edited by
2,870 views

1 Answer

Best answer
5 votes
5 votes
a) (x+x').(x+y) =x+y
b) (x.x')+(x.y) =xy
c) x'.z(y'+y)+x.y' = x'z+xy'
d) x.y+x'.z+y.z.1
   =xy+x'z+yz(x+x')
   =xy+x'z+yzx+yzx'
   =xy+xyz+x'z+x'yz
   =xy(1+z)+x'z(1+y)
   =xy.1+x'z.1
   =xy+x'z
e) same as option d this could be solved and
   result will be (x+y).(x'+z)
selected by

Related questions

6 votes
6 votes
7 answers
1
Payal Rastogi asked Dec 25, 2015
5,100 views
Q.86 The number of possible boolean functions that can be defined for $n$ boolean variables over $n$-valued boolean algebra is(a) $2^{2^n}$(b) $2^{n^2}$(c) $n^{2^n}$(d) $...