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Tugas Rangkaian Logika

Nama  : Alvian Ali Maskuri
Kelas  : I/A

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Hukum asosiatif:
(i)         a + (b + c) = (a + b) + c

A
B
C
a + (b + c)
0
0
0
0
0
0
1
1
0
1
0
1
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
A
B
C
(a + b) + c
0
0
0
0
0
0
1
1
0
1
0
1
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1





(ii)         a.(b.c) = (a.b). c

A
B
C
a.(b.c)
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
0
1
1
0
0
1
1
1
1
A
B
C
(a.b). c
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
0
1
1
0
0
1
1
1
1





Hukum distributif:

(i)              a + (b.c) = (a + b) (a + c)

A
B
C
a + (b.c)
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
A
B
C
(a + b) (a + c)
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1











(ii) a.(b + c) = a.b + a.c

A
B
C
a.(b + c)
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
1
1
1
0
1
1
1
1
1
A
B
C
a.b + a.c
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
1
1
1
0
1
1
1
1
1

           






 

 


Sederhanakan dengan cara Aljabar
1.        f(x, y, z) = xyz + xyz + xy
= xz(y’ + y) + xy
= xz + xy

2.        f(x,y,z)   = xy + x¢z + yz
= xy + x’z + yz (x + x’)
= xy + x’z + xyz + x’yz
= xy + xyz = x’z + x’zy
= xy (1 + z ) + x’z (1 + y)
= xy + x’z

3.        f(x,y,z)   = (x + y)(x¢ + z)(y + z)
= (x + y)(x¢ + z)(y + z) + (y’+z’)
= (x + y) (x’ + z) (y + y’) + (y + z’) + (z + y’) + (z + z’)
= (x + y) (x’ + z)


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