Hukum Aljabar Boolean & Tabel Kebenarannya
A) A + B = B + A
A | B | A+B | B+A |
0 | 0 | 0 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 |
B) A B = B A
A | B | A B | B A |
0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 |
1 | 0 | 0 | 0 |
1 | 1 | 1 | 1 |
T2. Hukum Asosiatif
A) (A + B) + C = A + (B + C)
A | B | C | A+B | (A+B)+C | 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 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 |
B) (A B) C = A (B C)
A | B | C | A B | (A B) C | B C | A (B C) |
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 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 1 | 1 |
A) A (B + C) = A B + A
A | B | C | B+C | A(B+C) | A B | AB+A |
0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 |
B) A + (B C) = (A + B) (A + C)
A | B | C | B C | A+(BC) | 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 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
A) A + A = A
A | A+A |
0 | 0 |
1 | 1 |
B) A A = A
A | AA |
0 | 0 |
1 | 1 |
T5.
A) AB + AB’ = A
A | B | B’ | A B | A B’ | AB+AB’ |
0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 |
B) (A+B) (A+B’) = A
A | B | B’ | A+B | A+B’ | (A+B)(A+B’) |
0 | 0 | 1 | 0 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 1 |
T6. Hukum Redudansi
A) A + A B = A
A | B | A B | A+AB |
0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 |
1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 |
B) A (A + B) = A
A | B | A+B | A(A+B) |
0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 |
T7.
A) 0 + A = A
A | 0 | 0+A |
0 | 0 | 0 |
1 | 0 | 1 |
B) 0 A = 0
A | 0 | 0A |
0 | 0 | 0 |
1 | 0 | 0 |
T8.
A) 1 + A = 1
A | 1 | 1+A |
0 | 1 | 1 |
1 | 1 | 1 |
B) 1 A = A
A | 1 | 1 A |
0 | 1 | 0 |
1 | 1 | 1 |
T9.
A) A’ + A = 1
A | A’ | A’+A |
0 | 1 | 1 |
1 | 0 | 1 |
B) A’ A = 0
A | A’ | A’ A |
0 | 1 | 0 |
1 | 0 | 0 |
T10.
A) A + A’ B = A + B
A | B | A’ | A’ B | A+A’B | A+B |
0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 1 |
B) A ( A’ + B) = A B
A | B | ‘A’ | A’+B | A(A’+B) | AB |
0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 1 |
T11.Theorema De Morgan's
A) ( A + B)’ = A’ B’
A | B | A’ | B’ | (A+B)’ | A’B’ |
0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 0 | 0 |
B) ( A B )’ = A’ + B’
A | B | A’ | B’ | (A B)’ | A’+B’ |
0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 |
Tugas 4.B
1. Give the relationship that represents the dual of the Boolean property A + 1 = 1?
(Note: * = AND, + = OR and ' = NOT)
2. A * 0 = 0 ( Jawabannya )
2. Give the best definition of a literal?
2. The complement of a Boolean variable ( Jawabannya )
3. Simplify the Boolean expression (A+B+C)(D+E)' + (A+B+C)(D+E) and choose the best answer.
1. A + B + C ( Jawabannya )
4. Which of the following relationships represents the dual of the Boolean property x + x'y = x + y?
1. x'(x + y') = x'y' ( Jawabannya )
5. Given the function F(X,Y,Z) = XZ + Z(X'+ XY), the equivalent most simplified Boolean representation for F is:
2. Z + XYZ ( Jawabannya )
6. Which of the following Boolean functions is algebraically complete?
1. F = xy ( Jawabannya )
7. Simplification of the Boolean expression (A + B)'(C + D + E)' + (A + B)' yields which of the following results?
2. A'B' ( Jawabannya )
8. Given that F = A'B'+ C'+ D'+ E', which of the following represent the only correct expression for F'?
5. F'= (A+B)CDE ( Jawabannya )
9. An equivalent representation for the Boolean expression A' + 1 is
3. 1 ( Jawabannya )
10. Simplification of the Boolean expression AB + ABC + ABCD + ABCDE + ABCDEF yields which of the following results?
2. AB ( Jawabannya )