Senin, 19 April 2010

Tugas 4.A

1. Hukum Komutatif

a. A+B=B+A

A

B

A+B

B+A

A+B=B+A

0

0

1

1

0

1

0

1

0

1

1

1

0

1

1

1





b. AB=BA

A

B

AB

BA

AB=BA

0

0

1

1

0

1

0

1

0

0

0

1

0

0

0

1





2. Hukum Asosiatif

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

A

B

C

A+B

B+C

(A+B)+C

A+(B+C)

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

0

0

0

0

1

1

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

0

1

1

1

1

1

1

0

1

1

1

0

1

1

1

0

1

1

1

1

1

1

1

0

1

1

1

1

1

1

1









b. (AB)C = A(BC)

A

B

C

AB

BC

(AB)C

A(BC)

(AB)C = A(BC)

0

0

0

0

1

1

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

0

0

0

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

1








3. Hukum Distributif

a. A(A+C) = AB+AC

A

B

C

BC

A+B

AC

A(B+C)

AB+AC

0

0

0

0

1

1

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

1

1

1

0

1

1

1

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

0

1

1

1

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

A

B

C

BC

A+B

A+C

A+(BC)

(A+B)(A+C)

0

0

0

0

1

1

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

0

0

1

0

0

0

1

0

0

1

1

1

1

1

1

0

1

0

1

1

1

1

1

0

0

0

1

1

1

1

1

0

0

0

1

1

1

1

1

4. Hukum Identity

a. A+A = A

A

A+A

A+A=A

0

0

1

1

0

0

1

1

b. AA = A

A

AA

AA=A

0

0

1

1

0

0

1

1

5. a. AB+AB = A

A

B

B(invers)

AB

AB(inverse)

A

0

0

1

1

0

1

0

1

1

0

1

0

0

0

0

1

0

0

1

0

0

0

1

1

b. (A+B)(A+B) = A

A

B

B(invers)

A+B

A+B(inverse)

A

0

0

1

1

0

1

0

1

1

0

1

0

0

1

1

1

1

0

1

1

0

0

1

1

6. Hukum Redudansi

a. A+AB = A

A

B

AB

A+AB

0

0

1

1

0

1

0

1

0

1

1

1

0

0

1

1

b. A+AB = A

A

B

A+B

A(A+B)

0

0

1

1

0

1

0

1

0

1

1

1

0

0

1

1

7. a. 0+A = A

A

0+A

0

0

1

1

0

0

1

1

b. 0A = 0

A

0A

0

0

0

1

1

0

0

0

0

0

0

0

0

8. a. 1+A = 1

A

1+A

1

0

0

1

1

1

1

1

1

1

1

1

1


b. 1A = A

A

1A

0

0

1

1

0

0

1

1

9. Theorema De Morgan

a.

A

B

A(invers)

B(inverse)

A+B

(A+B)inverse

A(inverse)B(inverse)

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(inverse)

AB

(AB)inverse

A(inverse)+B(inverse)

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



0 komentar:

Posting Komentar