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How to prove that $$ C_1(f) = \min\{k : f \text{ is a $k$-DNF}\} $$ and $$ C_0(f) = \min \{ k: f \text{ is a $k$-CNF}\} $$ fulfill $$ C_1(f) = \max\{C(f,x):f(x)=1\}, \qquad C_0(f)=\max\{C(f,x):f(x)=0\} $$ where $C$ is the certificate complexity of a function $f\colon \{0,1\}^n \to \{0,1\}$.

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(plus: thx @martini) –  meshuai Nov 20 '12 at 12:47
    
Duplicate of math.stackexchange.com/questions/241229/… –  Joel Reyes Noche Nov 20 '12 at 12:48
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I see that you deleted your earlier question, as well as the comments made by others. Some will consider this rude. In the future, please refrain from deleting your question and reposting it again. Editing it is fine. –  Joel Reyes Noche Nov 20 '12 at 12:51
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@Joel Reyes Noche thx –  meshuai Nov 20 '12 at 12:54

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