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ketz

f=x'y'z+(x'y'z)'. simplify the boolean expression. urgent

  • 11 months ago
  • 11 months ago

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  1. hartnn
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    you use the identity, A+A' = 1 so, f=x'y'z+(x'y'z)' = 1 (where x'y'z can be considered as A)

    • 11 months ago
  2. ketz
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    but if i proceed by expanding the bracket! I'm not getting the answer!!!!

    • 11 months ago
  3. hartnn
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    expanding the bracket ? like x'y'z +xyz' ? no need for that, you can use A=x'y'z and that A+A'=1 you can also verify it using truth table.

    • 11 months ago
  4. ketz
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    yeh i know but i want to proceed by expanding the bracket using de morgan's laws. and am not getting the answer. can anyone try it by this method!!

    • 11 months ago
  5. hartnn
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    let me try on, left of + x'y' = (x+y)' ---> x'y'z =(x+y)'z on right of + (x'y'z)' = (x+y+z') hmm

    • 11 months ago
  6. ganeshie8
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    f=x'y'z+(x'y'z)' x'y'z + x + y + z' y'z + x + y + z' x + y'z + y + z' x + y + z + z' x + y + 1 1

    • 11 months ago
  7. ganeshie8
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    in 2ns line use this rule : x+x'y = x+y

    • 11 months ago
  8. ketz
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    Can you prove the law x+x'y=x+y???

    • 11 months ago
  9. ketz
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    By Boolean Algebra!

    • 11 months ago
  10. ganeshie8
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    did u try to prove already... it wont be difficult i think

    • 11 months ago
  11. ketz
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    No

    • 11 months ago
  12. ketz
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    How can you prove or simplify it by boolean algebra??

    • 11 months ago
  13. ganeshie8
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    do u knw distributive property for AND logic ?

    • 11 months ago
  14. ganeshie8
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    x + (y.z) = (x+y).(x+z)

    • 11 months ago
  15. ganeshie8
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    u familiar wid that prop ? u can use that to prove x + x'y = x + y very easily

    • 11 months ago
  16. ketz
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    ok got it!

    • 11 months ago
  17. ganeshie8
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    great :)

    • 11 months ago
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