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ketz
 one year ago
f=x'y'z+(x'y'z)'. simplify the boolean expression. urgent
ketz
 one year ago
f=x'y'z+(x'y'z)'. simplify the boolean expression. urgent

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hartnn
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.0but if i proceed by expanding the bracket! I'm not getting the answer!!!!

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1expanding 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.

ketz
 one year ago
Best ResponseYou've already chosen the best response.0yeh 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
 one year ago
Best ResponseYou've already chosen the best response.1let 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
 one year ago
Best ResponseYou've already chosen the best response.3f=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
 one year ago
Best ResponseYou've already chosen the best response.3in 2ns line use this rule : x+x'y = x+y

ketz
 one year ago
Best ResponseYou've already chosen the best response.0Can you prove the law x+x'y=x+y???

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3did u try to prove already... it wont be difficult i think

ketz
 one year ago
Best ResponseYou've already chosen the best response.0How can you prove or simplify it by boolean algebra??

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3do u knw distributive property for AND logic ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3x + (y.z) = (x+y).(x+z)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3u familiar wid that prop ? u can use that to prove x + x'y = x + y very easily
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