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ketz

  • 3 years ago

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

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  1. hartnn
    • 3 years ago
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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)

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

  3. hartnn
    • 3 years ago
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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.

  4. ketz
    • 3 years ago
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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!!

  5. hartnn
    • 3 years ago
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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

  6. ganeshie8
    • 3 years ago
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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

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

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

  9. ketz
    • 3 years ago
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    By Boolean Algebra!

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

  11. ketz
    • 3 years ago
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    No

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

  13. ganeshie8
    • 3 years ago
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    do u knw distributive property for AND logic ?

  14. ganeshie8
    • 3 years ago
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    x + (y.z) = (x+y).(x+z)

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

  16. ketz
    • 3 years ago
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    ok got it!

  17. ganeshie8
    • 3 years ago
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    great :)

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