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ketz

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

  • 11 months ago
  • 11 months ago

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  1. Hoa
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    is it boolean?

    • 11 months ago
  2. Hoa
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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 :( , :) )

    • 11 months ago
  3. perl
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    yes it looks boolean

    • 11 months ago
  4. perl
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    Hoa, wnat to check over a solution of mine?

    • 11 months ago
  5. Hoa
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    sure.

    • 11 months ago
  6. perl
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    http://openstudy.com/study#/updates/517bc95ce4b0249598f7a8c1

    • 11 months ago
  7. Hoa
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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?

    • 11 months ago
  8. perl
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    yes we can simplify this

    • 11 months ago
  9. perl
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    We can use Demorgan's Theorem: ( x + y ) ' = x'y' (xy)' = x' + y'

    • 11 months ago
  10. perl
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    (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )

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

    • 11 months ago
  12. ketz
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    its boolean

    • 11 months ago
  13. perl
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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

    • 11 months ago
  14. ketz
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    but it seems the answer is: z(x'+y)!

    • 11 months ago
  15. Hoa
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    @ketz I am with perl. I don't get yours

    • 11 months ago
  16. ketz
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    I think the answer needs to be converted to MMF

    • 11 months ago
  17. Hoa
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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

    • 11 months ago
  18. perl
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    you claim that (x'z'+xy')' = z ( x' +y ) ?

    • 11 months ago
  19. perl
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    i think thats false, plug in values for x,y,z

    • 11 months ago
  20. ketz
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    but How to convert it to Minimal Multiplicative Form (MMF)? any idea guys?

    • 11 months ago
  21. perl
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    can you double check you copied problem correctly

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

    • 11 months ago
  23. perl
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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

    • 11 months ago
  24. perl
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    ketz theres a typo somewhere

    • 11 months ago
  25. Hoa
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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

    • 11 months ago
  26. perl
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    MMF is product of sums so (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )

    • 11 months ago
  27. Hoa
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    yeap

    • 11 months ago
  28. perl
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    i dont know why hes not accepting it?

    • 11 months ago
  29. perl
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    (x'z'+xy')' =/= z(x'+y)

    • 11 months ago
  30. perl
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    =/= means not equal

    • 11 months ago
  31. Hoa
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    because there is another way to get that form .

    • 11 months ago
  32. perl
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    they are not equal, i proved it

    • 11 months ago
  33. perl
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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

    • 11 months ago
  34. Hoa
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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

    • 11 months ago
  35. Hoa
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    I'm with you, you are 100% right. just misunderstand together with him

    • 11 months ago
  36. perl
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    ketz writes: but it seems the answer is: z(x'+y)!

    • 11 months ago
  37. Hoa
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    ok, let's go to his new one

    • 11 months ago
  38. perl
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    wait

    • 11 months ago
  39. perl
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    so theres two forms, a product of sums, or a sum of products

    • 11 months ago
  40. perl
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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_

    • 11 months ago
  41. perl
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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)

    • 11 months ago
  42. perl
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    sorry, i think wolfram has a boolean calculator, so let me find their website

    • 11 months ago
  43. perl
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    here http://www.wolframalpha.com/widgets/view.jsp?id=a52797be9f91295a27b14cb751198ae3

    • 11 months ago
  44. perl
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    so this is an interesting question. im going to post a new one

    • 11 months ago
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