Re: boolean algebra

On Oct 10, 1:49 am, William Elliot <ma...@xxxxxxxxxxxxxxxxxx> wrote:
On Thu, 9 Oct 2008, Mitch wrote:
On Oct 9, 8:46 am, William Elliot <ma...@xxxxxxxxxxxxxxxxxx> wrote:
On Thu, 9 Oct 2008, Frederick Williams wrote:
"Greg.Shumaker" wrote:

Given that:
xy' + x'y = z

show that
xz' + x'z = y

If z is the symmetric difference of x and y, then
y is the symmetric difference of x and z.  Interesting.

And also x = (y != z). And also the triple identity for the dual

Huh?  You're not talking algebra.

Sorry, mixed languages. By 'y != z', I was abbreviating 'y or z but
not both', the exclusive-or (symmetric difference in sets). How does
'y xor z' sound? Or is that logic to you? If so, what is the algebra
way of saying that?

Using the language of the original poster (where '+' is 'or' and
concatenation is 'and'):

The triple identity is that any one of the following implies the other
two:
xy' + x'y = z
xz' + x'z = y
yz' + y'z = x

and the dual is:

any one of the following implies the other two:
(x+y')(x'+y) = z
(x+z')(x'+z) = y
(y+z')(y'+z) = x

Does that make sense?

Mitch
.

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