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anonymous

  • 4 years ago

Boolean Logic: reduce to a minimum sum-of-product form: X'Y'Z' + X'YZ + XYZ X'Y'Z' + X'Y'Z + XY'Z + XYZ' for the first one, I have it down to X'Y'Z' + YZ, but is there anything more I can do that I'm not seeing?

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  1. anonymous
    • 4 years ago
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    |dw:1327917442864:dw| X'Y'Z' + YZ no other solution for sure ... you are welcome

  2. anonymous
    • 4 years ago
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    2nd one = X'Y'+X(Y'Z+YZ') = X'Y'+Y'Z+YZ' = X'Y' + (Y xor Z)

  3. anonymous
    • 4 years ago
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    Chris ... we have not x'y' ... please check your answer . I'm sure its not right answer

  4. anonymous
    • 4 years ago
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    What do you mean by "we have not x'y'"? The (2nd) given formula was X'Y'Z' + X'Y'Z + XY'Z + XYZ' and there you can easily see, that the first two terms lead to X'Y'(Z'+Z) = X'Y'. Or maybe I misunderstood you?

  5. anonymous
    • 4 years ago
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    look to original Boolean equ X'Y'Z' + X'YZ + XYZ X'Y'Z + XY'Z + XYZ' yours do you have any difference before simplification

  6. anonymous
    • 4 years ago
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    ?

  7. anonymous
    • 4 years ago
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    I'm all the time talking about the 2nd (second, the lower, the latter, the last) of the two expressions that were mentioned in the original post. (At least, as far as I can understand, the two lines with two expressions, one consisting of three, the other of four terms, are two independent expressions, each of he has to reduce.) Maybe I see a different posting, but for me the 2nd expressions reads: \[X'Y'Z' + X'Y'Z + XY'Z + XYZ'\].

  8. anonymous
    • 4 years ago
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    in second term you consider y' while its not ..

  9. anonymous
    • 4 years ago
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    any way .. logic is simple .. and simplification is easy . its just need to take it easy and step by step .. I'm sure that you can solve harder than this, I like your way of deffence because that is mean ... You are completely understand logic ... respect for you buddy .. see u around

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