<s>[INST] Self dual functions My book says that $f(x,y,z) = xy + yz + zx$ is self dual function but $(x+y)(y+z)(z+x)$ is not.
I understood how $xy + yz + zx$ is self dual but I think $(x+y)(y+z)(z+x)$ is also self dual because in k-map if we represent $(x+y)(y+z)(z+x)$ then it's minimized SOP is $xy + yz + zx$ and if we take dual of this, then we get original $(x+y)(y+z)(z+x)$. I think self dual is commutative.
Am I correct?
 [/INST] For all boolean expression $e(x_{1},\ldots,x_{n})$, let us represent the dual expression of $e(x_{1},\ldots,x_{n})$ by $e(x_{1},\ldots,x_{n})^{D}$. 
Expression $xy+yz+zx$ is self dual. In effect, in virtue of elementals properties of boolean algebra, the following equalities hold:
\begin{align*}
(x+y)(y+z)(z+x)&=x(y+z)(z+x)+y(y+z)(z+x)\\
               &=(xy+xz)(z+x)+(yy+yz)(z+x)\\
               &=(xy+xz)z+(xy+xz)x+(y+yz)z+(y+yz)x\\
               &=xyz+xzz+xyx+xzx+yz+yzz+yx+yzx\\
               &=xyz+xz+xy+xz+yz+yz+yx+yzx\\
               &=xz+xy+yz+xyz\\
               &=(xy+xyz)+yz+xz\\
               &=xy+yz+zx
\end{align*}
Therefore:
\begin{align*}
(xy+yz+zx)^{D}&=(x+y)(y+z)(z+x)\\
              &=xy+yz+zx
\end{align*}
Moreover:
\begin{align*}
((x+y)(y+z)(z+x))^{D}&=xy+yz+zx\\
              &=(x+y)(y+z)(z+x)
\end{align*}
and so $(x+y)(y+z)(z+x)$ is too self dual. Does this answer your question?
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