T1. Hukum Komutatif
(a). A+B = B+A
(b). A.B = B.A
T2. Hukum Asosiatif
a. (A+B)+C = A+(B+C)
b. (AB) C = A (BC)
T3. Hukum Distributif
a. A(B+C) = AB + AC
b. A +(BC) = (A+B) (A+C)
T4. Hukum Identity
(a) A + A = A
(b) A A = A
T5. (a) AB+AB'=A
(b) (A+B)(A+B')= A
T6. Hukum Redudansi
(a) A + A B = A
(b) A (A + B) = A
(b) A (A + B) = A
T7
(a) 0 + A = A
(b) 0 A = 0
(b) 0 A = 0
T8
(a) 1 + A = 1
(b) (b) 1 A = A
T9
(a) A' +A=1
(b) A' A = 0
(b) A' A = 0
T10
(a) A+A' B= A+B
(b) A(A' + B) = AB
(b) A(A' + B) = AB
T11. TheoremaDe Morgan's
(a) (A+B)' A' B'
(b) (AB)' = A' + B'
(b) (AB)'
- Give the relationship that represents the dual of the Boolean property A + 1 = 1?
(Note: * = AND, + = OR and ' = NOT)
SEBAB :
A | 1 | A+1 | A*0 | 0 |
0 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 0 | 0 |
SEBAB : A literal = 1 atau 2
Sebab : (A+B+C)(D+E)+(A+B+C)(D+E) = A+B+C
Sebab : x+x’y =x+y ekuivalen dengan x(x’+y)=xy
- Given the function F(X,Y,Z) = XZ + Z(X'+ XY), the equivalent most simplified Boolean representation for F is:
Sebab : fungsi yang tepat adal F(X,Y,Z)=Z
Sebab : F(x,y)= x+y’ bisa menggunakan fungsi AND, OR, NOT . F ini adalah aljabar sempurna. Contoh : NOT: F(0,B)=B’ || OR : F(A, F(0,B))=A+B || AND:F(0,F(F(0,A),B))=A*B
- Simplification of the Boolean expression (A + B)'(C + D + E)' + (A + B)' yields which of the following results?
Dengan contoh x + xy = x, dengan x=(A+B)’=A’ B’ dan y=(C+D+E)’ = C‘D’E’
- Given that F = A'B'+ C'+ D'+ E', which of the following represent the only correct expression for F'?
Sebab : persamaan diatas menggunakan persamaan DEMORGAN
Alasan : Dari contoh boolean A+1 = 1, A=A.
- Simplification of the Boolean expression AB + ABC + ABCD + ABCDE + ABCDEF yields which of the following results?
Alasan mengulang dari teori ASOSIATIF x+xy= x.
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