How do I write a rigorous proof of the following problem: $(A \triangle B) \cup C \neq (A \cup C) \triangle (B \cup C)$ Question: How do I write a rigorous proof of the following problem:
Does the following equality hold true for any sets $A$, $B$ and $C$:
$$
(A \triangle B) \cup C = (A \cup C) \triangle (B \cup C)
$$
where $\triangle$ denotes a symmetric difference in the book I'm reading.
The above I easily see with Venn diagrams and understand that both sides represent different sets. Writing it down though, rigorously at that, is a big challenge for me.
Here's how I've written it down. I decided to simplify the RHS and prove that both sets are not quite the same set.
$$
(A \cup C) \triangle (B \cup C) = 
$$
$$
((A \cup C) \backslash (B \cup C) \cup ((B \cup C) \backslash (A \cup C)) = _*
$$
$$
(A \backslash (B \cup C)) \cup (B \backslash (A \cup C)) = _*
$$
$$
(A \backslash B) \backslash C \cup (B \backslash A) \backslash C = _*
$$
$$
((A \backslash B) \cup (B \backslash A)) \backslash C =
$$
$$
(A \triangle B) \backslash C
$$
So I get the following equality:
$$ (A \triangle B) \cup C = (A \triangle B) \backslash C$$
Here I say that if $x \in C$, then $x \in$ the LHS set but $\notin$ the RHS set, which means that the two sets are different, so the initial equality doesn't hold true.
The $=_*$ symbol means that those transitions were intuitively made by me. They seem obvious to me by looking at their respective Venn diagrams but I'm not sure that this is enough to another reader. So, on separate sections of my paper I write the corresponding sets in set-builder notations and try to simplify them so that I get the set-builder notation of the next set, so that I continue my proof. I can also show them here if you want.
Background: I'm reading a book about Toplogy, which begins with elementary set theory topics. I have basic understanding of sets, so I'm trying to solve every problem presented along with the theory. With certain problems, however, I'm struggling with presenting rigorous proofs for what seem to be basic and rather elementary facts. And by rigorous I mean a proof that even I, when reading later, will have no difficulty understanding without making 0_o faces.
My proofs usually begin with one or two pages of trying different starting approaches, drawing Venn diagrams, etc. until I see the problem in its entirety. Since these proofs are not verified by any authority I have my own common sense to decide whether I have presented them right. And I want to learn to do them right.
 A: Your derivation is entirely correct. Moreover, it won't be hard to follow for anyone familiar with these elementary set operations. As long as you include the different steps, all should be fine.
Note, however, that this particular problem can be resolved more quickly.
Namely, from the nature of the symmetric difference, we see that $x \in C$ implies $x \notin (A \cup C)\triangle (B\cup C)$. It also implies that $x \in A \triangle B \cup C$. 
Therefore, the equality doesn't hold if $C \ne \varnothing$. On the other hand, it trivially holds if $C = \varnothing$. 
Therefore, we have completely characterised the solutions to the equality, without actually doing any set algebra.
A: If you can't find a counterexample to $(A \triangle B) \cup C = (A \cup C) \triangle(B \cup C)$, then you can transform the problem into one that is equivalent but easier to solve. So, here's how I would attack the problem. 
Suppose, for the sake of contradiction, that $(A \triangle B) \cup C = (A \cup C) \triangle(B \cup C)$. One can easily prove that $(A \triangle B) \cup C = (A \cup C) \triangle (B \backslash C)$, so $(A \cup C) \triangle(B \cup C) = (A \cup C) \triangle (B \backslash C)$ and therefore $B \cup C = B \backslash C$. But this is a contradiction, and thus it must be the case that $(A \triangle B) \cup C \neq (A \cup C) \triangle(B \cup C)$.
A: $$(A\triangle B)\cup C=$$
$$((A\cap B^c)\cup(B\cap A^c))\cup C \;^{(*)}$$ 
$$\;$$
$$(A\cup C)\triangle (B\cup C)=$$
$$((A\cup C)\cap (B\cup C)^c)\cup((B\cup C)\cap (A\cup C)^c)=$$ $$((A\cup C)\cap B^c\cap C^c)\cup((B\cup C)\cap A^c\cap C^c)=$$ $$(A\cap B^c\cap C^c)\cup(B\cap A^c\cap C^c)=$$ $$((A\cap B^c)\cup(B\cap A^c))\cap C^c$$
$$(A\triangle B)\cap C^c \;^{(**)}$$
Look at $^{(*)}$ and $^{(**)}$. You have a union with C and an intersection with its complement. Hope it helps although I've seen this five years later.
