Boolean Algebra Help

Discussion in 'Homework Help' started by Thirsty, Nov 18, 2006.

  1. Thirsty

    Thread Starter New Member

    Aug 9, 2006
    5
    0
    Hi, can anyone help me take this further please....

    (a+b)(a+c)(a+d)

    (aa+ac+ba+bc)(a+d)

    aaa+aad+aca+acd+baa+bad+bca+bcd
     
  2. Papabravo

    Expert

    Feb 24, 2006
    9,912
    1,724
    I'm not sure what you mean by "take it further". Could you elaborate?
     
  3. Thirsty

    Thread Starter New Member

    Aug 9, 2006
    5
    0
    Sorry, I should have mentioned I need (A+B)(A+C)(A+D) = A+BCD
     
  4. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Remember:

    A.A = A

    So you should be able to simplify your final line to:

    a+ad+ac+acd+ba+bad+bca+bcd

    a(d+c+cd+b+bd+bc) + bcd

    De Morgans:

    C.D = C' + D'

    a(d+c+c'+d'+b+b'+d'+b'+c') + bcd

    And:

    B' + B' = B'

    So simplifying:

    a(d+c+c'+d'+b+b') + bcd

    And:

    C + C' = 1

    Giving:

    a + bcd

    Dave
     
  5. shivani

    New Member

    Mar 31, 2010
    2
    0
    (a + b)(a + c)(a + d)
    (a + bc)(a + d) [Distributive]
    (a + bcd) [Distributive]
     
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