Results 1 to 4 of 4

Math Help - Boolean algebra simplification help

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    Boolean algebra simplification help

    Simplify the ff. Boolean function using De morgan's identity and or boolean identity and or special logic functions:
    note ' is the complemented value

    a.)F(w,x,y,z) = [(y+x)(w+z)']' ==> Ans: (w'x' + w + z)
    = (y+x)' + (w+z) de morgan
    = y'x' + w + z ????

    b.)F(X,Y,Z) = [(X + Y' + Z')(X + Z')]' ==>Ans: X'YZ
    = (X + Y' + Z')' + (X + Z')' de morgan
    = X'(Y'+Z')' + (X'Z) de morgan
    = X'YZ + X'Z
    = X'Z(Y+1)
    = X'Z ????

    c.) F(x,y,z) = x'yz' + x'yz + (xyz' + xyz)' ==> Ans: y
    = x'y(z' + z) + (xy(z' + z))'
    = x'y + (xy)'
    = x'y + x' + y'
    = x' + y' ?????

    are the answers obtainable through simplifying or is it just a typo?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Simplify the ff. Boolean function using De morgan's identity and or boolean identity and or special logic functions:
    note ' is the complemented value

    a.)F(w,x,y,z) = [(y+x)(w+z)']' ==> Ans: (w'x' + w + z)
    = (y+x)' + (w+z) de morgan
    = y'x' + w + z ????

    b.)F(X,Y,Z) = [(X + Y' + Z')(X + Z')]' ==>Ans: X'YZ
    = (X + Y' + Z')' + (X + Z')' de morgan
    = X'(Y'+Z')' + (X'Z) de morgan
    = X'YZ + X'Z
    = X'Z(Y+1)
    = X'Z ????

    c.) F(x,y,z) = x'yz' + x'yz + (xyz' + xyz)' ==> Ans: y
    = x'y(z' + z) + (xy(z' + z))'
    = x'y + (xy)'
    = x'y + x' + y'
    = x' + y' ?????
    that's weird, I think you're correct for all of them
    Or I'm mistaking all the way, because 3 typos may be too much...
    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 ^_^Engineer_Adam^_^ View Post
    Simplify the ff. Boolean function using De morgan's identity and or boolean identity and or special logic functions:
    note ' is the complemented value

    a.)F(w,x,y,z) = [(y+x)(w+z)']' ==> Ans: (w'x' + w + z)
    = (y+x)' + (w+z) de morgan
    = y'x' + w + z ????
    The answer is obviously a typo. You can easily see that if you substitute w=z=x=0 and y=1. However it does not mean your answer is right.... But you have justified every step correctly.. so you are right

    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    b.)F(X,Y,Z) = [(X + Y' + Z')(X + Z')]' ==>Ans: X'YZ
    = (X + Y' + Z')' + (X + Z')' de morgan
    = X'(Y'+Z')' + (X'Z) de morgan
    = X'YZ + X'Z
    = X'Z(Y+1)
    = X'Z ????
    Again a typo... Choose x=y=0 and z=1.

    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    c.) F(x,y,z) = x'yz' + x'yz + (xyz' + xyz)' ==> Ans: y
    = x'y(z' + z) + (xy(z' + z))'
    = x'y + (xy)'
    = x'y + x' + y'
    = x' + y' ?????

    are the answers obtainable through simplifying or is it just a typo?
    Try y=0 and x=1. Clearly your original expression is 1. But their answer says 0.

    As a side note, this type of expressions functional verification is automated these days. Programs like Verilog can automatically test with certain inputs. If you are wondering how i generated those combination of values.... I was just trying to give inputs that makes the expected answer differ from your answer. Which is the same as saying I was trying to make expected answer as the complement of your answer or in other words, expected answer XOR your answer = 1.

    For example in the second problem:

    (X'YZ) XOR (X'Z) = 1 is the same as X'Z (Y XOR 1) = 1 is the same as X'Y'Z = 1. Thus I chose X=Y=0 and Z=1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    these are the 3 out of 20 problems im getting confused at im sure it were typos

    thanks alot for the help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Boolean expression simplification
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 1st 2010, 08:23 AM
  2. [SOLVED] Boolean Simplification
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 2nd 2010, 04:19 AM
  3. Boolean Algebra Simplification
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 18th 2010, 07:02 PM
  4. Boolean algebra help
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 20th 2009, 04:13 AM
  5. Simplification of boolean algebra
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: January 12th 2009, 01:05 AM

Search Tags


/mathhelpforum @mathhelpforum