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Problem with boolean algebra!

  1. Sep 19, 2005 #1
    Hello everyone i'm having problems proving this:
    note: A' will stand for complemented;

    AB + BC'D' + A'BC + C'D = B + C'D
    B(A+C'D')
    B(A+C'+D')
    BA + BC' + BD' + A'BC + C'D
    BA + BD' + A'BC + C'(B+D)
    B(A + D' + A'C) + C'(B+D)
    now i'm stuck, i don't see how that is going to work out. Any help would be great.
     
  2. jcsd
  3. Sep 19, 2005 #2

    berkeman

    User Avatar

    Staff: Mentor

    It works on a Karnaugh map. Try mapping it out to help you with the algebra. Look for combinations on the map that you can use as tricks in the algebraic manipulations.
     
  4. Sep 19, 2005 #3
    OK this is gonna be long but i but a lot off steps for explanation so you can skip some
    steps

    AB+BC'D'+A'BC+C'D
    B(A+A'C)+BC'D'+C'D
    B((A+A').(A+C))+BC'D'+C'D
    B((1).(A+C))+BC'D'+C'D
    AB+BC+BC'D'+C'D
    B(C+C'D')+AB+C'D
    B((C+C').(C+D'))+AB+C'D
    BC+BD'+AB+C'D
    BC.(D+D')+BD'+AB+C'D
    BCD+BCD'+BD'+AB+C'D
    BCD+BD'+AB+C'D
    D(C'+BC)+BD'+AB
    D((C'+C).(C'+B))+BD'+AB
    D((1).(C'+B))+BD'+AB
    CD'+BD+BD'+AB
    C'D+B+AB
    B(A+1)+C'D
    B+C'D = RHS
     
  5. Sep 19, 2005 #4
    awesome, thanks for the help guys!!
     
  6. Sep 20, 2005 #5
    Your welcome
     
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