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 one year ago
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
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??

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0by 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

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

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.0i 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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.00 represents F, and 1 represents T

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1here'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 ?

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

RolyPoly
 one year ago
Best 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')'

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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.

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

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

RolyPoly
 one year ago
Best 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!!!
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