AXIOMA
pada aljabar Boole
A1: Idempotent : a.a = a ; a + a = a
A2: Commutative: a.b = b.a ; a + b = b + a
A3: Associative : a.(b.c) = (a.b).c; a + (b+c) = (a+b) + c
A4: Absorbtive :a.(a + b) = a ; a + (a.b) = a A5: Distributive : a.(b + c) = (a.b) + (a.c) ;
a + (b.c) = (a + b).(a + c)
A6: Elemen Unik : a.1 = 1.a = a ; a + 0 = 0 + a = a
a.0 = 0.a = 0 ; a + 1 = 1 + a = 1
MINIMISASI SUATU
RANGKAIAN DIGITAL
dengan
Nomor urut decimal
Buat persamaan boolean dari PK
berikut
0 1 0 1 0 1
0 1 0 1 0 1 1
1 1 1 1 1 1 1 1
0 1 0 1 0 1
0 1 1 0 1 0 1
1 1 1 1 1 1
0 1 0 1 0 1
0 0 0 1
1 1 1 1 1
Y =
Y =
Y = Y = Y =
Y = Y =
SISTEM DIGITAL- GPO 6
Nomor urut decimal
SISTEM DIGITAL- GPO 8
Tata letak
variabel
00 01 11 10 00 01 11 10
0 1 1 0 1 1
1 1 1 1 1 1
Y = Y =
00 01 11 10 00 01 11 10
0 1 1 0 1 1
1 1 1 1 1 1
Y = Y =
00 01 11 10 00 01 11 10
0 0 1 1 1 1
1 1 1 1 1 1
Buat persamaan boolean dari PK
berikut:
00 01 11 10 00 01 11 10
0 1 1 0 1 1 1 1
1 1 1 1 1 1 1 1
Y = Y =
00 01 11 10 00 01 11 10
0 1 1 1 1 0 1 1
1 1 1 1 1 1 1 1
Y = Y =
00 01 11 10 00 01 11 10
0 1 1 1 0 1 1 1 1
1 1 1 1 1
SISTEM DIGITAL- GPO 10
Buat persamaan boolean dari PK
berikut:
00 01 11 10 00 01 11 10
0 1 0 1 1 1
1 1 1 1 1 1
Y = Y =
00 01 11 10 00 01 11 10
0 1 1 1 0 1 1
1 1 1 1 1
Y = Y =
00 01 11 10 00 01 11 10
0 1 1 0 1 1 1
1 1 1 1 1
Buat persamaan boolean dari PK
berikut:
Y
= =Y
00 01 11 10 00 01 11 10
0 1 0 1 1 1
1 1 1 1 1 1 1 1 1
Y = Y =
00 01 11 10 00 01 11 10
0 1 1 0 1 1
1 1 1 1 1 1 1
Y = Y =
00 01 11 10 00 01 11 10
0 1 1 0 1 1 1
