Fuzzy logic de morgan

Question

i should show, that de morgan's law is also correct in the fuzzy logic:

$\neg(A\vee B)$ could be written as: $1-max(a,b)$

$(\neg A) \land (\neg B)$ as: $min(1-a, 1-b)$

But how could I show, that both is the same?

Thanks for your help!

Answer 1

All you need is:

$$\begin{align}1-\max (a,b) & = 1+\min(-a, -b) \\ &= \min(1-a, 1-b) \end{align}$$

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Why is this possible? I do not understand this solution. – Babs

Reason: $$\begin{align} 1-\max(a, b) & =1 +(-1)\times\begin{cases} a & : a\geq b\\b & : a< b\end{cases} \\ & = 1+\begin{cases}-a & : a\geq b\\-b& : a<b\end{cases} \\ & = 1+\begin{cases}-a & : -a\leq -b\\-b & : -a<-b\end{cases} & = 1+\min(-a,-b) \\ & = \begin{cases}1-a & : -a\leq -b\\1-b & : -a<-b\end{cases} \\ & = \min(1-a, 1-b) \end{align}$$