Question

Simplify:

f=(x'z'+xy')'

Answer 1

is it boolean?

Answer 2

I see it as boolean, not derivative, If it 's so, let me know. I don't want go tooo far on the wrong way, (may get ticket from 911 :( , :) )

Answer 3

yes it looks boolean

Answer 4

Hoa, wnat to check over a solution of mine?

Answer 5

sure.

Answer 6

http://openstudy.com/study#/updates/517bc95ce4b0249598f7a8c1

Answer 7

I'll be there now. but since we are here to help him/her. Would you please give out the answer before leaving? don't let him/her wait while we can, ok?

Answer 8

yes we can simplify this

Answer 9

We can use Demorgan's Theorem:
( x + y ) ' = x'y'
(xy)' = x' + y'

Answer 10

(x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )

Answer 11

one more step, friend, you have form of xx'

Answer 12

its boolean

Answer 13

(x'z'+xy')'
= (x'z')' (xy')'
= (x'' + z'' ) (x' + y'' )
= (x + z ) (x' + y )
= xx' +zx' +xy +zy
= zx' + xy + zy

Answer 14

but it seems the answer is: z(x'+y)!

Answer 15

@ketz I am with perl. I don't get yours

Answer 16

I think the answer needs to be converted to MMF

Answer 17

no need to get that form if the form is reducible . perl's answer is irreducible form that 's all you need

Answer 18

you claim that
(x'z'+xy')' = z ( x' +y ) ?

Answer 19

i think thats false, plug in values for x,y,z

Answer 20

but How to convert it to Minimal Multiplicative Form (MMF)? any idea guys?

Answer 21

can you double check you copied problem correctly

Answer 22

@ketz don't combine many problems into 1. that's another process, post a new one

Answer 23

Plug x=1, y=1, z = 0 into (x'z'+xy')'
(1' & 0' + 1&1')' = ( 0 &1 + 1 & 0 )' = ( 0 +0 )' = 0' = 1

Now plug x=1 , y = 1 , z = 0 into
z ( x' +y )
0 ( x' +y) = 0 , because 0*x = 0
clearly they are not equal

Answer 24

ketz theres a typo somewhere

Answer 25

@perl ketz asks about the way how to get the MMF form. not the original problem. All we need is a new problem such that we get the MMF form or we can manipulate the answer to get that form. However, by that step, we make the answer more complicated than it is.
We can separate it into new topic as:
convert zx' + xy + zy into MMF form

Answer 26

MMF is product of sums so
(x'z'+xy')'
= (x'z')' (xy')'
= (x'' + z'' ) (x' + y'' )
= (x + z ) (x' + y )

Answer 27

yeap

Answer 28

i dont know why hes not accepting it?

Answer 29

(x'z'+xy')' =/= z(x'+y)

Answer 30

=/= means not equal

Answer 31

because there is another way to get that form .

Answer 32

they are not equal, i proved it

Answer 33

Plug x=1, y=1, z = 0 into (x'z'+xy')'
(1' & 0' + 1&1')' = ( 0 &1 + 1 & 0 )' = ( 0 +0 )' = 0' = 1

Plug x=1 , y = 1 , z = 0 into
z ( x' +y )
0 ( x' +y) = 0 , because 0*x = 0
Therefore they are not equal

Answer 34

by multiply (x + x') into the term which doesn't have x,
(y + y') into the term which doesn't have y and so on

Answer 35

I'm with you, you are 100% right. just misunderstand together with him

Answer 36

ketz writes:
but it seems the answer is: z(x'+y)!

Answer 37

ok, let's go to his new one

Answer 38

wait

Answer 39

so theres two forms, a product of sums, or a sum of products

Answer 40

(x'z'+xy')'
= (x'z')' (xy')'
= (x'' + z'' ) (x' + y'' )
= (x + z ) (x' + y ) <---- Product of Sums
= xx' +zx' +xy +zy
= zx' + xy + zy

somehow we can get
xy + x'z <--- sum of products (with no terms repeating_

Answer 41

This is a Boolean Calculator
http://calculator.tutorvista.com/math/582/boolean-algebra-calculator.html
and i plugged in
~(~x & ~z or x & ~y)

Answer 42

sorry, i think wolfram has a boolean calculator, so let me find their website

Answer 43

here
http://www.wolframalpha.com/widgets/view.jsp?id=a52797be9f91295a27b14cb751198ae3

Answer 44

so this is an interesting question. im going to post a new one