## RolyPoly 2 years 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??

1. satellite73

since this is completely symmetric in \(x, y, z\) you can work by cases

2. satellite73

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

3. satellite73

then for the "if" part, take \(x=1,y=0,z=0\) and show it is not true and also in the case \(x=y=z=0\)

4. RolyPoly

But then, it would be similar to drawing the truth table...

5. mathsmind

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

6. mathsmind

0 represents F, and 1 represents T

7. hartnn

here's what we can do without using truth table. Double complement F , F = (xy+yz+xz) '' =[ (xy+yz+xz)']' now use De-morgan's law on (xy+yz+xz) ' what you get ?

8. hartnn

where a' means complement of a

9. RolyPoly

(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')'

10. hartnn

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.

11. hartnn

because if say you have only x=1 then (x'y' + x'z' + y'z')' = (0+0+1)' = 0

12. hartnn

so, atleast 2 variables have to be 1 to get (x'y' + x'z' + y'z')' =1

13. RolyPoly

(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!!!

14. hartnn

welcome ^_^