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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?

Zephyr
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  • If, as is customary, multiplication and addition refer to the Boolean meet and join operations, then not only is $(x+y)(y+z)(z+x)$ self-dual, but it's the same function as $xy+yz+zx$. – Andreas Blass Oct 28 '17 at 16:39
  • @AndreasBlass addition is just normal OR operation and multiplication is AND operation. – Zephyr Oct 28 '17 at 16:40
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    Since AND is Boolean meet and OR is Boolean join, my previous comment applies. – Andreas Blass Oct 28 '17 at 20:05

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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?

M. García
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