Let $f(x,y,z,w)=zw+z'w'+xy'z'w+xyz'w$
Disjunctive normal form
$zw(x'+x)(y'+y)+z'w'(x'+x)(y'+y)+xy'z'w+xyz'w=(zwx'+zwx)(y'+y)+(z'w'x'+z'w'x)(y+y')+xy'z'w+xyz'w$
$DNF=zwx'y'+zwx'y+zwxy'+zwxy+z'w'x'y'+z'wx'y'+z'w'xy'+z'w'xy+xy'z'w+xyz'w$
Simplified form
$ \begin{matrix} & y & y' & y' & y \\ x & \color{red}{1} & \color{red}{1} & 0 & 0 & z \\ x' & \color{red}{1} & \color{red}{1} & 0 & 0 & z \\ x' & 0 & 0 & \color{green}{1} & \color{green}{1} & z' \\ x & \color{blue}{1} & \color{blue}{1} & \color{green}{1} & \color{green}{1} & z' \\ & w & w & w' & w' \\ \end{matrix} $
Where $L_1, L_2, L_3$ are red, green and blue, respectively.
So (...)
$L_1=wz$
$L_2=w'z'$
$L_3=wz'$
$SF = L_1+L_2+L_3 =wz+w'z'+wz'=w(z+z')+w'z'=w+w'z'$
My question is: this is well done?
Not enough reputation to comment:
How come the $L_3 = wz'$
To my knowledge, if we combine the entire row not only the blues but also part of reds then $L_3 = xz'$
Or if we combine with a part of reds in the first rows then it would $L_3 = xw$