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Hoa
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1Hoa, wnat to check over a solution of mine?

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

Hoa
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.1We can use Demorgan's Theorem: ( x + y ) ' = x'y' (xy)' = x' + y'

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

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

perl
 one year ago
Best 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

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

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

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

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

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

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

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

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

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

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

Hoa
 one year ago
Best 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

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

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

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

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

perl
 one year ago
Best ResponseYou've already chosen the best response.1Plug 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

Hoa
 one year ago
Best ResponseYou've already chosen the best response.0by multiply (x + x') into the term which doesn't have x, (y + y') into the term which doesn't have y and so on

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

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

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

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

perl
 one year ago
Best 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_

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

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

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

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