Results 1 to 4 of 4

Math Help - Boolean expression simplification

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    62

    Boolean expression simplification

    I know this stuff should be way easy once I figure it out, but right now I am literally pulling out my hair trying to do so. The original expression is:

    A`D` + A`BC + A` + AB` + A`BC + ABC`

    I see that A and A` are common factors, so I pull them out:

    A`(D` + BC + 1 + BC) + A(B` + BC`)

    The fact that there is a "1" in the first parentheses makes everything in the parentheses 1, right?

    A`(1) + A(B` + BC`)

    X and 1 = X

    A` + A(B` + BC`)

    The only thing I can think to do is distribute the (B` + BC`) but I don't know if that is the right thing to do...

    A` + A(B+B`)(B`+C`)

    Now, (B+B`) is equal to 1. A and 1 is equal A.

    A` + A(B`+C`)

    Now I am stuck. I have no way to check my answer, but Wolfram says the simplified answer is:

    A`+B`+C`

    Can you help me finish/correct the above work?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    You're ok up until the moment you distribute. I don't think the way you did that was valid. If you multiply back out again your factorization, you end up with

    A(BB` + BC` + B`B` + B`C`) = A(0 + BC` + B` + B`C`) = A((B + B`)C` + B`) = A(C` + B`). This is not equal to A(B` + BC`). I don't think there's any way of simplifying any further. That is, A` + A(B` + BC`) is as far as you get.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by Ackbeet View Post
    You're ok up until the moment you distribute. I don't think the way you did that was valid. If you multiply back out again your factorization, you end up with

    A(BB` + BC` + B`B` + B`C`) = A(0 + BC` + B` + B`C`) = A((B + B`)C` + B`) = A(C` + B`). This is not equal to A(B` + BC`). I don't think there's any way of simplifying any further. That is, A` + A(B` + BC`) is as far as you get.
    Actually B` + C` = B` + BC`.

    To the original poster, use the identity A + A`B = A + B on your last step and you will get "Wolfram"'s identity.

    Additionally, whenever you wonder if two expression F and G are equal, I encourage you to write the truth tables of F and G and check if they are equal.
    (A truth table is just a fancy name for testing the functions on all possible values in the domain. In Boolean cases, we only have to try 1 and 0 for each variable and thus testing equality in Boolean logic for small number of variables is quite simple).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Actually B' + C' = B' + BC'.
    Hmm. You're right. Maybe it's been too long since I've done boolean simplifications!

    Going off of Isomorphism's post, and your last step

    A` + A(B`+C`),

    if it's true that x' + xy' = x' + y', then can you see that you only have one step from

    A` + A(B`+C`)

    to

    A` + B`+C`?
    Last edited by Ackbeet; September 1st 2010 at 09:24 AM. Reason: Added quote.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Boolean Expression
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: January 25th 2011, 12:51 AM
  2. [SOLVED] Boolean Simplification
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 2nd 2010, 05:19 AM
  3. Boolean Algebra Simplification
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 18th 2010, 08:02 PM
  4. Boolean algebra simplification help
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: January 15th 2009, 07:16 AM
  5. Simplification of boolean algebra
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: January 12th 2009, 02:05 AM

Search Tags


/mathhelpforum @mathhelpforum