ketz 2 years ago f=x'y'z+(x'y'z)'. simplify the boolean expression. urgent

1. 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)

2. ketz

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

3. 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.

4. 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!!

5. 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

6. 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

7. ganeshie8

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

8. ketz

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

9. ketz

By Boolean Algebra!

10. ganeshie8

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

11. ketz

No

12. ketz

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

13. ganeshie8

do u knw distributive property for AND logic ?

14. ganeshie8

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

15. ganeshie8

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

16. ketz

ok got it!

17. ganeshie8

great :)