# How to simplify the expression -- Boolean algebra

1. Apr 13, 2015

### physics=world

1. prove that:
X'Y'Z + X'YZ' + XY'Z' + XYZ = (X⊕Y)⊕Z

2. Relevant equations
Use postulates and theorems.

3. The attempt at a solution
X'Y'Z + X'YZ' + XY'Z' + XYZ (original expression)
X'Y'Z + X'YZ' + X(Y'Z' + YZ) (distributive)
X'Y'Z + X'YZ' + X.1 (complement)
X'Y'Z + X'YZ' + X (identity)

need help.
also may you expand the expression
: (X⊕Y)⊕Z

I think it is like this: (XY' + X'Y)(ZXY' +Z'X'Y)
but i do not know if it's right

2. Apr 13, 2015

### BvU

Hi,

Not familiar with the ' notation, but I suppose X' means $\neg X$

When you expand $(X\oplus Y) \oplus Z$ you just write it out using $(A\oplus B) = A'B + AB'$ twice.

For me the easiest part is the second term $(X'Y + XY') Z' = X'Y Z'+ XY' Z'$ which are the 2nd and 3rd in the left hand side of your original staement.

Leaves you to prove $(X'Y + XY')' Z = X'Y' Z + XY Z$; not that hard, I hope ?