Boolean reduction clarification

Discussion in 'Homework Help' started by snowrei, Sep 21, 2009.

  1. snowrei

    Thread Starter New Member

    Sep 21, 2009
    2
    0
    OK so maybe I'm just an idiot but I have a question regarding the following problem.

    (x || y || !z) & (!x || !y || z)

    Now am I getting this incorrect by assuming that that reduces down to just true? It doesn't seem to work out so any help would be appreciated.
     
    Last edited: Sep 21, 2009
  2. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    snowrei,

    Why don't you post your problems with notation like this. (x+y+z')(x'+y'+z) ?

    What is your question? What is your input?

    Restate your problem more clearly.

    Ratch
     
  3. snowrei

    Thread Starter New Member

    Sep 21, 2009
    2
    0
    Sorry, I'm trying to reduce the equation down to a minimal form, and if I'm reading the distributive property properly, I can reduce (x+y+z')(x'+y'+z) to xx'+yy'+z'z which reduces to 1.

    I don't think that's correct and I'm trying to find the error in my logic.
     
  4. Accipiter

    New Member

    Sep 20, 2009
    9
    0
    xx'+yy'+z'z reduces to zero, not one.

    Also, I think you have misunderstood the distributive property. You need to distribute each term in (x+y+z') to each term in (x'+y'+z).
     
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