[Math] simplify boolean expression: xy + xy’z + x’yz’

Question

As stated in the title, I'm trying to simplify the following expression: $xy + xy'z + x'yz'$ I've only gotten as far as step 3:

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

$=x(y+y’z) + x’(yz’)$

$=x(y+y’z)+x(y’+z)$

But I don't know where to go from this step, I'm not sure if I'm allowe to rewrite y'z as y+z' (I'm not even sure if that would help)

Answer 1

After figuring things out with a Karnaugh map, we see that splitting up the $xy$ term will be helpful: \begin{align*} xy + xy'z + x'yz' &= xy(1) + xy'z + x'yz' \\ &= xy(z + z') + xy'z + x'yz' \\ &= (xyz + xyz') + xy'z + x'yz' \\ &= (xyz + xy'z) + (xyz' + x'yz') \\ &= xz(y + y') + yz'(x + x') \\ &= xz(1) + yz'(1) \\ &= xz + yz' \end{align*}