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anonymous

  • one year ago

Boolean Algebra help

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  1. anonymous
    • one year ago
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    How can I simplify this to get WY'X + WY'Z' + W'X'Y + W'X'Z as an answer? \[(W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\]

  2. anonymous
    • one year ago
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    I tried distributing but it just make it worse and lengthy

  3. anonymous
    • one year ago
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    (WW′+WY′+X′W′+X′Y′+ZW′+ZY′)(W′+X+Z′)(W+X′)(W+Y+Z)

  4. anonymous
    • one year ago
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    Yes, there are some theorems here that can help simplify it faster but I am unsure as to how to proceed with this problem :( ( http://ece224web.groups.et.byu.net/reference/boolean_algebra.pdf )

  5. anonymous
    • one year ago
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    nope, X*X = 0

  6. anonymous
    • one year ago
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    I mean X*X = X and X*X'=0

  7. anonymous
    • one year ago
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    lool

  8. anonymous
    • one year ago
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    It's different than the other math you are thinking :[

  9. anonymous
    • one year ago
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    the dot which looks like multiplication is not multiplication. Stands for AND

  10. anonymous
    • one year ago
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    I mean OR! and the + stands for AND

  11. anonymous
    • one year ago
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    :(

  12. anonymous
    • one year ago
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    I guess I'll take the long path

  13. anonymous
    • one year ago
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    (WW′+WY′+X′W′+X′Y′+ZW′+ZY′)(W′+X+Z′)(W+X′)(W+Y+Z)

  14. zzr0ck3r
    • one year ago
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    Expand it and then use that identity rule. Did that not cut many things down?

  15. anonymous
    • one year ago
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    Do you want me to distribute it all? (working on it)

  16. zzr0ck3r
    • one year ago
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    I mean I would wolfram...I don't think your teacher will mind. At this level we know how to distribute.

  17. anonymous
    • one year ago
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    must show all work

  18. ganeshie8
    • one year ago
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    are you allowed to use kmap ?

  19. anonymous
    • one year ago
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    Professor hasn't taught it yet, only theorems http://ece224web.groups.et.byu.net/reference/boolean_algebra.pdf

  20. ganeshie8
    • one year ago
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    Alright, its going to be tricky, lets see...

  21. anonymous
    • one year ago
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    (WW'W'+WW'X+WW'Z'+WY'W'+WY'X+WY'Z'+X'W'W'+X'W'X+X'W'Z'+X'Y'W'+X'Y'X+X'Y'Z'+ZW'W'+ZW'X+ZW'Z'+ZY'W'+ZY'X+ZY'Z')(W+X')(W+Y+Z)

  22. anonymous
    • one year ago
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    W'(WW')=0 (WW'X+WW'Z'+WW'Y+WXY'+WY'Z'+W'W'X+W'X'X+W'X'Z'+W'X'Y'+X'XY+X'Y'Z'+W'W'Z+W'XZ+W'Z'Z+W'Y'Z+XY'Z)(W+X')(W+Y+Z)

  23. ganeshie8
    • one year ago
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    WY'X + WY'Z' + W'X'Y + W'X'Z step1 : group first two terms and last two terms WY'(X+Z') + W'X'(Y+Z)

  24. anonymous
    • one year ago
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    How did you get that?

  25. anonymous
    • one year ago
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    oh wait you are looking at the answer

  26. anonymous
    • one year ago
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    I am trying to obtain WY'X + WY'Z' + W'X'Y + W'X'Z from (W+X′+Z′)(W′+Y′)(W′+X+Z′)(W+X′)(W+Y+Z)

  27. anonymous
    • one year ago
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    So the problem I'm simplifying is (W+X′+Z′)(W′+Y′)(W′+X+Z′)(W+X′)(W+Y+Z)

  28. ganeshie8
    • one year ago
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    Ohk.. I thought your started with WY'X + WY'Z' + W'X'Y + W'X'Z lets start over

  29. ganeshie8
    • one year ago
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    \((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\) step1 : rearrange the product \((W+X'+Z')(W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\) step2 : use 10D on first two terms \((W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)

  30. anonymous
    • one year ago
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    How did you applied X ( X + Y ) = X on step 2?

