Selasa, 20 April 2010

Tugas 4

Tugas 4.A

Hukum Aljabar Boolean & Tabel Kebenarannya


T1. Hukum Komutatif


(a) A + B = B + A


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

A

B

A B

B A

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)

A

B

C

A+B

(A+B)+C

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

1

1

1

1

1

1

1

1

1

1

1

1


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

A

B

C

A B

(A B) C

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

0

1

0

1

0

1

0

0

0

0

1

1

1

1

1

1

1


T3. Hukum Distributif


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



A

B

C

B+C

A(B+C)

A B

AB+A

0

0

0

0

0

0

0

0

0

1

1

0

0

0

0

1

0

1

0

0

0

0

1

1

1

0

0

0

1

0

1

1

1

0

1

1

1

1

1

1

1

1


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

A

B

C

B C

A+(BC)

A+B

A+C

(A+B)(A+C)

0

0

0

0

0

0

0

0

0

0

1

0

0

0

1

0

0

1

0

0

0

1

0

0

0

1

1

1

1

1

1

1

1

0

1

0

1

1

1

1

1

1

1

1

1

1

1

1

T4. Hukum Identity

(a) A + A = A

A

A+A

0

0

1

1

(b) A A = A


A

AA

0

0

1

1

T5.

(a) AB + AB’ = A

A

B

B’

A B

A B’

AB+AB’

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



A

B

B’

A+B

A+B’

(A+B)(A+B’)

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



A

B

A B

A+AB

0

0

0

0

0

1

0

0

1

0

0

1

1

1

1

1


(b) A (A + B) = A




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


A


0+A

0

0

0

1

0

1

(b) 0 A = 0


A


0 A

0

0

0

1

0

0


T8.

(a) 1 + A = 1



A


1+A

0

1

1

1

1

1


(b) 1 A = A


A


1 A

0

1

0

1

1

1

T9.


(a) A’ + A = 1


A

A’

A’+A

0

1

1

1

0

1


(b) A’ A = 0



A

A’

A’ A

0

1

0

1

0

0


T10.


(a) A + A’ B = A + B


A

B

A’

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


A

B

‘A’

A’+B

A(A’+B)

AB

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.Theorema De Morgan's


(a) ( A + B)’ = A’ B’


A

B

A’

B’

(A+B)’

A’B’

0

0

1

1

1

1

0

1

1

0

0

0

1

0

0

1

0

0

1

1

0

0

0

0


(b) ( A B )’ = A’ + B’


A

B

A’

B’

(A B)’

A’+B’

0

0

1

1

1

1

0

1

1

0

1

1

1

0

0

1

1

1

1

1

0

0

0

0



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