(x’y’+z)’ + z + xy + wz = xz' +yz' +z+xy+wz ; first term by DeMorgan's theorem
xz'+yz'+z+xy+wz = xz'+yz'+xy+z(1+w) = xz'+yz'+xy+z ; now only three terms
xz'+yz'+xy+z = x(y+y')z'+(x+x')yz'+xy(z+z')+(x+x')(y+y')z ;expand out the terms
= xyz'+xy'z'+xyz'+x'yz'+xyz+xyz'+xyz+xy'z+x'yz+x'y'z
=x'y'z+x'yz'+x'yz+xy'z'+xy'z+xyz'+xyz ;eliminating duplicate terms
(xy'z'+xy'z+xyz'+xyz)+(x'y'z+x'yz+xy'z+xyz)+(x'yz'+x'yz+xyz'+xyz) ;rearranging terms, we can duplicate terms as many times as we need to do so
(xy'z'+xy'z+xyz'+xyz)=xy'(z'+z)+xy(z'+z)=xy'+xy=x(y+y')=x ;reducing the first bracketed term
(x'y'z+x'yz+xy'z+xyz) = z reducing the second term is just as easy
(x'yz'+x'yz+xyz'+xyz) = y and finally the third term is reduced