\begin{array}{c|c|c|c} x&y&z&f(x,y,z)\\\hline 0&0&0&0\\0&0&1&0\\0&1&0&1\\0&1&1&1\\1&0&0&1\\1&0&1&1\\1&1&0&0\\1&1&1&1 \end{array}
I think I have solved by half, but I can not completely simplify
$$x′yz′ + x′yz + xy′z′ + xy′z + xyz\\ x′y(z′+z) + xy′z′ + xz(y′+y)\\ x′y + xy′z′ + xz$$
Could you help me?
Instead of pulling $xz$ out of $xy'z + xyz$, I would suggest pulling $xy'$ out of $xy'z' + xy'z$. Then you end up with an expression equivalent to ($x$ XOR $y$) OR ($x$ AND $y$ AND $z$). I don't think you can do more to simplify that.
Yes, you can do a little bit more:
$$x′y + xy′z′ + xz=$$
$$x′y + x(y′z′ + z)=$$
$$x′y + x(y′+z)(z′ + z)=$$
$$x′y + x(y′+z)1=$$
$$x′y + x(y′+z)=$$
$$x′y + xy′+xz$$
In fact, there is an advanced equivalence principle called:
Generalized Reduction
$PQ+PQ'R = PQ+PR$
(that is, in the context of $PQ$, the term $PQ'R$ reduces to $PR$. Indeed, the special case $Q+Q'R=Q+R$ is known as Reduction)
Applying this to your expression, we can thus reduce $x'yz'$ in the context of $xz$ immediately to $xy'$ in a single step!
