# expressing union of sets using set differences

How to prove the following equality

Let A, B and C be sets. Prove that
$$\hspace{1cm}A \cup B \cup C\:= (A -B) \cup (B-C) \cup (C-A) \cup (A \cap B \cap C).$$
using a chain of if and only if statements.

I tried to express set equality formula using propositional variables and connectives, solve it and then use in the solution of above equality, but I failed.

By the De Morgan's laws we obtain: $$\sum_{cyc}(A-B)+ABC=\sum_{cyc}(A(-B)+ABC)=\sum_{cyc}A(-B+BC)=$$ $$=\sum_{cyc}A(-B+B)(-B+C)=\sum_{cyc}A(-B+C)=$$ $$=\sum_{cyc}(A(-B)+AC)=\sum_{cyc}(A(-B)+AB)=\sum_{cyc}A(-B+B)=\sum_{cyc}A.$$ I used the cyclic summation.
For example, $$\sum_{cyc}(A\setminus B)=(A\setminus B)+(B\setminus C)+(C\setminus A)=(A\setminus B)\cup(B\setminus C)\cup(C\setminus A),$$ $$\sum_{cyc}(A(-B)+ABC)=$$ $$=((A\cap\overline{B})\cup (A\cap B\cap C))\cup((B\cap\overline{C})\cup (B\cap C\cap A))\cup((C\cap\overline{A})\cup (C\cap A\cap B)).$$
EDIT: Oh I see you wanted if and only if statements. I guess need to say $$x\in A\cup B\cup C \iff x\in A\vee x\in B\vee x\in C \iff \ldots$$. But the idea can be the same.
Here is a similar proof for only two sets: $$\begin{split} (A\setminus B) \cup (B\setminus A) \cup (A\cap B) &= (A\cap B^c) \cup (B\cap A^c) \cup (A\cap B) \\ &= (A \cap (B \cup B^c)) \cup (B\cap A^c) \\ &= A \cup (B\cap A^c) \\ &= (A\cup B)\cap (A\cup A^c) \\ &= A\cup B \end{split}$$ Line 2 is distributivity in reverse; line 4 is distributivity.