4.2.2.1 Peraturan Boolean Algebra.
Ekspresi Boolean boleh di ringkaskan dan
boleh di manipulasi.Jadual 2 menunjukkan peraturan asas Boolean Algebra dapat
membantu memanipulasi dan menyelesaikan persamaan logik.
Jadual 4
– 2 : Peraturan Asas Boolean Algebra.
|
|
AND Form
|
OR
Form
|
|
Identity
Law
|
A.1 = A
|
A + 0 = A
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Zero
and One Law
|
A.0 = 0
|
A + 1 = 1
|
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Inverse
Law
|
A.
= 0
|
A +
= 1
|
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Idempotent
Law
|
A.A = A
|
A + A = A
|
|
Commutative
Law
|
A.B = B.A
|
A + B = B + A
|
|
Associative
Law
|
A.( B.C ) = ( A.B ).C
|
A + ( B + C ) = ( A+B ) + C
|
|
Distributive
Law
|
A
+ ( B.C ) = (A + B ).( A + C )
|
A.(
B + C ) = ( A.B ) + ( A.C )
|
|
Absorption
Law
|
A(
A + B) = A
|
A
+ A.B = A
A
+ A’B = A + B
|
|
DeMorgan’s
Law
|
(
) =
+
|
(
) =
.
|
|
Double
Complement Law
|
= X
|
= X
|
DERIVATION
Absorption Law Derivation
A( A+B ) = A ( 1 + B ) à1
+ B = 1
= A(1)
à A . 1 = A
= A
Absorption Law Derivation
A( A + B ) = AA + AB à
A = A . A
= A + AB
àA ( 1 + B ) = A (1)
= A
Absorption Law Derivation
A + A’B = ( A + AB) + A’B à
A = A .A
= ( AA + AB ) + A’B à
( AA + BB ) + A’B = AA + ( AB + A’B )
= AA + AB + A’B
= ( A + A’ ) + ( A + B) à
( A + A’) (A + B) = AA +AB + A’B
= 1. ( A + B ) à
A + A’ = 1
= ( A + B )
Distributive Law Reverse Derivation
( A + B ) . ( A + C ) = AA + AC + AB +
BC à
AA = A
= A + AC +AB
+BC à
A ( 1 + C ) = A( 1) = A
= A + AB + BC à
A (1 + B ) = A ( 1) = A
= A ( 1 + B ) +
BC
= A . 1 + BC
= A + BC
4.2.2.1.1 De Morgan’s Law
“ jika garisan di putuskan , maka
tandanya akan berubah”
=
+
=
.
