Basic Boolean Algebra Proof: Validity of the Cancellation Law?

[XcKyle93]

Homework Statement

Prove that in a Boolean algebra the cancellation law does not hold; that is, show that, for every x, y, and z in a Boolean algebra, xy = xz does not imply y = z.

Homework Equations

The 6 postulates of a Boolean Algebra

The Attempt at a Solution

I am uncertain as to whether or not what I have done is a valid proof. Plus, is there a way to do this using solely the postulates? That's what I initially tried to do, but I drew a blank. I honestly needed a hint for the below.

Suppose that for a Boolean algebra, x = 0, y = 1, z = 0. Then, xy = yz becomes:
0*1 = 1*0
0 = 0
Thus, xy = yz does not imply that y = z, and the cancellation law of multiplication does not hold for a Boolean Algebra.

 

[I like Serena]

Welcome to PF, XcKyle93!

You have just completed the proof.
It's a proof by counterexample.
If you can find an example in which a proposition does not hold, that is enough to proof it false.

To do it by the postulates, you need to consider what left cancellation means.
What you actually do, is multiply the left with the inverse of x.
However, in a boolean algebra not every element has an inverse, so cancellation is not allowed.

Note that "cancellation" is not an axiom, but a proposition.
It (only) follows if every element has an inverse (which is not the case here).

 

[DivisionByZro]

On one line you have:

show that, for every x, y, and z in a Boolean algebra, xy = xz does not imply y = z.

But on the other:

Suppose that for a Boolean algebra, x = 0, y = 1, z = 0. Then, xy = yz becomes:
0*1 = 1*0
0 = 0

Surely you meant to plug in values to "xy = xz"?

 

[XcKyle93]

Whoops, my bad. I did mean xy = xz, and 0 * 1 = 0 * 0, 0 = 0. That was a bad typo.