TUGAS 4

HukumAljabar Boolean

T1. HukumKomutatif

(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. HukumAsosiatif

(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	A B	B C	(A B) C	A (B C)
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. HukumDistributif

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

Pembuktian:

A	B	C	B+C	A B	A C	A (B+C)	A B+A C
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	B C	A+B	A+C	A+(B C)	(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)   A B + AB = A (BENAR)

Pembuktian:

A	B	B(invers)	A B	A B(invers)	A B + AB = A
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)=A (BENAR)

Pembuktian:

A	B	B(invers)	A+B	A+B(invers)	( A+B) (A+B)=A
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. HukumRedudansi

(a) A + A B = A (BENAR)

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 (BENAR)

Pembuktian:

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

T7

(a) 0 + A = A (BENAR)

Pembuktian:

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

(b)0 A = 0 (BENAR)

Pembuktian:

A	0	0 A = 0
0	0	0
0	0	0
1	0	0
1	0	0

T8

(a) 1 + A = 1(BENAR)

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

(b) 1 A = A (BENAR)

Pembuktian:

A	1	1 A = A
0	1	0
0	1	0
1	1	1
1	1	1

T9

(a) A + A =1 (BENAR)

Pembuktian:

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

(b) A A = 0 (BENAR)

A	A	0	A 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(BENAR)

Pembuktian:

A	B	A	A B	A+B	A + A B =A + B
0	0	1	0	0	0
0	1	1	1	1	1
1	0	0	0	1	1
1	1	0	0	1	1

(b) A (A + B) = A B(BENAR)

Pembuktian:

A	B	A	A +B	A B	A ( A +B) = A 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	B	A+B	(A+B)	AB
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 (BENAR)

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