tugas 4A sistem logika

Hukum Aljabar Boolean

• Hukum Komutatif
• Hukum Asosiatif
• Hukum Distributif
• Hukum Identity
• Hukum Redudansi
• Theorem De Morgan's

Hukum Aljabar Boolean

T1. Hukum Komutatif

(a) A + B = B + A

Pembuktian:

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

(b) A B = B A

Pembuktian:

A	B	AB	BA	AB = BA
0	0	0	0	Ö
0	1	0	0	Ö
1	0	0	0	Ö
1	1	1	1	Ö

T2. Hukum Asosiatif

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

Pembuktian:

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

(b) (A B) C = A (B C)

Pembuktian:

A	B	C	A B	B C	(A B) C	A (B C)	(A B) C = A (B C)
0	0	0	0	0	0	0	Ö
0	0	1	0	0	0	0	Ö
0	1	0	0	0	0	0	Ö
0	1	1	0	1	0	0	Ö
1	0	0	0	0	0	0	Ö
1	0	1	0	0	0	0	Ö
1	1	0	0	0	0	0	Ö
1	1	1	1	1	1	1	Ö

T3. Hukum Distributif

(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	A + A = A
0	0	Ö
0	0	Ö
1	1	Ö
1	1	Ö

(b) A A = A

Pembuktian:

A	A A	A A = A
0	0	Ö
0	0	Ö
1	1	Ö
1	1	Ö

T5.

(a)

Pembuktian:

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

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)

Pembuktian:

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

(b)

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

T10

(a)

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)

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(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(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