Boolean Algebra Proof (Distribution and XOR)

[PhotonSSBM]

Homework Statement

Use the definition of exclusive or (XOR), the facts that XOR commutes and
associates (if you need this) and all the non-XOR axioms and theorems you know
from Boolean algebra to prove this distributive rule:

A*(B (XOR) C) = (A*B) (XOR) (A*C)

Homework Equations

All the boolean algebra theorems

The Attempt at a Solution

So here's what I've tried

A*(B (XOR) C)
=A*B*C' + A*B'*C
=A*X+A*X'
=A*1
=A*(X (XOR) X')

And

(A*B) (XOR) (A*C)
=A*B*(A*C)' + (A*B)'*B*C
=A*B*(A' + C') + (A' + B')*B*C
=A*B*A' + A*B*C' + A'*B*C + B'*B*C
=A*B*C' + A'*B*C
=B*(A (XOR) C)
I feel like I go in circles no matter what I do.
Any clues?

 

[NascentOxygen]

=A*X+A*X'

This looks wrong.

=A*B*(A*C)' + (A*B)'*B*C

Mistake here, too.

 

[PhotonSSBM]

Oops, that's not how demorgans works...brb with corrections.

 

[PhotonSSBM]

Ok, the answer came out once I realized how to use DeMorgan's Rule for the right hand side:

(A*B) (XOR) (A*C)
=(A*B) * (A*C)' + (A*B)' * (A*C)
=A*B * (A' + C') + (A' + B') * A*C
=A*A'*B + A*B*C' + A'*A*C + B'*A*C
=A*B*C' + A*B'*C
=A*(B*C' + B'*C)
=A* [B (XOR) C]

Thanks for the help!

 

[Raimundo]

in the next step:
=A*A'*B + A*B*C' + A'*A*C + B'*A*C
=A*B*C' + A*B'*C

what happened with (A*A'*B + A'*A*C)?

 

[Master1022]

Hi,

what happened with (A*A'*B + A'*A*C)?

I think those go to zero as $A \cdot \bar A = 0$