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ketz

  • 2 years ago

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

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  1. Hoa
    • 2 years ago
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    is it boolean?

  2. Hoa
    • 2 years 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
    • 2 years ago
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    yes it looks boolean

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

  5. Hoa
    • 2 years ago
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    sure.

  6. perl
    • 2 years ago
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    http://openstudy.com/study#/updates/517bc95ce4b0249598f7a8c1

  7. Hoa
    • 2 years 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
    • 2 years ago
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    yes we can simplify this

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

  10. perl
    • 2 years ago
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    (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )

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

  12. ketz
    • 2 years ago
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    its boolean

  13. perl
    • 2 years 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
    • 2 years ago
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    but it seems the answer is: z(x'+y)!

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

  16. ketz
    • 2 years ago
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    I think the answer needs to be converted to MMF

  17. Hoa
    • 2 years 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
    • 2 years ago
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    you claim that (x'z'+xy')' = z ( x' +y ) ?

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

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

  21. perl
    • 2 years ago
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    can you double check you copied problem correctly

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

  23. perl
    • 2 years 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
    • 2 years ago
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    ketz theres a typo somewhere

  25. Hoa
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    yeap

  28. perl
    • 2 years ago
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    i dont know why hes not accepting it?

  29. perl
    • 2 years ago
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    (x'z'+xy')' =/= z(x'+y)

  30. perl
    • 2 years ago
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    =/= means not equal

  31. Hoa
    • 2 years ago
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    because there is another way to get that form .

  32. perl
    • 2 years ago
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    they are not equal, i proved it

  33. perl
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    I'm with you, you are 100% right. just misunderstand together with him

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

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

  38. perl
    • 2 years ago
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    wait

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

  40. perl
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    sorry, i think wolfram has a boolean calculator, so let me find their website

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

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

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