Reduce the following expressions to a minimum SOP form.

\(\[f = x'y'z + w'xz + wxyz' + wxz + w'xyz\]\) (to 3 terms, 7 literals)

My final answer was x'y'z + xz + xy
The book gives y'z + xz + wxy, which I think is irreducible.

My question is, for minimum SOP, do I have to find the solution that is irreducible (and also matches the condition, 3 terms & 7 literals)?

I basically applied twice ab + ab' = a (adjacency)

first time, wxyz' + w'xyz (where xy is a, b = wz')
then this reduces to

f = x'y'z + w'xz + xy + wxz
and again, w' is the b, and xz is the a, so we have x'y'z + xz + xy

It's true the first and last terms can be reduced. But this form is also 3 terms and 7 literals.

Thank you for input!

 

I'm afraid you have a mistake. You have included at least minterm #12 in your map. That is visible by the grouping XY you have done. 1100 satisfies XY but it doesn't satisfy any of the terms of the f as described above. Your book has the right answer, which is also the only one.

 

yes, you are right, using the K-map I will get the right answer.

But say I need to prove both.
Say I am doing the simplification using only switching algebra.
How are my steps wrong?

I appreciate your input, Thank you sir!

 

I basically applied twice ab + ab' = a (adjacency)

first time, wxyz' + w'xyz (where xy is a, b = wz')
then this reduces to

f = x'y'z + w'xz + xy + wxz

Well, there's your problem! Notice that in order to reduce a(b+b') to a, the two terms in the parenthesis must be complementary. But complementary the wz' and w'z are not. (wz')'=w'+z.

I say it all the time to inquirers of this forum and it's still not enough. When you are a starter in the Boolean algebra, don't assume anything. Prove everything over and over because things don't work exactly like the regular algebra.

 

Yes you are right. Now I know why I always end up in a wrong expression because I forgot that it's a product of two variables, and the complement is the sum of product like you said.

Thank you!