I have the foolowing statement:
$$\bar{X_1}\cup(\bar{X_2}\setminus(X_3\cap\bar{X_1}))=\bar{X_2}\cup\bar{X_1}\cup X_3 \Leftrightarrow X_1 \cap X_2 \cap X_3= \emptyset,$$
so what I did was an introduction of shortcuts:
$$Z_A \equiv X_1 \cap X_2 \cap X_3 $$ $$Z_B \equiv \bar{X_2}\cup\bar{X_1}\cup X_3$$ $$Z_c = \bar{X_1}\cup\bar{X_2}$$
The last one after some simolification
And applying boolean algebra rules I got:
$$Z_A= y_1 \wedge y_2 \wedge y_3$$ $$Z_B = \bar{y_2}\vee\bar{y_1}\vee y_3$$ $$Z_C = \bar{y_1}\vee\bar{y_2}$$
Finally I got something like:
$$Z_C = Z_B \Leftrightarrow Z_A = \emptyset$$
The only thing came to my mind: $$F=\overline{y_1\vee y_2} = \overline{y_2\vee y_1}\vee y_3 \Leftrightarrow y_1\wedge y_2\wedge y_3 = 0$$
I feel my conversion/simplification is pretty wrong (I have to get simplified as much as possible boolean functiom from set statements above and then write up the truth table...)