Minimizing basic boolean function

Question

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Given the function $f(x,y,z) = y'z'+x'y+x'yz+xyz'$
(where ' means the NOT operator), I need to transfer this function to its basics. The possible answers are:

1. $x'y+y'z'$
2. $xy+z'$
3. $x'y'+z'$
4. $x'y+z'$

This is what I've done. I can't seem to figure out what's wrong. It's not in the answers....

$\begin{align} F(x\,,y\,,z)&=\overline{y}\overline{z}+\overline{x}y+\overline{x}yz+xy\overline{z}\\ &=\overline{y}\overline{z}+\overline{x}y(1+z)+xy\overline{z}\\ &=\overline{y}\overline{z}+\overline{x}y+xy\overline{z}\\ &=(x+\overline{x})\overline{y}\overline{z}+\overline{x}y(z+\overline{z})+xy\overline{z}\\ &=x\overline{y}\overline{z}+\overline{x}\overline{y}\overline{z}+\overline{x}yz+\overline{x}y\overline{z}+xy\overline{z}\\ &=x\overline{z}+\overline{x}\overline{z}+\overline{x}yz\\ &=\overline{z}+\overline{x}yz \end{align}$

Answer 1

$$\begin{align}\\ & y'z'+x'y+x'yz+xyz'\\ &=y'z'+x'y(1+z)+xyz'\\ &=y'z'+x'y+xyz'\\ &=y'z'+(x'+xz')y\\ &=y'z'+(x'+z')y\\ &=y'z'+x'y+yz'\\ &=(y+y')z'+x'y\\ &=z'+x'y \end{align}$$

Answer 2

You are very close! To finish up, note that $$\bar z=\bar z(1+\bar x y),$$ from which you should see that answer 4 is correct.