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*+ BComplement 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 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*)CIdempotent, 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. 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 + A *T*+ ABCommutative, 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.