1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Boolean Algebra problem

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Minimize the following using boolean identities
    1. AB'CD+(ABC')'+ABCD'



    2. Relevant equations
    Identity 1A=A 0+A = A
    Null (or Dominance) Law 0A = 0 1+A = 1
    Idempotence Law AA = A A+A = A
    Inverse Law AA = 0 A+A = 1
    Commutative Law AB = BA A+B = B+A
    Associative Law (AB)C = A(BC) (A+B)+C = A+(B+C)
    Distributive Law A+BC = (A+B)(A+C) A(B+C) = AB+AC
    Absorption Law A(A+B) = A A+AB = A
    DeMorgan's Law (AB) = A+B (A+B) = A B


    3. The attempt at a solution
    I'm going to use lower case letters now.


    f=ab'cd+(abc')'+abcd'
    f'=(ab'cd+(abc')+abcd')'
    =(ab'cd)'(abc')''(abcd')'
    =(a'+b+c'+d')(abc')(a'+b'+c'+d)
    =(aa'bc'+abbc'+abc'c'+abc'd')(a'+b'+c'+d)
    =(abc'+abc'+abcd')(a'+b'+c'+d)
    =(abc'+abcd')(a'+b'+c'+d')
    =(0+0+0+0+abc'c'+abc'c'd'+abc'd+0)
    =abc'+abc'd'+abc'd
    =abc'+abc'(d'+d)
    =abc'+abc'(1)
    =abc'
    f'=abc'
    f=(abc')'

    --> f=a'+b'+c

    Do this look correct? If so is there a shorter way to minimize it? Is there a way to minimize without using DeMorgan's theorem at the top? Thanks




     
  2. jcsd
  3. Apr 22, 2015 #2
    I didn't read you calculation, but it seem clear that a and c must both be true, sp your final answer must be wrong. Try actually thinking about the logic, & afterward pick the formal identities to back up your intuition
     
  4. Apr 22, 2015 #3

    berkeman

    User Avatar

    Staff: Mentor

    Impressive. I got the same answer via the K-map. :smile:
     
  5. Apr 22, 2015 #4
    sorry my mistake, i misread the question. your answer is correct. i would expand the middle term, and then show that it dominates.
     
  6. Apr 22, 2015 #5
    Great. Thank you. I found a much quicker way.

    f=ab'cd+(abc')'+abcd'
    =ab'cd+a'+b'+c+abcd'
    =(acd+1)b'+a'+c(1+abd')
    =(1)b'+a'+c(1)
    =a'+b'+c
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Boolean Algebra problem
  1. Boolean algebra (Replies: 2)

Loading...