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ketz
 2 years ago
Simplify:
f=(x'z'+xy')'
ketz
 2 years ago
Simplify: f=(x'z'+xy')'

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Hoa
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Hoa, wnat to check over a solution of mine?

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

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

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

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

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

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

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

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

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

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

Hoa
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1ketz theres a typo somewhere

Hoa
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1i dont know why hes not accepting it?

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

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

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

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

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

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

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

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

perl
 2 years 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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