COS1521
EXAM
PACK
,Boolean Expression Simplification
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 is the
list of simplification rules.
Simplify: C + BC:
Expression Rule(s) Used
C + BC Original Expression
C + (B + C) DeMorgan's Law.
(C + C) + B Commutative, Associative Laws.
T+B Complement Law.
T Identity Law.
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
Distributive law. This step uses the fact that or distributes over and. It can look a
A + BB 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.
You can also use distribution of or over and starting from A(A+C)+C to reach the same result by another
route.
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.
Complement, then Identity. (Strictly speaking, we also used the
AB + (B + A)A + (B + A)B Commutative Law for each of these applications.)
AB + BA + AA + BB + AB Distributive, two places.
Idempotent (for the A's), then Complement and Identity to remove
AB + BA + A + AB 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.
EXAM
PACK
,Boolean Expression Simplification
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 is the
list of simplification rules.
Simplify: C + BC:
Expression Rule(s) Used
C + BC Original Expression
C + (B + C) DeMorgan's Law.
(C + C) + B Commutative, Associative Laws.
T+B Complement Law.
T Identity Law.
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
Distributive law. This step uses the fact that or distributes over and. It can look a
A + BB 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.
You can also use distribution of or over and starting from A(A+C)+C to reach the same result by another
route.
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.
Complement, then Identity. (Strictly speaking, we also used the
AB + (B + A)A + (B + A)B Commutative Law for each of these applications.)
AB + BA + AA + BB + AB Distributive, two places.
Idempotent (for the A's), then Complement and Identity to remove
AB + BA + A + AB 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.
