Boolean Expression Simplifying explanation

Question

Currently have worked

xz' + x'y + (yz)' Down to z' + x'y + y'

Is this its simplest form?

METHOD: xz' + x'y + (yz)' -> De-Morgan on (yz)'
xz' + x'y + y' + z' -> Commutative
xz' + z' + x'y + y' -> Simplifying xz'+ z'
z'(x + 1) + x'y + y' -> Boundedness x+1 = 1
z'(1) + x'y + y' -> Identity z' * 1 = z'
z' + x'y + y'

Answer 1

You could simplify further by proving $x'y + y' = x' + y'$. The simplest answer should be $(xyz)'$.