  31. ganeshie8
    • one year ago
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    call it U(U+V) = U U = W+X' V = Z'

  32. anonymous
    • one year ago
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    ooooo magic

  33. ganeshie8
    • one year ago
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    not really we're just using the given theorems

  34. anonymous
    • one year ago
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    (W+X′)(W′+Y′)(W′+X+Z′)(W+Y+Z)

  35. anonymous
    • one year ago
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    (W'+Y')(W'+X+Z')(W+X')(W+Y+Z) (brain still loading trying to figure out next step)

  36. ganeshie8
    • one year ago
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    \((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\) step1 : rearrange the product \((W+X'+Z')(W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\) step2 : use 10D on first two terms \((W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\) step3 : rearrange the product \((W+X')(W+Y+Z)(W'+Y')(W'+X+Z')\) step4 : use 8D in reverse on first two terms \([W+X'(Y+Z)](W'+Y')(W'+X+Z')\) step5 : use 8D in reverse on last two terms \([W+X'(Y+Z)][W'+Y'(X+Z')]\)

  37. anonymous
    • one year ago
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    (W+X′)(W+Y+Z) if 8D = X + YZ = ( X + Y ) ( X + Z ) U+BC = (U+B)(U+C) U= W B=X' C=(Y+Z)?

  38. anonymous
    • one year ago
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    (W′+Y′)(W′+X+Z′) U+BC = (U+B)(U+C) U=W' B=Y' C=X+Z' W'+Y'(X+Z')

  39. ganeshie8
    • one year ago
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    Yes

  40. ganeshie8
    • one year ago
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    finally use theorem 16

  41. anonymous
    • one year ago
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    W+X′(Y+Z) = (W+X')(Y+Z) right?

  42. anonymous
    • one year ago
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    since W+X'(Y+Z) is way different and idk how to apply theorem 16 to it :[

  43. ganeshie8
    • one year ago
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    \((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\) step1 : rearrange the product \((W+X'+Z')(W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\) step2 : use 10D on first two terms \((W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\) step3 : rearrange the product \((W+X')(W+Y+Z)(W'+Y')(W'+X+Z')\) step4 : use 8D in reverse on first two terms \([W+X'(Y+Z)](W'+Y')(W'+X+Z')\) step5 : use 8D in reverse on last two terms \([W+X'(Y+Z)][W'+Y'(X+Z')]\) step6 : use 16 \(WY'(X+Z') + W'X'(Y+Z)\)

  44. anonymous
    • one year ago
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    How did you applied it? (( X + Y ) ( X' + Z ) = X Z + X' Y)

  45. ganeshie8
    • one year ago
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    (U+V)(U'+T) = UT + U'V U = W V = X'(Y+Z) T = Y'(X+Z')

  46. anonymous
    • one year ago
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    What's your thought process when figuring out what theorem to use? I'm having a hard time figuring out what theorem to apply.

  47. ganeshie8
    • one year ago
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    you need to understand well why those theorems work and convince everything in terms of conjunctions and disjunctions

  48. ganeshie8
    • one year ago
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    for example, can you explain why below holds ? X + X' = 1

  49. anonymous
    • one year ago
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    |dw:1441253021508:dw| my brain went completely blank when I was doing that

  50. anonymous
    • one year ago
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    no, im unsure

  51. ganeshie8
    • one year ago
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    you're confusing between AND gate and OR gate

  52. ganeshie8
    • one year ago
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    X + X' is read as X or X' |dw:1441253525157:dw|

  53. anonymous
    • one year ago
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    ow yes, my bad

  54. ganeshie8
    • one year ago
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    it spits out 1 when at least one of its inputs is 1

  55. ganeshie8
    • one year ago
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    easy to see that one of the X or X' is always 1

  56. ganeshie8
    • one year ago
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    because X, X' are opposites of each other if X is 0, then X' will be 1 and viceversa

  57. anonymous
    • one year ago
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    Why is it that X+0=X is not = 1?

  58. ganeshie8
    • one year ago
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    X + 0 let X = 0 what does previous expression evaluate to

  59. anonymous
    • one year ago
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    If X = 0 then we have 0 + 0 which is = 0

  60. ganeshie8
    • one year ago
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    so X + 0 = 1 does not always hold

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