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HoaBest ResponseYou've already chosen the best response.0
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 :( , :) )
 one year ago

perlBest ResponseYou've already chosen the best response.1
Hoa, wnat to check over a solution of mine?
 one year ago

perlBest ResponseYou've already chosen the best response.1
http://openstudy.com/study#/updates/517bc95ce4b0249598f7a8c1
 one year ago

HoaBest ResponseYou've already chosen the best response.0
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?
 one year ago

perlBest ResponseYou've already chosen the best response.1
We can use Demorgan's Theorem: ( x + y ) ' = x'y' (xy)' = x' + y'
 one year ago

perlBest ResponseYou've already chosen the best response.1
(x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )
 one year ago

HoaBest ResponseYou've already chosen the best response.0
one more step, friend, you have form of xx'
 one year ago

perlBest ResponseYou've already chosen the best response.1
(x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y ) = xx' +zx' +xy +zy = zx' + xy + zy
 one year ago

ketzBest ResponseYou've already chosen the best response.0
but it seems the answer is: z(x'+y)!
 one year ago

HoaBest ResponseYou've already chosen the best response.0
@ketz I am with perl. I don't get yours
 one year ago

ketzBest ResponseYou've already chosen the best response.0
I think the answer needs to be converted to MMF
 one year ago

HoaBest ResponseYou've already chosen the best response.0
no need to get that form if the form is reducible . perl's answer is irreducible form that 's all you need
 one year ago

perlBest ResponseYou've already chosen the best response.1
you claim that (x'z'+xy')' = z ( x' +y ) ?
 one year ago

perlBest ResponseYou've already chosen the best response.1
i think thats false, plug in values for x,y,z
 one year ago

ketzBest ResponseYou've already chosen the best response.0
but How to convert it to Minimal Multiplicative Form (MMF)? any idea guys?
 one year ago

perlBest ResponseYou've already chosen the best response.1
can you double check you copied problem correctly
 one year ago

HoaBest ResponseYou've already chosen the best response.0
@ketz don't combine many problems into 1. that's another process, post a new one
 one year ago

perlBest ResponseYou've already chosen the best response.1
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
 one year ago

perlBest ResponseYou've already chosen the best response.1
ketz theres a typo somewhere
 one year ago

HoaBest ResponseYou've already chosen the best response.0
@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
 one year ago

perlBest ResponseYou've already chosen the best response.1
MMF is product of sums so (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )
 one year ago

perlBest ResponseYou've already chosen the best response.1
i dont know why hes not accepting it?
 one year ago

HoaBest ResponseYou've already chosen the best response.0
because there is another way to get that form .
 one year ago

perlBest ResponseYou've already chosen the best response.1
they are not equal, i proved it
 one year ago

perlBest ResponseYou've already chosen the best response.1
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
 one year ago

HoaBest ResponseYou've already chosen the best response.0
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
 one year ago

HoaBest ResponseYou've already chosen the best response.0
I'm with you, you are 100% right. just misunderstand together with him
 one year ago

perlBest ResponseYou've already chosen the best response.1
ketz writes: but it seems the answer is: z(x'+y)!
 one year ago

HoaBest ResponseYou've already chosen the best response.0
ok, let's go to his new one
 one year ago

perlBest ResponseYou've already chosen the best response.1
so theres two forms, a product of sums, or a sum of products
 one year ago

perlBest ResponseYou've already chosen the best response.1
(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_
 one year ago

perlBest ResponseYou've already chosen the best response.1
This is a Boolean Calculator http://calculator.tutorvista.com/math/582/booleanalgebracalculator.html and i plugged in ~(~x & ~z or x & ~y)
 one year ago

perlBest ResponseYou've already chosen the best response.1
sorry, i think wolfram has a boolean calculator, so let me find their website
 one year ago

perlBest ResponseYou've already chosen the best response.1
here http://www.wolframalpha.com/widgets/view.jsp?id=a52797be9f91295a27b14cb751198ae3
 one year ago

perlBest ResponseYou've already chosen the best response.1
so this is an interesting question. im going to post a new one
 one year ago
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