A set theory proving question with symmetrical difference. I have this task and i'm quite stuck.
Prove that if $A \cap B = \emptyset $, $(A \Delta C) \cup (B \Delta C) = A \cup B \cup C$.
Currently i am  trying to show that $(A \Delta C) \cup (B \Delta C) \subseteq A \cup B \cup C$  by breaking down the left hand side.
iv'e came to the conclusion that for a given $x$,
$[x \in (A \cup C) \land x \notin (A \cap C)] \lor [x \in (B \cup C) \land x \notin (B \cap C)] $
all using logical bi-conditionals and the definitions. I know i need to somehow show the $x$ is part of the right hand side and currently stuck. Afterwards of course i'll need to show the opposite way as it's part of $=$ definition. I'm sorry if the questions doesn't follow the site guidelines / i offended someone. this is my first time here and this is my first question.
Thank you!
 A: To show LHS $\subseteq$ RHS, you could show:

*

*If $x \in A \Delta C$, then $x$ is contained in at least one of $A, B, C$.

*If $x \in B \Delta C$, then $x$ is contained in at least one of $A, B, C$.

(Do you see why those statements would prove this direction?)
To show RHS $\subseteq$ LHS, you should show:

*

*If $x \in A$, then $x$ is contained in at least one of $A \Delta C$ or $B \Delta C$.

*If $x \in B$, then $x$ is contained in at least one of $A \Delta C$ or $B \Delta C$.

*If $x \in C$, then $x$ is contained in at least one of $A \Delta C$ or $B \Delta C$.

(Again, do you see why those statements together would imply the result you want?)

To prove these individual statements, the key idea is to break into cases based on whether $x$ is in $A$, $B$, and/or $C$. I'll show how to prove one of the statements as an example, and then you could try to prove the others yourself. I'll prove: If $x \in A$, then $x$ is contained in at least one of $A \Delta C$ or $B \Delta C$.
Well, $x \in A$, so we know $x \not \in B$ by the initial assumption $A \cap B = \emptyset$. If $x \in C$ then we have $x \in (B \cup C)$ and $x \not \in (B \cap C)$, which means $x \in (B \Delta C)$ using the definition you gave. On the other hand, if $x \not \in C$ then we have $x \in (A \cup C$ and $x \not \in (A \cap C)$, which means $x \in (A \Delta C)$. So either way, we get $x \in (A \Delta C) \cup (B \Delta C)$.
