The No.1 Website for Pro Audio
 Search This Thread  Search This Forum  Search Reviews  Search Gear Database  Search Gear for sale  Search Gearslutz Go Advanced
boolean
Old 10th September 2011
  #1
Lives for gear
boolean

A'+B'+(ABC')

can this be simplified ?
maybe A'B'C' ?

thx
Old 10th September 2011
  #2
Lives for gear
 
duggabax's Avatar
I've never seen that notation before. Is this propositional logic or some kind of electronics thing? I could help you if we could put it in terms of negations, conjunctions, disjunctions.
Old 10th September 2011
  #3
Gear Head
 

Quote:
Originally Posted by robertshaw View Post
A'+B'+(ABC')

can this be simplified ?
maybe A'B'C' ?

thx
Assuming A' = NOT[A], etc, then ( expanding by "ORing" out)

A'+B'+(ABC') = (A'+B'+A)(A'+B'+B)(A'+B'+C') = (True)(True)(A'+B'+C')
since A'+A = True, True+B' = True, etc

So, A'+B'+(ABC') = A'+B'+C'

in long-hand:
NOT[A] OR NOT[B] OR (A AND B AND NOT[C]) = NOT[A] OR NOT[B] OR NOT[C]

If you apply DeMorgan's Theorem to this (for three variables), then
A'+B'+C' = (ABC)', ie NOT[A AND B AND C]

I think this is correct, it's awhile since I did boolean arithmetic and you could always check the answer using a truth table.

Dr F.
Old 10th September 2011
  #4
Lives for gear
Quote:
Originally Posted by Dr F. View Post
Assuming A' = NOT[A], etc, then ( expanding by "ORing" out)

A'+B'+(ABC') = (A'+B'+A)(A'+B'+B)(A'+B'+C') = (True)(True)(A'+B'+C')
since A'+A = True, True+B' = True, etc

So, A'+B'+(ABC') = A'+B'+C'

in long-hand:
NOT[A] OR NOT[B] OR (A AND B AND NOT[C]) = NOT[A] OR NOT[B] OR NOT[C]

If you apply DeMorgan's Theorem to this (for three variables), then
A'+B'+C' = (ABC)', ie NOT[A AND B AND C]

I think this is correct, it's awhile since I did boolean arithmetic and you could always check the answer using a truth table.

Dr F.
thx DR. F I used a kmap but I wanted to see it in algebra
much appreciated. I forget all this stuff
kmaps are so much easier
and yeah the truth table gives me the same answer
Post Reply

Welcome to the Gearslutz Pro Audio Community!

Registration benefits include:
  • The ability to reply to and create new discussions
  • Access to members-only giveaways & competitions
  • Interact with VIP industry experts in our guest Q&As
  • Access to members-only sub forum discussions
  • Access to members-only Chat Room
  • Get INSTANT ACCESS to the world's best private pro audio Classifieds for only USD $20/year
  • Promote your eBay auctions and Reverb.com listings for free
  • Remove this message!
You need an account to post a reply. Create a username and password below and an account will be created and your post entered.


 
 
Slide to join now Processing…
Thread Tools
Search this Thread
Search this Thread:

Advanced Search
Forum Jump
Forum Jump