10. Simplify the following functional expressions using Boolean algebra and its identities. List the identity used at each step.
a) F(x,y,z) = x’y+xyz’+xyz
= x’y+xy(z+z’) {identity :- z+z’=1 }
= x’y+xy.1
= y(x+x’) { again applying the same identity}
= y
b) F(w,x,y,z) = (xy’+w’z)(wx’+yz’)
=(xx’y’w+xy’yz’+w’wx’z+w’zyz’) {Apply identity :-}x.x’=0}
=0+ 0 + 0+0
= 0
c) F(x,y,z)= (x+y)’(x’+y’)’ {By applying DeMorgans law (x+y)’=x’.y’ }
= (x’.y’)[(x’)’.(y’)’] {Apply identities x.x’=0 and (x’)’=x}
= (x’.y’)x.y
=0