11. Simplify the following functional expressions using Boolean algebra and its identities. List the identity used at each step.
a) = x’yz+xz
= z(x’y+x) {Apply identity :- A+BC=(A+B)(A+C)}
= z[(x’+x)(y+x)] {Apply identity:- x’+x=1}
= z(y+x)
= xz+yz
b) = (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) =(x’(x’)’y)’ {By applying DeMorgans law (A.B)’=A’+B”}
= (x’)’+((x’)’)+y’ {By applying identity :- (X’)’=X and ((x’)’)’=x’}
=x+x’+y’ {By applying identity:- x+x’=1}
= 1+y’ { By applying identity :- 1+x=1}
=1