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

Question

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

Answer 1

(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)

Answer 2

Your notation is unfamiliar to me, but I'm reading the first formula as:

(not ( x and y )) or ( y and z )

and the second as:

x and y

That seems consistent with your application of de Morgan's.

In that case, observe that when y is false, the first formula evaluates to true and the second evaluates to false, regardless of the values of x and z. Thus, the first formula does not imply the second. The simplification you are looking for cannot (validly) be performed.