Tugas 4 A " Sistem Digital "

Tugas 4 A “ Sistem Digital ”

Hukum Aljabar Boolean

T1. Hukum Komutatif

(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. Hukum Asosiatif

(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	AB	BC	(AB)C	A(BC)
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. Hukum Distributif

(a) A (B + C) = A B + A C

Pembuktian:

A	B	C	B +C	AB	AC	A(B+C)	(AB)+(AC)
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	BC	A+B	A+C	A+(BC)	(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.

(a) AB + A B’

Pembuktian:

A	B	B(invers)	A B	A B(invers)	AB+AB(infers)
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’)

Pembuktian:

A	B	B(invers)	A+B	A+B(invers)
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. Hukum Redudansi

(a) A + A B = A

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

Pembuktian:

A	B	A + B	A (A + B)
0	0	0	0
0	1	1	0
1	0	1	1
1	1	1	1

T7

(a) 0 + A = A

Pembuktian:

A	0 + A
0	0
0	0
1	1
1	1

(b) 0 A = 0

Pembuktian:

A	0 A	0
0	0	0
0	0	0
1	0	0
1	0	0

T8

(a) 1 + A = 1

A	1 + A	1
0	1	1
0	1	1
1	1	1
1	1	1

(b) 1 A = A

Pembuktian:

A	1 A
0	0
0	0
1	1
1	1

T9

(a) A’ + A = 1

Pembuktian:

A	A(invers)	A(infers)	1
0	1	1	1
0	1	1	1
1	0	1	1
1	0	1	1

(b) A’ A=0

A	A(invers)	A(invers)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

Pembuktian:

A	B	A(invers)	A(invers) B	A+B	A+A(invers) B
0	0	1	1	0	0
0	1	1	0	1	1
1	0	0	1	1	1
1	1	0	0	1	1

(b) A (A’ + B) = AB

Pembuktian:

A	B	A(invers)	A(invers)+B	A B	A(A(invers)+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

(a) (A’+B’)= A’B’

A	B	A(invers)	B(invers)	A+B	(A+B)invers	A(invers) B(invers)
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’

A	B	A(invers)	B(invers)	A B	(AB)invers	A(invers)+B(invers)
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