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