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

- ketz

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

- schrodinger

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- anonymous

is it boolean?

- anonymous

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 :( , :) )

- perl

yes it looks boolean

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## More answers

- perl

Hoa, wnat to check over a solution of mine?

- anonymous

sure.

- anonymous

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?

- perl

yes we can simplify this

- perl

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

- perl

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

- anonymous

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

- ketz

its boolean

- perl

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

- ketz

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

- anonymous

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

- ketz

I think the answer needs to be converted to MMF

- anonymous

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

- perl

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

- perl

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

- ketz

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

- perl

can you double check you copied problem correctly

- anonymous

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

- perl

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

- perl

ketz theres a typo somewhere

- anonymous

@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

- perl

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

- anonymous

yeap

- perl

i dont know why hes not accepting it?

- perl

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

- perl

=/= means not equal

- anonymous

because there is another way to get that form .

- perl

they are not equal, i proved it

- perl

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

- anonymous

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

- anonymous

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

- perl

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

- anonymous

ok, let's go to his new one

- perl

wait

- perl

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

- perl

(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_

- perl

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

- perl

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

- perl

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

- perl

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

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