# SOP vs POS

## 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 dont cares.
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 wanna know the boolean algebra way

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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

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

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.

• kukumaluboy