anonymous
  • anonymous
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?
Computer Science
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1327917442864:dw| X'Y'Z' + YZ no other solution for sure ... you are welcome
anonymous
  • anonymous
2nd one = X'Y'+X(Y'Z+YZ') = X'Y'+Y'Z+YZ' = X'Y' + (Y xor Z)
anonymous
  • anonymous
Chris ... we have not x'y' ... please check your answer . I'm sure its not right answer

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anonymous
  • anonymous
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
?
anonymous
  • anonymous
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: \[X'Y'Z' + X'Y'Z + XY'Z + XYZ'\].
anonymous
  • anonymous
in second term you consider y' while its not ..
anonymous
  • anonymous
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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