1. Limited time only! Sign up for a free 30min personal 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 Simplification - Please help

  1. Mar 6, 2014 #1
    I'm trying to simplify 2 boolean expressions. Have I done it correctly and how would I simplify these further if possible. Thank you in advance! :)

    Relevant Equations:

    (Associativity of +)
    A+(B+C) = (A+B)+C​
    (Associativity of x)
    A*(B*C) = (A*B)*C​
    (Commutativity of +)
    A+B = B+A​
    (Commutativity of x)
    A*B = B*A​
    (Distributivity of x over +)
    A*(B+C) = (A*B)+(A*C)​
    (Identity for +)
    A+0 = A​
    (Identity for x)
    A*1 = A​
    (Annihilator for x)
    A*0 = 0​


    (Idempotence of +)
    A+A = A​
    (Idempotence of x)
    A*A = A​
    (Absorption 1)
    A*(A+B) = A​
    (Absorption 2)
    A+(A*B) = A​
    (Distributivity of + over x)
    A+(B*C) = (A+B)*(A+C)​
    (Annihilator for +)
    A+1 = 1​

    This is the first expression:
    E = A'B'C'D' + A'B'CD' + A'B'CD + A'BC'D' + A'BC'D + A'BCD' + A'BCD + ABC'D' + ABC'D + ABCD' + ABCD

    Here is my working:
    E = A'B'C'D' + A'B'CD' + A'B'CD + A'BC'D' + A'BC'D + A'BCD' + A'BCD + ABC'D' + ABC'D + ABCD' + ABCD
    E = A'B'C'D' + (D' + D) A'B'C + (D' + D)A'BC' + (D' + D)A'BC + (D' + D)ABC' + (D' + D)ABC
    E = A'B'C'D' + (1) A'B'C + (1)A'BC' + (1)A'BC + (1)ABC' + (1)ABC
    E = A'B'C'D' + A'B'C + A'BC' + A'BC + ABC' + ABC
    E = A'B'C'D' + (B' + B)A'C + (A' + A)BC' + ABC
    E = A'B'C'D' + (1)A'C + (1)BC' + ABC
    E = A'B'C'D' + A'C + BC' + ABC

    This is the second expression:
    E = A'B'C'D' + A'B'C'D + A'BC'D' + A'BC'D + A'BCD' + A'BCD

    Here is my working:
    E = A'B'C'D' + A'B'C'D + A'BC'D' + A'BC'D + A'BCD' + A'BCD
    E = B'(A'C'D' + A'C'D) + B(A'C'D' + A'C'D) + A'BCD' + A'BCD
    E = (B' + B)(A'C'D' + A'C'D) + A'BCD' + A'BCD
    E = (1)(A'C'D' + A'C'D) + A'BCD' + A'BCD
    E = (A'C'D' + A'C'D) + A'BCD' + A'BCD
    E = A'C'D' + A'C'D + (D' + D)A'BC
    E = A'C'D' + A'C'D + (1)A'BC
    E = A'C'D' + A'C'D + A'BC
    E = (D' + D)A'C' + A'BC
    E = (1)A'C' + A'BC
    E = A'C' + A'BC
    E = A'(C' + BC)
     
    Last edited: Mar 6, 2014
  2. jcsd
  3. Mar 6, 2014 #2
    Looks good to me.
     
  4. Mar 6, 2014 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    You are not finished yet! This can be further simplified.

    The same holds for your earlier working.
     
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 Simplification - Please help
  1. Boolean simplification (Replies: 1)

Loading...