# Boolean Equation Simplification

1. Aug 4, 2009

### Berbanog

1. The problem statement, all variables and given/known data
Q1: Show that
a) ~(~(A+B)~C.A).~(~(B.C)+A.B) = ~(~(C.~A)+~B.C)~(A.~B.~(C+~B))
a-2) *same as above* ~(~(A+B).C'.A).~(~(B.C)+A.B) = ~(~(C.A')+B'.C).~(A.B'.~(C+B'))
b) ~(A XOR B) = (~A) XOR B
c) A + A'B = A + B

Q2: Represent the following using only NAND gates, and only NOR gates
a) A.B + ~(A.C).~(B+C)
b) (A XOR B) + ~(A XOR B).(B + C)
c) (A+B).(~(A+C)+~(B.C))

2. Relevant equations
3. The attempt at a solution
a)
LHS = A + (~A)BC = A + BC
RHS = ~A + BC
*where did i do wrong?
b) Tried Truth Table
A B || ~(A XOR B)
---------------------
0 0 || 1
0 1 || 0
1 0 || 0
1 1 || 1

A B ~A|| (~A) XOR B
---------------------
0 0 1 || 0
0 1 1 || 1
1 0 0 || 1
1 1 0 || 0

Is this suppose to be correct? i find it confusing...to me
(~A) XOR B = A XOR B...but ofcourse the question is asking to prove
~(A XOR B) = (~A) XOR B!

c) i'm quite sure this was distributive law...but how do i prove this WITHOUT using Karnaugh map?

Q2

I : Trial and error solved the question backward!!!
II: I first simplied the question using theorems & K-map, then with the result figured out the NOR/NAND equation.
a) i got 2 solutions on this one..(???!!!!)
I : ~(~A+~B~(~A+~C)+~(~B.~C)) or in different form...~(A'+B'~(A'+C')+~(B'C'))
II: ~(A~BC) or in different format...~(AB'C)

Method used
b) dunno where i have to begin this since i ain't that great with XOR complements...If someone could give me any tips i'd appreciate it!
c) ~(~A+B(~B)+C) or in different form...~(A'+BB'+C)