Simplifying an Expression by Using Boolean Algebra

Question

I have this expression: X'YZ'+X'YZ+XY'Z'+XYZ'+XYZ (' means not) I know the answer is Y+XZ' but I am stuck in the last part. Can anyone give me a hand?

This is what I got so far:

X'YZ' + X'YZ + XY'Z' + XYZ' + XYZ
X'YZ' + X'YZ + XZ'(Y' + Y) + XYZ
X'YZ' + X'YZ + XZ' + XYZ
X'YZ' + X'YZ + XYZ + XZ'
Y(X'Z' + X'Z + XZ) + XZ'   
Y(1) + XZ'   # I am not sure if is there is a rule that makes (X'Z+X'Z+XZ)= 1

Thanks

Answer 1

The only solution that I can think of is this (i.e. using XYZ' twice):

X'YZ' + X'YZ + XY'Z' + XYZ' + XYZ
X'YZ' + X'YZ + XY'Z' + XYZ' + XYZ + XYZ'
X'YZ' + X'YZ + XZ'(Y' + Y) + XYZ + XYZ'
X'YZ' + X'YZ + XYZ + XYZ'+ XZ'
Y(X'Z' + X'Z + XZ + XZ') + XZ'
Y(1) + XZ'   
Y + XZ'

Answer 2

Beginning with X'YZ'+X'YZ+XY'Z'+XYZ'+XYZ

(¬X ∧ Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ Z) ∨ (X ∧ ¬Y ∧ ¬Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (X ∧ Y ∧ Z) ↔ ⊤

Construct a Karnaugh map http://en.wikipedia.org/wiki/Karnaugh_map

XY 00 01 11 10
Z 0 0  1  1  1
  1 0  1  1  0

Reduce to Y ∨ (X ∧ ¬Z) in two steps.