Results 1 to 2 of 2

Math Help - boolean algebra

  1. #1
    Newbie
    Joined
    Jul 2012
    From
    usa
    Posts
    1

    boolean algebra

    simplify the expression using boolean algebra: w'xy+wxz+wy'z+w'z' ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,527
    Thanks
    773

    Re: boolean algebra

    If you need the final expression to be a CNF, then I don't think you can simplify much. The minimal number of conjuncts that I can see is three, versus four in the original formula. Also, it is difficult to do the simplification without a Karnaugh map.

    Below, the first two steps split some conjuncts into two, and the rest of the steps recombine them in new ways. I write comments in curly braces.

    xyw' + xzw + y'zw + z'w' = {xyw' = xy(z + z')w' = xyzw' + xyz'w'}
    xyzw' + xyz'w' + xzw + y'zw + z'w' = {xzw = x(y + y')zw = xyzw + xy'zw}
    xyzw' + xyz'w' + xyzw + xy'zw + y'zw + z'w' = {xyz'w' + z'w' = (xy + 1)z'w' = z'w'}
    xyzw' + xyzw + xy'zw + y'zw + z'w' = {xy'zw + y'zw = (x + 1)y'zw = y'zw}
    xyzw' + xyzw + y'zw + z'w' = {xyzw' + xyzw = xyz(w + w') = xyz}
    xyz + y'zw + z'w'

    Edit: It is recommended to post questions about Boolean algebra in the Discrete Math subforum.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Boolean Algebra
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: May 27th 2012, 01:38 PM
  2. [SOLVED] boolean algebra help
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 10th 2011, 02:06 PM
  3. Boolean Algebra Help
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 3rd 2010, 04:16 AM
  4. Boolean algebra help
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 20th 2009, 04:13 AM
  5. Boolean Algebra
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: February 8th 2008, 07:55 AM

Search Tags


/mathhelpforum @mathhelpforum