106 points
In the following table ! preceeding an input specifies NOT
|
Laws Identity Null Itempotent Inverse Commutative Associative Distributive Absorption 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*(A+B)=A !(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+A*B=A !A*!B=!(A+B) |
General rule: When given a combination of ANDs, ORs and NOTs, NOTS applied to particular input are applied first, then ANDS, then ORs. Statements inside parens resolved before those outside.
All problems are to be done by hand with pencil. Typed assignments not accepted.
Simplification
For each of the following problems, simplify the statements using boolean algebra.
Example:
Problem: F=S*T+V*W+R*S*T
| F=S*T+R*S*T+V*W | - Commutative law |
| F=1*S*T+R*S*T+V*W | - Identity law |
| F=(1+R)*S*T+V*W | - Distributave(or) |
| F=1*S*T+V*W | - Null |
| F=S*T+V*W | - Identity law |
Part A. 31 pts. Give the rules for each step of solution.
Part B. (10 points ea.) Simplify the equation. Show your work and name the rules as they are applied. You may use any rule except absorption and you don't have to state the commutative and associative rules when applied. The solution for each problem is stated on the right of the =
| 1. !(!D*!E+E) = D*!E | 2. C*(A+B+!(!C+!D))*D = CD | ||
| 3. A+!A*B*C=A+B*C | 4. !(X*(X*!Y+!X*Y))=!X+Y | ||
| Hint : using distribution rule x+yz = (x+y)(x+z)
and substitution, it is possible to set this so that the inverse removes the !A Try treating !A*B or B*C as if they are single terms. |
Use Demorgans | ||
| 5. (A+B)*(A+C)*(!B+!C)=A*!(B*C) |