Proving $(A\triangle C)\cup(B\triangle C)=A\cup B\cup C$ 
Let $A,B,C$ be sets and $A\cap B=\emptyset$. Prove $(A\triangle C)\cup(B\triangle C)=A\cup B\cup C$.

My attempt:
Let $x\in (A\triangle C)\cup(B\triangle C)$ and from $\triangle$ definition:
$$\begin{align}&x\in(A\triangle C)\cup(B\triangle C)\\ & \equiv 
x\in(((A\cup C)\setminus (A\cap C))\cup ((B\cup C)\setminus (B\cap C)))\\
& \equiv 
((x\in(A\cup C)\wedge x\not\in(A\cap C))\vee (x\in(B\cup C)\wedge x\not\in (B\cap C)))\\\\
& \overset{*}\equiv
\color{green}{(x\in(A\cup C)\vee x\in(B\cup C))}\wedge(x\in(A\cup C)\vee x\not\in(B\cap C))\wedge(x\not\in(A\cap C)\vee x\in(B\cup C))\wedge\color {blue}{(x\not\in(A\cap C)\vee x\not\in(B\cap C))}
\end{align}$$
Blue part is equivalent to (using De Morgan): $(A\cap C)\cap C\overset{A\cap B=\emptyset}=\emptyset$
$*$ applying double distribution.
That's how far I got, I can see the goal in the green part, but I'm not sure what to do with the two other parts. I'd also like to know if there's a way that doesn't involve the double distribution.
Also, I can't simply discard the parts where $x\not \in X$ right?
 A: Show two inclusions: take $x \in (A \triangle C) \cup (B \triangle C)$, we want to see it is in $A \cup B \cup C$. This is quite obvious, as $A \setminus C \subseteq A$, $C \setminus A \subseteq C$, $B \setminus C \subseteq B$, and $C \setminus B \subseteq C$, the left hand sides are the different parts of the left hand set, and all right hand sets are substes of $A \cup B \cup C$.
So take $x \in A \cup B \cup C$. 
If $x \in A$, then $x \notin B$ (by the disjointness). Now, if $x \in C$, then $x \in C \setminus B \subseteq B \triangle C$, and if $x \notin C$, $x \in A \setminus C \subseteq A \triangle C$. So $x$ is in the left hand side, either way.
If $x \in B$, we have $x \notin A$ and we can have a similar argument, with two cases depending on $x \in C$ or not. 
So the last case is where $x \notin A, x \notin B, x \in C$. So $x \in C \setminus A \subseteq A \triangle C$ and again we are done. 
A: First consider the symmetric difference of two sets $A$ and $B$ which is defined as
$$
A\Delta B = (A\cap \overline{B})\cup (\overline{A}\cap B).
$$
The following is probably the easiest route to your answer:
\begin{align}
(A\Delta C)\cup (A\Delta B) &= [(A\cap\overline{C})\cup(\overline{A}\cap C)]\cup[(A\cap\overline{B})\cup(\overline{A}\cap B)]\tag{defns.}\\[0.5em]
                            &= [(A\cap\overline{C})\cup(A\cap\overline{B})] \cup [(\overline{A}\cap C)\cup(\overline{A}\cap B)]\tag{assoc.}\\[0.5em]
                            &= [A\cap(\overline{C}\cup\overline{B})]\cup[\overline{A}\cap(C\cup B)]\tag{distrib}\\[0.5em]
                            &= A\cup \overline{A}\\[0.5em]
                            &= U = A\cup B\cup C,
\end{align}
where I'm assuming the sets $A,B,C$ are the only sets you are dealing with. 
A: It follows from the definition that $A\mathbin{\triangle}B\subseteq A\cup B$, so
$$
(A\mathbin{\triangle}C)\cup(B\mathbin{\triangle}C)
\subseteq
(A\cup C)\cup(B\cup C)
=
A\cup B\cup C
$$
Thus we have to prove the reverse inclusion. Let $x\in A\cup B\cup C$.
Case 1: $x\in A$. Then $x\notin A\mathbin{\triangle}C$ implies $x\in C$; but, since $A\cap B=\emptyset$, we have $x\in B\mathbin{\triangle}C$.
Case 2: $x\in C$. Then $x\notin B\mathbin{\triangle}C$ implies $x\in B$; but, since $A\cap B=\emptyset$, we have $x\in A\mathbin{\triangle}C$.
Case 3: $x\in B$. Then $x\notin B\mathbin{\triangle}C$ implies $x\in C$; but, since $x\notin A$, $x\in A\mathbin{\triangle}C$.
A: Semantically, we have:
\begin{align*}
x \in (A \triangle C) \cup (B \triangle C) &\iff \text{$x$ is in precisely one of $A$ and $C$, or in precisely one of $B$ and $C$}\\
x \in A \cup B \cup C &\iff \text{$x$ is in one of $A, B$ and $C$}
\end{align*}
That the former implies the latter is trivial. Conversely, we note that if $x$ is in $C$, then as $A \cap B = \varnothing$, it must be in at least one of $A \triangle C$ and $B \triangle C$ (the former if $x \notin B$, the latter if $x \notin A$). If $x \notin C$, then it is automatically in one of the two sets as well. The result follows.
You can also make a kind of truth table on three options $x \in A, B,C$ and seeing what each of these eight possibilities means for $x \in (A \triangle C) \cup (B \triangle C)$ and $x \in A \cup B \cup C$. This is the formal analogue of the Venn diagram method.
