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)

this would start to sound a little like mortal combat