How to prove this set equation?

Question

I started as

$$\bar{X_1}\cup(\bar{X_2}\setminus(X_3\cap\bar{X_1})=\bar{X_1}\cup(\bar{X_2}\setminus X_3)\cup (\bar{X_2}\setminus \bar{X_1})=\bar{X_1}\cup(\bar{X_2}\setminus X_3)\cup \bar{X_2}=\bar{X_1}\cup \bar{X_2}$$

To simplify the left side, but I do not know how to proceed further with the proof itself :(

Answer 1

We can proceed with $\to$ first and then $←$.

For $→$ : assume that the equality holds and assume for contradiction that $X_1 ∩ X_2 ∩ X_3 \ne \emptyset$.

This means that, for some $z : z ∈ X_1,X_2,X_3$, and this in turn implies that:

$z ∈ X_3$ and $z ∉ \bar{X_1} ∪ \bar{X_2}$.

But thus : $z ∈ X_3 \cup \bar{X_1} \cup \bar{X_2}$.

In conclusion, we have that, for some $z : z ∉ \bar{X_1} ∪ \bar{X_2}$ and $z ∈ X_3 \cup \bar{X_1} \cup \bar{X_2}$, contradicting the fact that the two are equals.

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Similarly for $←$ : assume that $X_1 ∩ X_2 ∩ X_3 = \emptyset$ and assume for contradiction that $\bar{X_1} ∪ \bar{X_2} \ne X_3 \cup \bar{X_1} \cup \bar{X_2}$.

This means that there is a $z$ such that $z ∈ X_3 \cup \bar{X_1} \cup \bar{X_2}$ and $z ∉ \bar{X_1} ∪ \bar{X_2}$.

But $z ∉ \bar{X_1} ∪ \bar{X_2}$ means that $z \in X_1$ and $z \in X_2$.

And $z ∉ \bar{X_1} ∪ \bar{X_2}$, with the fact that $z ∈ X_3 \cup \bar{X_1} \cup \bar{X_2}$, means that $z \in X_3$.

Thus :

$z \in X_1,X_2,X_3$,

contradicting the fact that $X_1 ∩ X_2 ∩ X_3 = \emptyset$.

Answer 2

To continue from where you left off,
show the LHS is equivalent to both
$X_3 \subseteq \bar{X_1}\cup\bar{X_2}$ and $X_1 \cap X_2 \subseteq \bar{X_3}.$
The rest is now apparent.