Confusion in SOP and POS form

Question

Given

x’ + x(x + y’)(y + z’)

Turn it into SOP and POS form (not the Canonical ones) and without using a K-map

I have done as follows;

    = x’ + x(x + y’)(y + z’)
    = x’ + (x + xy’)(y+z’)
    = x’ + xy’z’+ xyy’ + xz’ + xy
    = x’ + xy’z’ + xz’ + xy
    = x’ + xz’(y’+1) + xy
    = x’ + xz’ + xy
    = (x’ + x)(x’ + z’) + xy
    = x’ + z’ + xy
    = (x’ + x)(x’ + y) + z’
    = x’ + y + z’

Now is x’ + y + z’ both the SOP and POS form of the given? Or should I have stopped midway and did another approach?

I see. I'm just a little bit confused as most of the materials always give an example of e.g. x + yz as an SOP as it is a sum of product/s and some such as z(x+y) for a POS — Tyro, Oct 25 '20 at 05:01
So I could just see this in this way: in SOP as if they are products of 1 and in POS as if the whole (x' + y + z') is a product of 1 so hence a product of sums as well — Tyro, Oct 25 '20 at 05:03

score 2 · Answer 1 · answered Oct 26 '20 at 16:32

You did everything correctly, and yes, the final form is both in SOP and POS, but here are two very useful principles that will make your boolean-algebra-life a lot easier:

Absorption

$x + xy = x$

and its dual:

$x(x+y)=x$

Reduction

$x+x'y=x+y$

and its dual:

$x(x'+y)=xy$

With those:

$x'+x(x+y')(y+z')\overset{Absorption}{=}$

$x'+x(y+z')\overset{Reduction}{=}$

$x'+y+z'$

Done!

Confusion in SOP and POS form

1 Answers1