Boolean Algebra question

gdbugger · 10-22-2003, 05:56 PM

I don't know of any other website/forum where I could post it. I guess you can post anything here excluding homework. So here goes:

A.B + A.B'.C.D + A'.C.D = A.B + C.D

I use A' to denote negation of A (or A bar). I can work it out using a Karnaugh Map, but not using the basic properties of Boolean Algebra. Can you help?

"Do not expect LQ members to do your homework - you will learn much more by doing it yourself." - http://www.linuxquestions.org/rules.php

This is not homework I just stumbled upon it while doing BDDs. I converted AB+CD to a ROBDD and when I convert it back to Boolean equation this is what I get A.B+A.B'.C.D+A'.C.D

Stack · 10-22-2003, 06:32 PM

Code:

AB+AB'CD+A'CD
AB+CD(AB'+A')
AB+CD((A'+A)(A'+B'))
AB+CD(A'+B')
AB+CD(AB)'
(AB+(AB)')(CD+AB)
AB+CD

gdbugger · 10-22-2003, 07:19 PM

Thank you.
I could manage only this much:

AB+AB'CD+A'CD
AB+CD(AB'+A')
AB+CD((A'+A)(A'+B'))
AB+CD(A'+B')

BTW can anyone suggest a place where I could post such questions. I actually didn't like posting this question at LQ

js_530 · 10-22-2003, 08:17 PM

Quote:

BTW can anyone suggest a place where I could post such questions. I actually didn't like posting this question at LQ

google newsgroups?

gdbugger · 10-22-2003, 08:37 PM

Thank you too.

sk8guitar · 10-23-2003, 06:54 PM

oh man, nightmares of my digital logic design class just came flooding back. karnaugh maps are the only way to go. i hated boolean algebra proofs. i hated doing combined variable karnaugh maps, where like instead of doing a 4 variable one you'd do a 3 variable one and use a bunch of random ass crap to figure it out. what a pain in the ass. i got a freakin C in that class.