ketz
  • ketz
f=x'y'z+(x'y'z)'. simplify the boolean expression. urgent
Mathematics
chestercat
  • chestercat
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hartnn
  • hartnn
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)
ketz
  • ketz
but if i proceed by expanding the bracket! I'm not getting the answer!!!!
hartnn
  • hartnn
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.

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ketz
  • ketz
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!!
hartnn
  • hartnn
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
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
in 2ns line use this rule : x+x'y = x+y
ketz
  • ketz
Can you prove the law x+x'y=x+y???
ketz
  • ketz
By Boolean Algebra!
ganeshie8
  • ganeshie8
did u try to prove already... it wont be difficult i think
ketz
  • ketz
No
ketz
  • ketz
How can you prove or simplify it by boolean algebra??
ganeshie8
  • ganeshie8
do u knw distributive property for AND logic ?
ganeshie8
  • ganeshie8
x + (y.z) = (x+y).(x+z)
ganeshie8
  • ganeshie8
u familiar wid that prop ? u can use that to prove x + x'y = x + y very easily
ketz
  • ketz
ok got it!
ganeshie8
  • ganeshie8
great :)

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