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Math Help - Simplifying circuit using Boolean Algebra

  1. #1
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    Simplifying circuit using Boolean Algebra

    Hopefully I've uploaded the circuit properly
    Simplifying circuit using Boolean Algebra-untitled.png

    Using an apostrophe to denote the over-bar (Not), "+" is or, and "." is and, I wrote the initial circuit as:

    r = (p'.(p.q).q)'
    which I then simplified as follows:
    r = p'' + (p.q)' + q'
    = p + p' + q + q'
    = 1

    However, I'm not convinced this is correct. Help please.
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  2. #2
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    The line

    = p + p' + q + q' should be

    = p + p' + q' + q',

    which does not reduce to 1.

    [EDIT]: See below for a correction.
    Last edited by Ackbeet; March 29th 2011 at 05:50 AM.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    The line

    = p + p' + q + q' should be

    = p + p' + q' + q',

    which does not reduce to 1.
    Hmm, I think it still does.
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  4. #4
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    p + p' + q' + q'
    = 1 + q' + q'
    = q' + q'
    = q', right?

    [EDIT]: See below for a correction.
    Last edited by Ackbeet; March 29th 2011 at 05:50 AM.
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  5. #5
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    Thanks for your help.

    When you think about it logically, it appears q' is the right answer.

    To draw the simplified circuit, is it simply q, going into a "not" semi-circle thing and then to r, can p be ignored?
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    p + p' + q' + q'
    = 1 + q' + q'
    = q' + q'
    = q', right?
    I think the Earth is round or there is a mistake here.
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  7. #7
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    Oh, now I'm with you. Clever pointing out of my mistake. So, 1 + x = 1, no matter what x is. Gotcha.
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    Oh, now I'm with you. Clever pointing out of my mistake. So, 1 + x = 1, no matter what x is. Gotcha.
    Right so the answer is one then after all?
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  9. #9
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    Yes. This could also be seen without applying the De Morgan's law (though it's not much of a change). In the original formula, (p'.(p.q).q)', conjunction is associative and p' . p is always false, so p' . p . q . q is false as well. Therefore, its negation is always true.
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  10. #10
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    Quote Originally Posted by emakarov View Post
    Yes. This could also be seen without applying the De Morgan's law (though it's not much of a change). In the original formula, (p'.(p.q).q)', conjunction is associative and p' . p is always false, so p' . p . q . q is false as well. Therefore, its negation is always true.
    Cool, that way's simpler to be fair. Does no circuit exist then?
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  11. #11
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    If you don't have a gate for 1, you can take p + p'.
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