PERTEMUAN 6

Buktikan bahwa untuk sembarang elemen a dan b dari aljabar Boolean :
(i). a+a'b=a+b
(ii).a(a'+b)=ab
(iii).a+1=1
(iv).(ab)'=a'+b'
Jawab:
(i). a+a'b=(a+ab)+a'b    penyerapan
              =a+(ab+a'b)    Asosiatif
              =a(a+a')b        Distributif
              =a+1.b            Komplemen
              =a+b               Identitas

(ii). a(a'+b)=aa'+ab      distributif
                  =0+ab        Komplemen
                  =ab            Identitas

(iii). a+1=a+(a+a')    Komplemen
              =(a+a)+a'    Asosiatif
              =a+a'           Idempoten
              =1               Komplemen

(iv). (ab)'=ab.a'+abb'      Dsitributif
              = 0.b+a.0          Komplemen
              = 0+0                Dominansi
              = 0                    Identitas


Cari Komplemen Dari:
1. f(x,y,z)=x'(yz'+y'z)
2. f(x)=x
3. f(x,y)=x'y+xy'+y'
4. f(x,y)=x'y'
5. f(x,y)=(x+y)'
6. f(x.y,z)=xyz'

jawab:
1. f'(x,y,z)= (x'(yz'+y'z))'
                = x+(yz'+y'z)'
                = x+(yz')'(y'z)'
                = x+(y'+z(y+z')

2. f'(x)=x'

3. f'(x,y)=(x'y)+(xy'+y')
=(x+y')(x'+y)+y

4. f'(x,y)=(x+y)'

5. f'(x,y)=(x)(y)

6. f'( x,y,z)=x'+y'+z