Use Boolean algebra to prove the following equations Describ

Use Boolean algebra to prove the following equations. Describe the properties that you use. x + yz = (x+y)(x+z) xy+yz + x\'z = xy + x\'z Use De Morgan\'s rule to find the complement of F = (w\'z + xy\') + (x + y) + (w + z)

Solution

a) (x+y)(x+z)= x.x+x.z+x.y+y.z      (distributive law)

                   =x+x.z+x.y+y.z ,      (a.a=a) boolean algebra rule

                   =x(1+z)+x.y+y.z = x.1+xy+yz,      (1+a=1)

                   = x(1+y)+yz=x+yz

thus LHS=RHS proved

b) xy+yz+x\'z=xy+x\'z

xy+(x+x\')yz+x\'z=xy+xyz+x\'yz+x\'z     (x=x\'=1 and distibutive law)

=xy(1+z)+x\'z(1+y)=xy+x\'z proved

F=(w\'z+xy\')+(x+y\')+(w+z)

=((w\'z+xy\')+(x+y\')+(w+z))\'

let, w\'z+xy\'=A, x+y\'=B, w+z=C

applying demorgans theorem(A+B+C)\'=A\'.B\'.C\'

=(w\'z+xy\')\' (x+y\')\' (w+z)\'

=(w\'z)\'.(xy\')\' (x\'.y\'\') (w\'.z\')       (A.B.C)\'=A\'+B\'+C\'

=(w\'\'+z\') (x\'+y\'\') (x\'.y)(w\'.z\')

=(w+z\')(w\'.z\')(x\'+y)(x\'.y)    

=(w.w\'.z\'+w\'.z\'.z\')(x\'.x\'.y+x\'.y.y)    (distributive law)

=(0+w\'.z\')(x\'y+x\'y)                (A.A\'=0,A.A=A,)

=(w\'z\')(x\'y)                             (A+A=A,A+0=A)