hi, guys....i need some help with these question...i will be grateful if u could help me...

simplification using boolean algebra only..

A) F1 = w'x' + (w'x')'

b) F2 = w(w + w'xy)

c) F3 = xy(v.v')

d) F4 = (v + (wy)')'

e) F(v,w,x,y) = F1+F2+F3+F4


thank you in advance

 

i have tried to do them not sure if they r right?
w'x' + (w'x')'
=w'x' + wx

 

if ' means bar then solution might be

f1=w'x'+(w'x')'
for this now let w'x' be a then
f1=a+a' that is f1=1;
let me explain this
suppose a=1 then a'=0 in this case f1=1;
and if a=0 then a'=1 in this case also f1=1;
hence f1=1 always...

now going for f2=w(w+w'xy)
or f2=ww+ww'xy and as ww=w and ww'=0 always
therefore f2=w....

now f3=x.y(v.v') then v.v'=0....
therefore f3=0....

now F4 = (v + (wy)')'
using de'morgans law
f4=(v+w'+y')'..
and again we will use de'morgans law..
f4=v'wy..

therefore..

F(v,w,x,y) = F1+F2+F3+F4=1+ w+0+v'wy..

i think i have tried my level best...

 

thanks alot for the help...really appreciate...

 

wow! boolean algebra!! need ko po ng tutorial dyan assignment ko po kasi eh... pls post naman po ng link about boolean
ty po...