Boolean Algebra - Simplification

Discussion in 'Homework Help' started by alb_ross, Jul 21, 2018.

  1. alb_ross

    Thread Starter New Member

    Jul 21, 2018
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    Hi All

    I've seen multiple threads on this, but I'm having a bit of trouble remembering / getting back into the topic of boolean algebra. After taking note of all the essential rules, I'm not trying to simplify some equations I've found online.

    This one seems extremely easy, but I just can't seem to wrap my mind around the boolean algebra identities to solve this:
    (xy)+(xz)+(yz')

    Solution for simplification is supposed to be:
    (xz)+(yz')

    I've tried to use the distributive and absorption law, but all I get is getting back to the original equation...

    Could you give me a hint?

    Thank you in advance!
     
  2. WBahn

    Moderator

    Mar 31, 2012
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  3. MrAl

    AAC Fanatic!

    Jun 17, 2014
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    Hi,

    Also interesting is if you take the two term solution and invert it and simplify, then invert that also and simplify, you get the original three term expression which includes the term xy which shows that the term xy is superfluous.
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    Expand each of the original three terms:

    xy = xyz + xyz'
    xz = xyz + xy'z
    yz' = xyz' + x'yz'

    You will observe that the expansion of (xy) is repeated in the other two expansions.

    As a check, draw the Karnaugh map and then it will become obvious that (xy) is redundant.
    However, it becomes important if one needed to prevent glitches in an encoder circuit.
     
  5. MrAl

    AAC Fanatic!

    Jun 17, 2014
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    Hi,

    Could you elaborate a little with that?
    I just woke up so maybe that's why i am not following your logic :)
     
  6. MrChips

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    Oct 2, 2009
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    What part do you not follow?
     
  7. MrAl

    AAC Fanatic!

    Jun 17, 2014
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    Hi,

    What you mean by the last sentence that xy is repeated or something.
    I dont see it in the last expression for example.

    What exactly were you trying to show?
     
  8. MrChips

    Moderator

    Oct 2, 2009
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    xy can be expanded into xyz + xyz'.

    xyz is also contained in xz.
    xyz' is also contained in yz'.

    Hence we can eliminate xy.

    Summary here:
    xy = xyz + xyz'
    xz = xyz + xy'z
    yz' = xyz' + x'yz'
     
  9. MrChips

    Moderator

    Oct 2, 2009
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    Karnaugh map:

    Karnaugh_map.jpg
     
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