Due 13 Feb. 2008 in class at beginning of class.
In the following table ! preceeding an input specifies NOT
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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.
Part 1 - Absorption 5 points
Prove the AND form of absorption using the other rules. Do NOT use the OR form of absorption as a shortcut.
Part 2 - Simplification 5 points each
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 |
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 =
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A. A*B*C+A*B+C = A*B+C B. A*(B+A+C) = A C. A*C+C(!A+A*B) = C |
D. (A+B*C*D+E)*E*D = D*E E. F=(A+B)(A+C)(!A+C) = (A+B)*C F. C=!(A*B)*(A+B) = A!B + B!A |
To determine how DeMorgan is applied when a NOT is applied to a mixed collection of gates, start by grouping the ANDs inside parenthesis then apply DeMorgans. Keep the added parethesis in place while distributing. Re-evaluate the equation until NOT fully distributed.
A. !(A*B+C) = !((A*B)+C) = !(A*B)*!C = (!A+!B)*!C = !A!C + !B!C -------- -----------
Part 4 - Using Truth Tables; 5 points each
Give the truth table for each statement. With the help of the truth table, see if there is a simpler form. If possible, simplify the statement using the boolean rules. Show your work and name the rules as they are applied. confirm the simplification. The link below will give you a prepared set of truth tables for the full expressions. Distribution as a first step may prove helpful for several of these.
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A. A+!A+B B. A*(!A+B) C. !A+A*B |
D. !A+A*!B (try cancelation and deMorgan) E. A*(!A*B) F. !A*(A+!B) |
To prove the above statement is true, fill in the truth tables for the starting and ending statements. Blank tables are provided on the same sheet as the previous problem.
Part 5 - More DeMorgan 5 Points each
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.
A. !(!D*!D+E)
B. C*(A+B+!(!C+!D))*D
C. !(!(A*B)*(A+B)*C)
D. !(!(A*B)*(A+B)+C)
E. !((A*B+B*!C)*A)