second question

Discussion in 'Homework Help' started by urchik, Mar 8, 2007.

  1. urchik

    Thread Starter Member

    Mar 8, 2007
    12
    0
    also, can someone explain or direct me into understanding bool identities, I need to prove that xz = ( x + y ) ( x+y' ) ( x' + z )

    so I did the truth table as follows:

    XZ =
    X Z X&Z
    0 0 0
    0 1 0
    1 0 0
    1 1 1

    and ( x + y ) ( x+y' ) ( x' + z ) as follows:
    0 | 0 | 0 | 0 0 1 1 0
    0 | 0 | 1 | 0 0 1 1 0
    0 | 1 | 0 | 1 0 0 0 0
    0 | 1 | 1 | 1 0 0 0 0
    1 | 0 | 0 | 1 1 1 1 0
    1 | 0 | 1 | 1 1 1 1 1
    1 | 1 | 0 | 1 1 1 0 0
    1 | 1 | 1 | 1 1 1 0 1

    is that correct?

    Also the identities are as follows:
    XX+XX’+YX+YY’+YZ+XX’+XZ+Y’X’+Y’Z ;Distributive
    XX+(XX’)+YX+(YY’)+YZ+(XX’)+XZ+Y’X’+Y’Z ; Associative
    XX+YX+YZ+XZ+Y’X’+Y’Z ; Inverse

    and I'm stuck. Can anyone direct me to teach me how to do this... I read alot of articles but I dont understand much for it. Thanks!
     
  2. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    First things first. Can I ask to you clarify what you are doing here? I am struggling to follow what you have done. If need be scribble it down on paper, scan it into your computer and upload it to the forums.

    Dave
     
  3. urchik

    Thread Starter Member

    Mar 8, 2007
    12
    0
    Sorry, I should have been more clear.

    What I need to do is prove that xz = ( x + y ) ( x + y' ) ( x' + z )

    First part asks to prove it using a truth table. Now I constructed a truth table for both and here they are:

    for XZ:
    X Z X&Z
    0 0 0
    0 1 0
    1 0 0
    1 1 1

    and for ( x + y ) ( x + y' ) ( x' + z ):
    0 | 0 | 0 | 0 0 1 1 0
    0 | 0 | 1 | 0 0 1 1 0
    0 | 1 | 0 | 1 0 0 0 0
    0 | 1 | 1 | 1 0 0 0 0
    1 | 0 | 0 | 1 1 1 1 0
    1 | 0 | 1 | 1 1 1 1 1
    1 | 1 | 0 | 1 1 1 0 0
    1 | 1 | 1 | 1 1 1 0 1

    Last row being the Q or the output. They don't match as you see... am I doing something wrong here? The next step in this assignment is to prove the same problem but only using boolean identities which I'm stuck on as well, this is what I did so far:

    XX+XX’+YX+YY’+YZ+XX’+XZ+Y’X’+Y’Z ;Distributive
    XX+(XX’)+YX+(YY’)+YZ+(XX’)+XZ+Y’X’+Y’Z ; Associative
    XX+YX+YZ+XZ+Y’X’+Y’Z ; Inverse

    Can you guide me or help me in this problem.
    Thanks
     
  4. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    Ok, lets deal with proof by truth table first. I am still at a loss as to how you are getting the outputs as 5-bits long. For each input combination for x, y and z then the expression will evaluate to '0' or '1'.

    For example:

    ( x + y ) ( x + y' ) ( x' + z ) for x = 0, y = 0, z = 0

    ( 0 + 0 ) ( 0 + 1 ) ( 1 + 0 ) = 0

    You need to do this for each input combination. What you will see is that in the resulting truth table, you can ignore the variable Y, from which you can simplify the truth table to original truth table for XZ (I haven't done this, but this is how you would do it).

    I suggest you know together the truth table for ( x + y ) ( x + y' ) ( x' + z ) as I have shown above and post it up here, from there I can show you how interpret the truth table to prove the expression is equal to XZ.

    Dave
     
  5. urchik

    Thread Starter Member

    Mar 8, 2007
    12
    0
    sorry for the confusion, maybe this will help. The output (truth table) for the long expression is :

    [​IMG]

    hope this helps, I did this into a truth table program so you can see things clearly.

    I still don't see how xy can equal to that...the only thing I can think of is to simplify ( x + y ) ( x + y' ) ( x' + z ) into something that will equal to xy when constructing the truth table.

    After this, I have to prove the same thing but only using boolean identities.
     
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