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hartnnBest ResponseYou've already chosen the best response.1
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

ketzBest ResponseYou've already chosen the best response.0
but if i proceed by expanding the bracket! I'm not getting the answer!!!!
 11 months ago

hartnnBest ResponseYou've already chosen the best response.1
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

ketzBest ResponseYou've already chosen the best response.0
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

hartnnBest ResponseYou've already chosen the best response.1
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

ganeshie8Best ResponseYou've already chosen the best response.3
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

ganeshie8Best ResponseYou've already chosen the best response.3
in 2ns line use this rule : x+x'y = x+y
 11 months ago

ketzBest ResponseYou've already chosen the best response.0
Can you prove the law x+x'y=x+y???
 11 months ago

ganeshie8Best ResponseYou've already chosen the best response.3
did u try to prove already... it wont be difficult i think
 11 months ago

ketzBest ResponseYou've already chosen the best response.0
How can you prove or simplify it by boolean algebra??
 11 months ago

ganeshie8Best ResponseYou've already chosen the best response.3
do u knw distributive property for AND logic ?
 11 months ago

ganeshie8Best ResponseYou've already chosen the best response.3
x + (y.z) = (x+y).(x+z)
 11 months ago

ganeshie8Best ResponseYou've already chosen the best response.3
u familiar wid that prop ? u can use that to prove x + x'y = x + y very easily
 11 months ago
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