Simplify the following Boolean expressions to a minimum number of literals

a. xy+xy'

b. (x+y) (x+y')

c. (a+b+c') (a'b'+c)

d. a'bc+abc'+abc+a'bc'

e. (x+y) ' (x'+y') f. xy+x (wz+wz')

a. xy+xy' = x

b. (x+y) (x+y') = x

c. (a+b+c') (a'b'+c) = c (a + b) + a'b'c'

d. a'bc+abc'+abc+a'bc' = b

e. (x+y) ' (x'+y') = x'y'

f. xy+x (wz+wz') = x (y+w)

Explanation:

a) xy+xy'

= x (y + y′) / / taking x common

= x. 1 / / y + y' = 1

= x

b) (x+y) (x+y')

= xx + xy′ + yx + yy′ / /multiplying x+y with x+y'

= x + xy′ + xy + 0 / / xx = 1 yy' = 0

= x (1 + y′ + y) / / taking x common

= x. 1 / / y+y' = 1

= x

c) (a+b+c') (a'b'+c)

= aa'b' + ac + a'b'b + bc + a'b'c' + cc'

= 0 + ac + 0 + bc + a'b'c' + 0 / / aa' = 0 bb'=0 cc' = 0

= ac + bc + a'b'c'

= c (a + b) + a'b'c'

d) a'bc + abc' + abc + a'bc'

= b (a'c + ac' + ac + a'c') / /taking b common

= b (a'c + a'c' + ac' + ac)

= b (a' (c+c') + a (c+c') / / taking a common

= b (a' (1) + a (1)) / /c+c' = 1

= b (a'+a) / /a+a'=1

= b (1)

= b

e) (x+y) ' (x'+y')

= ((x+y) ' (x'+y'))

= ((x+y) + (x'+y') ') '

= (x + y + (x''y'')) '

= (x + y + (xy)) '

= (x + y + xy) '

= (x (1 + y) + y) '

= (x + y) '

= x'y'

OR it can be done as:

e) (x+y) ' (x'+y')

= x′y′ (x′+y′) / / (x+y) ' = x'y'

= x′y′ + x′y′ / / (xy) ' = x'+y'

= x′y′

f. xy+x (wz+wz')

= xy + xw (z+z′) / /taking w common

= xy + xw / / z+z'=1

= x (y+w)