Computing the dual of the boolean function

Question

Suppose, $f(A,B,C,D) = \sum m(0,2,4,6)$ is a boolean expression, expressed in minterm form. How can I calculate and express the dual of the function?

I have tried the method using the exact definition, but it turns out to be very cumbersome and prone to errors. Any other method?

Answer 1

$f(A,B,C,D)=\Sigma m(0,2,4,6)$

$f(A,B,C,D)= A'B'C'D'+A'B'CD'+A'BC'D'+A'BCD'$

Dual of $f$, 

$f_d(A,B,C,D)= (A'+B'+C'+D').(A'+B'+C+D').(A'+B+C'+D').(A'+B+C+D')$

$= M_{15}.M_{13}.M_{11}.M_9$

$=\Pi M(9,11,13,15)$

$= \Sigma m(0,1,2,3,4,5,6,7,8,10,12,14)$

That's it.

Note:

In Dual of $f$, that is $f_d$, we replace AND $(.)$ by OR $(+)$, OR $(+)$ by AND $(.)$, $0$ by $1$ and $1$ by $0$ ONLY.

In Complement of $f$, that is $f'$, we need to replace all variable, say $x$ , by it complement variables , say $x'$, also.

Comments

This method is much easier n simple too.. rahter thyan described above

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Canonical sum of products and canonical pos is said to be dual to each other, is this valid?

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yes https://en.wikipedia.org/wiki/Canonical_normal_form