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RolyPoly
Group Title
Boolean algebra
Show that F(x, y, z) = xy + xz + yz has the value 1 if and only if at least two of the variables x, y, and z have the value 1.
How can I show that apart from drawing the truth table??
 one year ago
 one year ago
RolyPoly Group Title
Boolean algebra Show that F(x, y, z) = xy + xz + yz has the value 1 if and only if at least two of the variables x, y, and z have the value 1. How can I show that apart from drawing the truth table??
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
since this is completely symmetric in \(x, y, z\) you can work by cases
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
by which i mean without loss of generality you can say \(x=1,y=1, z=1\) and get it, or \(x=1,y=1,z=0\) and also get it
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
then for the "if" part, take \(x=1,y=0,z=0\) and show it is not true and also in the case \(x=y=z=0\)
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
But then, it would be similar to drawing the truth table...
 one year ago

mathsmind Group TitleBest ResponseYou've already chosen the best response.0
i have attached a file of the truth table but it is not uploading for a reason probably due to a file format, however, all the cases are valid apart from 0 or 0 or 0 of your and statments
 one year ago

mathsmind Group TitleBest ResponseYou've already chosen the best response.0
0 represents F, and 1 represents T
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
here's what we can do without using truth table. Double complement F , F = (xy+yz+xz) '' =[ (xy+yz+xz)']' now use Demorgan's law on (xy+yz+xz) ' what you get ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
where a' means complement of a
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
(xy + yz +xz)'' = [(xy + yz +xz)']' = [(xy)' (yz)' (xz)']' = [(x'+y') (z'+y') (x'+z') ]' = [ (x'z' + y'z' + x'y' + y') ( x'+z') ]' = [ x'z' + x'y'z' + x'y' + y'z' ]' = [x'z' + x'y' (z'+1) + y'z' ]' = (x'y' + x'z' + y'z')'
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yes, with (x'y' + x'z' + y'z')' you can conclude that F has the value 1 if and only if at least two of the variables x, y, and z have the value 1.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
because if say you have only x=1 then (x'y' + x'z' + y'z')' = (0+0+1)' = 0
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
so, atleast 2 variables have to be 1 to get (x'y' + x'z' + y'z')' =1
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
(Testing) x=1, y=1 (x'y' + x'z' + y'z')' = (0 + 0 + 0)' = 1 x=1, z=1 (x'y' + x'z' + y'z')' = (0 + 0 +0)' = 1 y=1, z=1 (x'y' + x'z' + y'z')' = (0 + 0 + 0)' = 1 x=y=z=1 (x'y' + x'z' + y'z')' = (0 + 0+ 0)' =1 Wow!! Thanks!!!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 one year ago
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