Aljabar BOOLEAN

TEOREMA ALJABAR BOOLEAN

T1. COMMUTATIVE LAW :

a. A + B = B + A

b. A . B = B . A

T2. ASSOCIATIVE LAW :

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

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

T3. DISTRIBUTIVE LAW :

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

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

T4.IDENTITY LAW:

a. A + A = A 

b. A . A = A 

T5.NEGATION LAW:

a.( A’) = A’

b. ( A’’) = A

T6. REDUNDANCE LAW :

a. A + A. B = A

b. A .( A + B) = A

T8. :

a. A’+ A = 1

b. A’. A = 0

T9. :

a. A + A’. B = A + B

b. A.( A’+ B ) = A . B

T7. :

a. 0 + A = A 

b. 1 . A = A 

c. 1 + A = 1

d. 0 . A = 0

10. DE MORGAN’S THEOREM:

a. (A + B ) = A . B 

b. (A . B ) = A + B

Contoh soal dan pembahasan

Contoh1 Sederhanakan  A . (A . B + C)

Penyelesaian

A . (A . B + C)

= A . A . B + A . C (T3a)

= A . B + A . C       (T4b)

= A . (B + C)          (T3a)

Contoh 2  Sederhanakan  A’. B + A . B + A’. B’

Penyelesaian

A’. B + A . B + A’. B’

= (A’+ A) . B + A’. B’ (T3a)

= 1 . B + A’. B’           (T8a)

= B + A’. B’                (T7b)

= B + A’                     (T9a)

Contoh3 Sederhanakan  A + A . B’+ A’. B

Penyelesaian

A + A . B’+ A’. B

= (A + A . B’) + A’. B

= A + A’. B                 (T6a)

= A + B                       (T9a)