There are two equivalent boolean functions: $$f_{1}(x,y,z)=\overline{z}\wedge(\overline{x}\vee x\wedge\overline{y}\wedge\overline{z})$$ and $$f_{2}(x,y,z)=(\overline{x}\wedge\overline{z})\vee(\overline{y}\wedge\overline{z})$$
The problem is that I don't understand which laws of boolean algebra I should apply to get from $f_1$ to $f_2$. How can I get $f_2$ from $f_1$?
You may go via transforming both expressions into their disjoint normal form.
I use $+,\cdot$ instead of $\vee,\wedge$ for easier readability:
$$f_{1}(x,y,z)=\bar{z}(\bar{x}+ x\bar{y}\bar{z}) = \bar x \bar z + x\bar y\bar z = \bar x(y+\bar y) \bar z + x\bar y\bar z = \bar x y \bar z + \bar x \bar y \bar z + x\bar y\bar z $$ $$f_{2}(x,y,z) = \bar x \bar z + \bar y\bar z = \bar x (y+\bar y)\bar z + (x+\bar x)\bar y\bar z = \bar x y \bar z + \bar x \bar y \bar z + x\bar y\bar z + \bar x\bar y\bar z \stackrel{a+a = a}{=} $$ $$\bar x y \bar z + \bar x \bar y \bar z + x\bar y\bar z $$ So, both disjoint normal forms are equal and you can transform one expression into the other one that way.
It is not the most elegant way but it surely works in any case.
