Deriving SOP and POS Forms of a 4-Var Kmap Function

[kukumaluboy]

Homework Statement

Hi my question is whether an SOP = POS for a given function F(A, B, C, D). Or is the SOP a complement of POS?Tasked to convert derive a simplfied SOP from a 4 var Kmap with don't cares.
My answer was,
F = C'.D' + A'.B'.C which is correct.

Then for part B they asked for the simplified POS form.

The Attempt at a Solution

F = C'.D' + A'.B'.C
F' = (C'.D' + A'.B'.C)' (Using Dmorg)
F' = ((c+d)' + (a+b+c')')' (Using Dmorg)
F' = (c+d)'.(a+b+c')'

F = ((c+d)'.(a+b+c')')'
F = (c+d) +(a+b+c') (Stuck here)If i used Kmap i got the answer which is
D.(B'+C').(A'+C')

I want to know the boolean algebra way

 

[cpscdave]

There is a modification to KMaps to get POS instead of SOP.
It involves instead of circling the 1's you circle the 0's
Prehaps look for it in your textbook? I only used it on the 1 assignment where we had to do it and then promptly forgot it.
The other way is to use applications of DeMorgans to get there. But that seems like a lot of work :) the POS KMap was much easier

 

[kukumaluboy]

Yea i got the KMap part. I want know the demorgans part

 

[Korisnik]

I'm not sure what you actually want. Sum of minterms can be converted to product of maxterms and that is equivalent.
The easy way to convert them is using a dual form (twice), so, you simply switch * and + (and 1 and 0) to obtain a dual function. Then multiply everything in order to again get a sum of minterms and use a dual.
In your case: f = C'D' + A'B'C
f_d = (C' + D') (A' + B' + C) = C'A' + C'B' + C'C + D'A' + D'B' + D'C = C'A' + C'B' + D'A' + D'B' + D'C
(f_d)_d = f = (C' + A')(C' + B)(D' + A')(D' + B')(D' + C')
If that's what you wanted. I don't know how you got: C'.D' + A'.B'.C == D.(B'+C').(A'+C'), because I can't get that. I can simplify my own form though, but I don't get your result.