Boolean Algebra - Binary logic

In the following table ! preceeding an input specifies NOT

Laws
Identity
Null
Itempotent
Inverse
Commutative
Associative
Distributive
De Morgan's
Cancelation
  AND
1*A=A
0*A=0
A*A=A
A*!A=0
A*B=B*A
(A*B)*C=A*(B*C)
A+B*C=(A+B)*(A+C)
!(A*B)=!A+!B
!!A=A
  OR
0+A=A
1+A=1
A+A=A
A+!A=1
A+B=B+A
(A+B)+C=A+(B+C)
A*(B+C)=A*B+A*C
!A*!B=!(A+B)

General rule: When given a combination of ANDs, ORs and NOTs,

Part A. Simplify the following :

1. !(!A*!B+B)    2. !(A*!B+B)    3. (A+B)*(A+C)*(!B+!C)
4. C*(A+B+!(!C+!D))*D    5. !A+A*!B
Give the rules for each step of solution.
Worksheet1
Worksheet2

Part B. Draw the gate representation for the initial boolean statement for each of the problems in A.

Part C. Give the truth tables for both the initial boolean statement and simplified staement in Part A.