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04.50 |


·        Quiz Aljabar Boolean
1.Give the relationship that represents the dual of the Boolean property A + 1 = 1?
(Note: * = AND, + = OR and ' = NOT)
3.XZ
1.A
2.A'
3.1Ö
4.0
2.AB



Tugas 4 A Sistemlogika
HukumAljabar Boolean






HukumAljabar Boolean
T1. HukumKomutatif
(a) A + B = B + A, pembuktian:
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, pembuktian:
A
B
AB
BA
0
0
0
0
0
1
0
0
1
0
0
0
1
1
1
1

 
T2. HukumAsosiatif
(a) (A + B) + C = A + (B + C),pembuktian:
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),pembuktian:
A
B
C
A B
B C
(A 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
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

 
T3. HukumDistributif
(a) A (B + C) = A B + A C,pembuktian:
A
B
C
B+C
A B
A C
A (B+C)
A B+A C
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

(b) A + (B C) = (A + B) (A + C),pembuktian:
A
B
C
B C
A+B
A+C
A+(B C)
(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,pembuktian: 
A
A + A
0
0
0
0
1
1
1
1

 
(b) A A = A,pembuktian:
A
A A
0
0
0
0
1
1
1
1
T5.
1.      A B + AB = A (BENAR),pembuktian:
A
B
B(invers)
A B
A B(invers)
 A B + AB = A
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 (BENAR),pembuktian:
A
B
B(invers)
A+B
A+B(invers)
 ( A+B) (A+B)=A
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. HukumRedudansi
(a) A + A B = A (BENAR), Pembuktian:
A
B
A B
A + A B
0
0
0
0
0
1
0
1
1
0
0
1
1
1
1
1

(b) A (A + B) = A (BENAR),pembuktian:
A
B
A + B
A (A + B) = A
0
0
0
0
0
1
1
0
1
0
1
1
1
1
1
1
T7
(a) 0 + A = A (BENAR), Pembuktian:

A
0
0 + A = A
0
0
0
0
0
0
1
0
1
1
0
1
(b)0 A = 0 (BENAR), Pembuktian:

A
0
0 A = 0
0
0
0
0
0
0
1
0
0
1
0
0
T8
(a) 1 + A = 1(BENAR)
A
1
1 + A
0
1
1
0
1
1
1
1
1
1
1
1

(b) 1 A = A (BENAR), Pembuktian:

A
1
1 A = A
0
1
0
0
1
0
1
1
1
1
1
1

 

 
T9
(a.) A + A =1 (BENAR)
A
A
1
A + A =1
0
1
1
1
0
1
1
1
1
0
1
1
1
0
1
1

(b) A A = 0 (BENAR)
A
A
0
A A = 0
0
1
0
0
0
1
0
0
1
0
0
0
1
0
0
0

 T10
(a)A + A B =A + B(BENAR)
A
B
A
A B
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(BENAR)
A
B
A
A +B
A B
A ( A +B) = A B
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. TheoremaDe Morgan's
1.      (A + B) = A B
A
B
A
B
A+B
(A+B)
AB
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 (BENAR)
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


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