Simplify the following functional expression using Boolean a

Simplify the following functional expression using Boolean algebra. (x + y)\'(x\' + y\')\'

Solution

given expression is = (x+y)\'(x\'+y\')\'

=(x+y)\'((x\')\'.(y\')\') [ since formula demorgans law (x+y)\'=x\'.y\']

=(x+y)\'(xy) [ since formula (x\')\'=x; (y\')\'=y]

=(x\'.y\')(xy) [since demorgans law (x+y)\'=x\'.y\']

=x\'.y\' xy [ demorgans law(x+y)\'=x\'.y\']

=(x.x\')(y\'.y)

=(0).(0) [ since formula x.x\'=0 ; y.y\'=0]

=0

Therefore (X+Y)\'(X\'+Y\')\'=0