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ketz Group Title

Simplify: f=(x'z'+xy')'

  • one year ago
  • one year ago

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  1. Hoa Group Title
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    is it boolean?

    • one year ago
  2. Hoa Group Title
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    I see it as boolean, not derivative, If it 's so, let me know. I don't want go tooo far on the wrong way, (may get ticket from 911 :( , :) )

    • one year ago
  3. perl Group Title
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    yes it looks boolean

    • one year ago
  4. perl Group Title
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    Hoa, wnat to check over a solution of mine?

    • one year ago
  5. Hoa Group Title
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    sure.

    • one year ago
  6. perl Group Title
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    http://openstudy.com/study#/updates/517bc95ce4b0249598f7a8c1

    • one year ago
  7. Hoa Group Title
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    I'll be there now. but since we are here to help him/her. Would you please give out the answer before leaving? don't let him/her wait while we can, ok?

    • one year ago
  8. perl Group Title
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    yes we can simplify this

    • one year ago
  9. perl Group Title
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    We can use Demorgan's Theorem: ( x + y ) ' = x'y' (xy)' = x' + y'

    • one year ago
  10. perl Group Title
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    (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )

    • one year ago
  11. Hoa Group Title
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    one more step, friend, you have form of xx'

    • one year ago
  12. ketz Group Title
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    its boolean

    • one year ago
  13. perl Group Title
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    (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y ) = xx' +zx' +xy +zy = zx' + xy + zy

    • one year ago
  14. ketz Group Title
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    but it seems the answer is: z(x'+y)!

    • one year ago
  15. Hoa Group Title
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    @ketz I am with perl. I don't get yours

    • one year ago
  16. ketz Group Title
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    I think the answer needs to be converted to MMF

    • one year ago
  17. Hoa Group Title
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    no need to get that form if the form is reducible . perl's answer is irreducible form that 's all you need

    • one year ago
  18. perl Group Title
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    you claim that (x'z'+xy')' = z ( x' +y ) ?

    • one year ago
  19. perl Group Title
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    i think thats false, plug in values for x,y,z

    • one year ago
  20. ketz Group Title
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    but How to convert it to Minimal Multiplicative Form (MMF)? any idea guys?

    • one year ago
  21. perl Group Title
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    can you double check you copied problem correctly

    • one year ago
  22. Hoa Group Title
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    @ketz don't combine many problems into 1. that's another process, post a new one

    • one year ago
  23. perl Group Title
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    Plug x=1, y=1, z = 0 into (x'z'+xy')' (1' & 0' + 1&1')' = ( 0 &1 + 1 & 0 )' = ( 0 +0 )' = 0' = 1 Now plug x=1 , y = 1 , z = 0 into z ( x' +y ) 0 ( x' +y) = 0 , because 0*x = 0 clearly they are not equal

    • one year ago
  24. perl Group Title
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    ketz theres a typo somewhere

    • one year ago
  25. Hoa Group Title
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    @perl ketz asks about the way how to get the MMF form. not the original problem. All we need is a new problem such that we get the MMF form or we can manipulate the answer to get that form. However, by that step, we make the answer more complicated than it is. We can separate it into new topic as: convert zx' + xy + zy into MMF form

    • one year ago
  26. perl Group Title
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    MMF is product of sums so (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )

    • one year ago
  27. Hoa Group Title
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    yeap

    • one year ago
  28. perl Group Title
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    i dont know why hes not accepting it?

    • one year ago
  29. perl Group Title
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    (x'z'+xy')' =/= z(x'+y)

    • one year ago
  30. perl Group Title
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    =/= means not equal

    • one year ago
  31. Hoa Group Title
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    because there is another way to get that form .

    • one year ago
  32. perl Group Title
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    they are not equal, i proved it

    • one year ago
  33. perl Group Title
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    Plug x=1, y=1, z = 0 into (x'z'+xy')' (1' & 0' + 1&1')' = ( 0 &1 + 1 & 0 )' = ( 0 +0 )' = 0' = 1 Plug x=1 , y = 1 , z = 0 into z ( x' +y ) 0 ( x' +y) = 0 , because 0*x = 0 Therefore they are not equal

    • one year ago
  34. Hoa Group Title
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    by multiply (x + x') into the term which doesn't have x, (y + y') into the term which doesn't have y and so on

    • one year ago
  35. Hoa Group Title
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    I'm with you, you are 100% right. just misunderstand together with him

    • one year ago
  36. perl Group Title
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    ketz writes: but it seems the answer is: z(x'+y)!

    • one year ago
  37. Hoa Group Title
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    ok, let's go to his new one

    • one year ago
  38. perl Group Title
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    wait

    • one year ago
  39. perl Group Title
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    so theres two forms, a product of sums, or a sum of products

    • one year ago
  40. perl Group Title
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    (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y ) <---- Product of Sums = xx' +zx' +xy +zy = zx' + xy + zy somehow we can get xy + x'z <--- sum of products (with no terms repeating_

    • one year ago
  41. perl Group Title
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    This is a Boolean Calculator http://calculator.tutorvista.com/math/582/boolean-algebra-calculator.html and i plugged in ~(~x & ~z or x & ~y)

    • one year ago
  42. perl Group Title
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    sorry, i think wolfram has a boolean calculator, so let me find their website

    • one year ago
  43. perl Group Title
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    here http://www.wolframalpha.com/widgets/view.jsp?id=a52797be9f91295a27b14cb751198ae3

    • one year ago
  44. perl Group Title
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    so this is an interesting question. im going to post a new one

    • one year ago
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