Can any one help me solving this expression using boolean Algebra F=x'(y.z'+y'.z)+x.y'(y+x.z')

Question

I have solved some expression but can't proceed further

F=x'(y.z'+y'.z)+x.y'(y+x.z')

=x'y.z'+x'.y'.z+x.y'.y+x.x.y'.z'(by distributive law a(b+c)=ab+ac)

=x'y.z'+x'.y'.z+x.x.y'.z'(by using rule y'.y=0)

=x'y.z'+x'.y'.z+x.y'.z'(by using rule x.x=x)

Answer 1

You can factor out $z$ and $z'$ in the first two terms:

$$ =x'yz' + x'yz + xy'z'$$

$$ = x'y (z'+z) + xyz' $$

$$ = x'y + xyz' $$