Question

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

Answer 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)

Answer 2

but if i proceed by expanding the bracket! I'm not getting the answer!!!!

Answer 3

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.

Answer 4

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!!

Answer 5

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

Answer 6

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

Answer 7

in 2ns line use this rule : x+x'y = x+y

Answer 8

Can you prove the law x+x'y=x+y???

Answer 9

By Boolean Algebra!

Answer 10

did u try to prove already... it wont be difficult i think

Answer 11

No

Answer 12

How can you prove or simplify it by boolean algebra??

Answer 13

do u knw distributive property for AND logic ?

Answer 14

x + (y.z) = (x+y).(x+z)

Answer 15

u familiar wid that prop ? u can use that to prove x + x'y = x + y very easily

Answer 16

ok got it!

Answer 17

great :)