XY’+Z’ is a minimal SoP expression which represents the function (X,Y,Z).

The expression XY’ + YZ’ + X’Y’Z’ can be reduced to XY’+Z’

XY’ + YZ’ + X’Y’Z

= Y’(X+X’Z’) + YZ

= Y’(X+Z’) + Y

= XY’ + Y’Z’ + YZ’

= XY’ + (Y’+Y)Z’

= XY’ + Z’.

The expression (X+Z’)(Y’+Z’) is a PoS expression which also represents the same function (X,Y,Z).

f(w,0,0,z)= 1 If x=y=0, then the sum of the corresponding minterms be 1. 

The minterms with literals x’ and y’ are wx’y’z(9), w’x’y’z(1), wx’y’z’(8), w’x’y’z’(0) . 

If x=y=0, then we get wz+w’z+wz’+w’z’ = 1. 

f(1,x,1,z)= x+z. 

The minterms with variables w and y in true form and x or z or both in true form. 

The corresponding minterms are wx’yz(11), wxyz’(14), wxyz(15) 

If w=1 and y=1, then we get x’z+xz’+xz= x+z. 

f(w,1,y,z)= wz+y

The corresponding minterms are w’xyz’(6), w’xyz(7), wxyz’(14), wxyz(15), wxy’z(13).

If x=1, then we get w’yz’ + w’yz+ wyz’ + wyz+ wy’z = y + wz

So, the function f(w,x,y,z)= Σ(0,1,6,7, 8,9, 11, 13, 14, 15,).

Therefore, the k-map will be:

Therefore, the minimal expression will be: X’Y’ + WZ + XY 

Thus, the number of literals will be 6.