A. Hukum Aljabar Boolean
1.Hukum komutatif
A+B = B+A
AB = BA
A | B | A+B | B+A | AB | BA |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 1 |
2. Hukum Asosiatif
(A+B)+C = A+(B+C)
(AB)C = A(BC)
A | B | C | A+B | B+C | AB | BC | (A+B)+C | A+(B+C) | (AB)C | A(BC) |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
3. Hukum Distributif
A(B+C) = AB+AC
A+(BC) = (A+B)(A+C)
A | B | C | B+C | A(B+C) | AB | AC | AB+AC | BC | A+(BC) | A+B | A+C | (A+B)(A+C) |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
4. Hukum Identity
A+A = A
AA = A
A | A+A | AA |
0 | 0 | 0 |
1 | 1 | 1 |
5.
AB+AB’ = A
(A+B)(A+B’) = A
A | B | B’ | AB | AB’ | AB+AB’ | A+B | A+B’ | (A+B)(A+B’) |
0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
6. Hukum Redudansi
A+AB = A
A (A + B) = A
A | B | AB | A+AB | A+B | A(A+B) |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 |
7.
0+A = A
0A = 0
A | 0+A | 0A |
0 | 0 | 0 |
1 | 1 | 0 |
8.
1+A = 1
1A = A
A | 1+A | 1A |
0 | 1 | 0 |
1 | 1 | 1 |
9.
A’+A = 1
A’A = 0
A | A’ | A’+A | A’A |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 |
10.
A+A’B = A+B
A(A’+B) = AB
A | B | A’B | A+A’B | A+B | A’ | A’+B | A(A’+B) | AB |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
11. TheoremaDe Morgan's
(A+B)’ = A’B’
(AB)’ = A’+B.
A | B | A+B | (A+B)’ | A’B’ | AB | (AB)’ | A’+B’ |
0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 |
B. Quiz aljabar boolean
1. Give the relationship that represents the dual of the Boolean property A + 1 = 1?
(Note: * = AND, + = OR and ' = NOT)
(Note: * = AND, + = OR and ' = NOT)
1. A * 1 = 1
2. A * 0 = 0
3. A + 0 = 0
4. A * A = A
5. A * 1 = 1
3. 1 or 2
1. A + B + C
2. D + E
3. A'B'C'
4. D'E'
4. Which of the following relationships represents the dual of the Boolean property x + x'y = x + y?
2. x(x'y) = xy
5. Given the function F(X,Y,Z) = XZ + Z(X'+ XY), the equivalent most simplified Boolean representation for F is:
1. Z + YZ
2. Z + XYZ
3. XZ
4. X + YZ
1. F = xy
2. F = x + y
3. F = x'
4. F = xy + yz
5. F = x + y'
7. Simplification of the Boolean expression (A + B)'(C + D + E)' + (A + B)' yields which of the following results?
1. A + B
2. A'B'
3. C + D + E
4. C'D'E'
5. A'B'C'D'E'
8. Given that F = A'B'+ C'+ D'+ E', which of the following represent the only correct expression for F'?
2. F'= ABCDE
5. F'= (A+B)CDE
1. A
2. A'
3. 1
4. 0
10. Simplification of the Boolean expression AB + ABC + ABCD + ABCDE + ABCDEF yields which of the following results?
1. ABCDEF
2. AB
3. AB + CD + EF
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