Can anyone simplify A'C'+ABC+AC' with explanation?

Printable View

- Feb 26th 2010, 01:11 PMSirhcSimplify boolean algebra
Can anyone simplify A'C'+ABC+AC' with explanation?

- Feb 26th 2010, 02:13 PMDrexel28
- Feb 26th 2010, 04:10 PMSirhc
I am serious, I really need help with this.

- Feb 26th 2010, 05:26 PMBacterius
What Drexel28 means is that you need to show your work, what you have attempted, where you are stuck ... you can't just drop your homework here and come back later, this is not how it works. So either you know what boolean algebra is, and you try to apply the methods you learnt onto this problem, either you don't have a clue what "boolean" means and then you should go and read lessons about it (from google or books).

- Feb 27th 2010, 05:56 PMSirhcBool. Alg.
That I think is fair and I now have an idea how it works. However this is my first time in one of these forums, so I assumed that it would have been real time conversations(I waited for a very looooong time b4 i see 1 reply). I even went to the chat room and same thing happened. So forgive my ignorance.

The meat of the matter. I know what boolean alg. is about and the rules involved BUT I just can't apply it ESPECIALLY when they are not as straight forward as the rules. So I was hoping someone could break it down for me (realtime).

So if you can i would greatly appreciate it thanks. - Feb 27th 2010, 06:30 PMMoo
If it's your first time, it's understandable then lol.

Basic rules : for any A, AA'=0 and A+A'=1. And de Morgan's laws (for that, you'll have to look a bit on your own)

After that, you can use all the basic properties of an addition and a multiplication.

Note that A'C' and AC' are similar. You can factorize by C' (Surprised) : A'C'+AC'=C'(A+A')=C'

So A'C'+ABC+AC'=C'+ABC

Then for that, there's a rule you should also know : A+A'B=A+B

which helps simplify what we have into ........... - Feb 27th 2010, 06:56 PMSirhc
....C'+AB! Thanks

another one:

**A'B(D'+C'D)+B(A+A'CD)**

A'BD'+A'BC'D+BA+BA'CD

A'BD'+BA+A'BC'D+BA'CD

B(A'D'+A)+A'BD(C'+C)

B(D'+A)+A'BD(1)

BD'+AB+A'BD

B(D'+A+AD)

B(D'+A+A)

B(D'+A)

**ANS=BD'+AB**

**WHAT DO YOU THINK, AM I ON THE RIGHT TRACK?** - Feb 27th 2010, 07:07 PMMoo
Oh yeah it's correct (Rock)

But note that you could've factorized by B at the very beginning (Smile) - Feb 27th 2010, 07:46 PMSirhc
Can you explain:

prove that X+YZ = (X+Y)(X+Z)

using boolean reductions as well as the law of logical equiv.

NOT SURE WHAT TO DO WITH THAT.