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moli1993 Group Title

simplify the following using Boolean algebra. (xy)'+xz'+yz a)y+z' b)x+y c)X'+y d)xz'

  • 4 months ago
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  1. moli1993 Group Title
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    c) x'+y

    • 4 months ago
  2. ybarrap Group Title
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    $$ \Large{ (xy)'+xz'+yz\\ =[(xy)(xz')'(yz)']'\\ =[(xy)(x'+z)(y'+z')]'\\ =[((xyx')+(xyz))(y'+z')]'\\ =[0+xyzy'+xyzz']'\\ =[0+0+0]'\\ =[0]'\\ =1 } $$ This is a tautology, because it is always true. Here is another method using a Truth Table |dw:1395590669537:dw| This also shows a tautology

    • 4 months ago
  3. ybarrap Group Title
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    Does this make sense?

    • 4 months ago
  4. AkashdeepDeb Group Title
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    (xy)' + xz' + yz We know that (xy)' = x' + y' [De morgan's laws] x' + y' + xz' + yz x' + z' + y' + z x' + y' + 1 = 1 None of the options are as 1 @ybarrap Is truth table the only method to find that out?

    • 4 months ago
  5. AkashdeepDeb Group Title
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    @moli1993 I guess you have to see which of the options also show Tautology [Always true].

    • 4 months ago
  6. ybarrap Group Title
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    The truth table is authoritative. This is a tautology. None of the options are a tautology so they are all incorrect.

    • 4 months ago
  7. moli1993 Group Title
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    I am so sorry , wrote the wrong question here :- it is (xz)'+xz'+yz and the options are same as given earlier !

    • 4 months ago
  8. ybarrap Group Title
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    It's the same process

    • 4 months ago
  9. moli1993 Group Title
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    process is same but it's not a tautology i suppose !

    • 4 months ago
  10. dumbcow Group Title
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    (xz)' = x' + z' z' +xz' = z'(1+x) = z' --> x' + z' + yz = x' +z' +z = x' + 1 = 1

    • 4 months ago
  11. moli1993 Group Title
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    how x' + z' + yz = x' +z' +z

    • 4 months ago
  12. AkashdeepDeb Group Title
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    (xz)'+xz'+yz We know that (xz)' = x' + z' [De morgan's laws] x' + z' + xz' + yz x' + (z' + xz') + yz [ In z' + xz' take z' common, z'(1+x) = z'(1) = z' ] x' + z' + yz [yz = zy] z' + zy + x' Using simplification law [A + A'B = A+B] we get, z' + y + x' Final answer should be x' + y + z'

    • 4 months ago
  13. moli1993 Group Title
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    but it doesnt matches any of the options given above

    • 4 months ago
  14. AkashdeepDeb Group Title
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    (xz)' + y I am pretty sure, this is the final answer. :/ @ybarrap @dumbcow ?

    • 4 months ago
  15. moli1993 Group Title
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    me too I think u r correct !

    • 4 months ago
  16. dumbcow Group Title
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    yep http://www.wolframalpha.com/input/?i=~%28x+and+z%29+or+%28x+and+~z%29+or+%28y+and+z%29

    • 4 months ago
  17. ybarrap Group Title
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    yes it is

    • 4 months ago
  18. moli1993 Group Title
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    Thank you all for helping me !

    • 4 months ago
  19. AkashdeepDeb Group Title
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    :)

    • 4 months ago
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