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- anonymous

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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- anonymous

- jamiebookeater

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- anonymous

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

- anonymous

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

- anonymous

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

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- anonymous

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

- anonymous

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

- anonymous

0 represents F, and 1 represents T

- 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 ?

- hartnn

where a' means complement of a

- anonymous

(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

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.

- hartnn

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

- hartnn

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

- anonymous

(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!!!

- hartnn

welcome ^_^

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