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ketz

  • one year ago

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

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  1. Hoa
    • one year ago
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    is it boolean?

  2. Hoa
    • one year ago
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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 :( , :) )

  3. perl
    • one year ago
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    yes it looks boolean

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

  5. Hoa
    • one year ago
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    sure.

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

  7. Hoa
    • one year ago
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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?

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

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

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

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

  12. ketz
    • one year ago
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    its boolean

  13. perl
    • one year ago
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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

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

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

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

  17. Hoa
    • one year ago
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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

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

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

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

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

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

  23. perl
    • one year ago
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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

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

  25. Hoa
    • one year ago
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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

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

  27. Hoa
    • one year ago
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    yeap

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

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

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

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

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

  33. perl
    • one year ago
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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

  34. Hoa
    • one year ago
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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

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

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

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

  38. perl
    • one year ago
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    wait

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

  40. perl
    • one year ago
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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_

  41. perl
    • one year ago
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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)

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

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

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

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