booleanboolean-expressionsimplify

How is this (xy)' + (yz) simplified into this (x’+y’)+z in boolean algebra?

I not understand how the first Boolean expression on the question can be simplified into the last. Please help me.

My attempt:

1. (xy)' + (yz)

2. (x' + y') + (yz) # Using de Morgan's law.

3. x' + (y' + yz) # First Distribution axiom.

4. x' + (y' + y) * (y'z) # Rewriting

5. x' + 1 * (y'z) # Law of Inverse.

6. x' + y' + z # 2nd Identitiy axiom.

But the final answer is supposed to be (x’+y’)+z !

I am using the following book:

Discrete Mathematics for Computing / Edition 3
by Peter Grossman
Solution 
  • (xy)’+yz

(x’ +y’)+yz (2nd de Morgan's axiom)

x’ + (y’ +yz) (1st Association axiom)

x’+ (y’+y)(y’+z) (1st Distribution axiom)

x’ + 1(y’+z) (1st Inverse axiom)

x’+ (y’+z) (1st Identity axiom)

(x’+y’)+z (1st Association axiom)