Boolean algebra simplification

Question

Is this the simplest form of the expression?

Given: $$x'y'z+x'yz'+x'yz$$ My work: $$x'y(z'+z)+x'yz'= x'y+x'yz= x'y(z)$$

Answer 1

The simplest form is: x'z+ x'y

Answer 2

You could use Karnaugh maps. Alternatively, notice that: \begin{align*} &x'y'z + x'yz' + x'yz \\ &= x'y'z + x'yz' + (x'yz + x'yz) &\text{Idempotent Law}\\ &= (x'y'z + x'yz) + (x'yz' + x'yz) &\text{Commutativity/Associativity Law}\\ &= x'z(y' + y) + x'y(z' + z) &\text{Distributivity Law}\\ &= x'z(1) + x'y(1) &\text{Inverse Law}\\ &= x'z + x'y &\text{Identity Law}\\ \end{align*}