To prove they are disjoint, suppose for the sake of contradiction that they aren't i.e., $\exists x: x\in (A\setminus B) \cap (A\cap B) \cap (B\setminus A) \iff x \in A\setminus B$ and $x\in A\cap B$ and $x\in B\setminus A \iff (x\in A \land x \notin B) \text{ and } (x\in A \land x\in B)\text{ and }(x \in B \land x\notin A)$ so $x\in A$ and $x \notin A$ and $x \in B$ and $x \notin B$ which is a contradiction. Thus by the definition of the empty set, $(A\setminus B) \cap (A\cap B) \cap (B\setminus A)=\emptyset$.
For the second one let $x\in (A\setminus B) \cup (A\cap B) \cup (B\setminus A) \iff x\in A\setminus B$ or $x \in A\cap B$ or $x \in B\setminus A \iff (x\in A \text{ and } x\notin B)$ or $(x\in A \text{ and } x\in B)$ or $(x \in B \text{ and } x\notin A)$.
If $x\notin A$ then $x\notin B$ or $x \in B$ or $x \in A \implies x\in A$ or $x\in B \implies x\in A\cup B$.
If $x\in A$ then $(x\in A \text{ and } x\notin B)$ or $(x\in A \text{ and } x\in B)$ or $x\in B \implies x\in A$ or $x\in B\implies x\in A\cup B$.
In both cases $x\in A\cup B$ thus $ (A\setminus B) \cup (A\cap B) \cup (B\setminus A) \subseteq A\cup B$.
Now suppose $x \in A\cup B$. Then $x\in A$ or $x\in B$ and I'm not sure how to continue here.
My questions are: is my first proof correct? Is the second proof up to that point correct? If so, how could I continue in order to show $ A\cup B \subseteq (A\setminus B) \cup (A\cap B) \cup (B\setminus A)$?
Edit: The first proof should be proving the sets are pairwise disjoint. Here is a post where this gets answered for one of these cases and the rest are similar. Proof that $A \cap B$ and $A \setminus B$ are disjoint.
Best Answer
According to the properties involved, one has that: \begin{align*} (A\backslash B)\cap(A\cap B) & = (A\cap B^{c})\cap(A\cap B)\\\\ & = A\cap(B^{c}\cap(A\cap B))\\\\ & = A\cap(B^{c}\cap(B\cap A))\\\\ & = A\cap ((B^{c}\cap B)\cap A)\\\\ & = A\cap (\varnothing\cap A)\\\\ & = A\cap\varnothing = \varnothing \end{align*}
Similarly, we prove that $A\cap B$ and $B\backslash A$ are disjoint. As to the third case, we may proceed as follows: \begin{align*} (A\backslash B)\cap(B\backslash A) & = (A\cap B^{c})\cap(B\cap A^{c})\\\\ & = A\cap(B^{c}\cap(B\cap A^{c}))\\\\ & = A\cap((B^{c}\cap B)\cap A^{c})\\\\ & = A\cap(\varnothing\cap A^{c})\\\\ & = A\cap\varnothing = \varnothing \end{align*}
As to the union, let us do it in two steps. First, it results that \begin{align*} (A\backslash B)\cup(A\cap B) & = (A\cap B^{c})\cup(A\cap B)\\\\ & = A\cap(B^{c}\cup B) = A \end{align*}
And finally we obtain the desired claim: \begin{align*} (A\backslash B)\cup(A\cap B)\cup(B\backslash A) & = A\cup(B\backslash A)\\\\ & = A\cup(B\cap A^{c})\\\\ & = (A\cup B)\cap(A\cup A^{c})\\\\ & = A\cup B \end{align*}
Hopefully this helps!