ketz
  • ketz
Simplify: f=(x'z'+xy')'
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
is it boolean?
anonymous
  • 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
  • perl
yes it looks boolean

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

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