
(b) 1 A = A
| 1 | A | 1 A |
| 1 1 | 0 1 | 0 1 |
T9
(a)A’ + A = I
| A | A’ | I | A’+A |
| 0 1 | 1 0 | 1 1 | 1 1 |
(b) A’ A = 0
| A | A’ | 0 | AA’ |
| 0 1 | 1 0 | 0 0 | 0 0 |
T10
(a ) A+ A’B = A+B
| A | B | A’ | A’B | A+B | A+A’B |
| 0 0 1 1 | 0 1 0 1 | 1 1 0 0 | 0 1 0 0 | 0 1 1 1 | 0 1 1 1 |
(b) A (A’+B) = A B
| A | B | A’ | A’+B | AB | A(A’+B) |
| 0 0 1 1 | 0 1 0 1 | 1 1 0 0 | 1 1 0 1 | 0 0 0 1 | 0 0 0 1 |
T11. TheoremaDe Morgan's
(a) (A’+B’)= A’B’
| A | B | A’ | B’ | A+B | (A’+B’) | A’ B’ |
| 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 |
(b) (A’B’) = A’ + B’
| A | B | A’ | B’ | A B | (A’B’) | A’+B’ |
| 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 |
