Simplification of Boolean Expressions
Here
are some examples of Boolean algebra simplifications. Each line gives a form of
the expression, and the rule or rules used to derive it from the previous one.
Generally, there are several ways to reach the result. Here are some of the examples for you as guideline :
- Simplify: AB(A + B)(B +
B):
Expression
|
Rule(s) Used
|
AB(A + B)(B + B)
|
Original Expression
|
AB(A + B)
|
Complement law, Identity law.
|
(A + B)(A + B)
|
DeMorgan's Law
|
A + BB
|
Distributive law. This step uses
the fact that or distributes over and. It can look a bit strange since
addition does not distribute over multiplication.
|
A
|
Complement, Identity.
|
- Simplify: (A + C)(AD + AD)
+ AC + C:
Expression
|
Rule(s) Used
|
(A + C)(AD + AD) + AC + C
|
Original Expression
|
(A + C)A(D + D) + AC + C
|
Distributive.
|
(A + C)A + AC + C
|
Complement, Identity.
|
A((A + C) + C) + C
|
Commutative, Distributive.
|
A(A + C) + C
|
Associative, Idempotent.
|
AA + AC + C
|
Distributive.
|
A + (A + T)C
|
Idempotent, Identity,
Distributive.
|
A + C
|
Identity, twice.
|
- Simplify: A(A
+ B) + (B + AA)(A + B):
Expression
|
Rule(s) Used
|
A(A + B) + (B + AA)(A + B)
|
Original Expression
|
AA + AB + (B + A)A + (B + A)B
|
Idempotent (AA to A), then
Distributive, used twice.
|
AB + (B + A)A + (B + A)B
|
Complement, then Identity.
(Strictly speaking, we also used the Commutative Law for each of these
applications.)
|
AB + BA + AA + BB + AB
|
Distributive, two places.
|
AB + BA + A + AB
|
Idempotent (for the A's), then
Complement and Identity to remove BB.
|
AB + AB + AT + AB
|
Commutative, Identity; setting up
for the next step.
|
AB + A(B + T + B)
|
Distributive.
|
AB + A
|
Identity, twice (depending how you
count it).
|
A + AB
|
Commutative.
|
(A + A)(A + B)
|
Distributive.
|
A + B
|
Complement, Identity.
|
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