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

Mathematics
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is it boolean?
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 :( , :) )
yes it looks boolean

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Other answers:

Hoa, wnat to check over a solution of mine?
sure.
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?
yes we can simplify this
We can use Demorgan's Theorem: ( x + y ) ' = x'y' (xy)' = x' + y'
(x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )
one more step, friend, you have form of xx'
its boolean
(x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y ) = xx' +zx' +xy +zy = zx' + xy + zy
but it seems the answer is: z(x'+y)!
@ketz I am with perl. I don't get yours
I think the answer needs to be converted to MMF
no need to get that form if the form is reducible . perl's answer is irreducible form that 's all you need
you claim that (x'z'+xy')' = z ( x' +y ) ?
i think thats false, plug in values for x,y,z
but How to convert it to Minimal Multiplicative Form (MMF)? any idea guys?
can you double check you copied problem correctly
@ketz don't combine many problems into 1. that's another process, post a new one
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
ketz theres a typo somewhere
@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
MMF is product of sums so (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )
yeap
i dont know why hes not accepting it?
(x'z'+xy')' =/= z(x'+y)
=/= means not equal
because there is another way to get that form .
they are not equal, i proved it
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
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
I'm with you, you are 100% right. just misunderstand together with him
ketz writes: but it seems the answer is: z(x'+y)!
ok, let's go to his new one
wait
so theres two forms, a product of sums, or a sum of products
(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_
This is a Boolean Calculator http://calculator.tutorvista.com/math/582/boolean-algebra-calculator.html and i plugged in ~(~x & ~z or x & ~y)
sorry, i think wolfram has a boolean calculator, so let me find their website
here http://www.wolframalpha.com/widgets/view.jsp?id=a52797be9f91295a27b14cb751198ae3
so this is an interesting question. im going to post a new one

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