Set theory: Prove that $C \subseteq A \Delta B \iff C \subseteq A \cup B \wedge A \cap B \cap C = \emptyset$ I have been attempting this problem for a presentation I am due to give. I am wondering if my proof is correct and how I should go about the presentation. 
My proof: 
Then let $c \in C \subseteq A\Delta B = (A\cup B) − (A\cap B)$
$c \in C$ and $c \in (A\cup B)$ and $c \notin (A\cap B)$.
It follows that every $c \in C$ satisfies $c \in A\cup B$.
    Hence, $C \subseteq A\cup B$.
Also, every $c \in C$ satisfies $c \notin A \cap B$.
    Hence, $C \cap A \cap B = \emptyset$.
Let $c \in C \subseteq A \cup B \wedge A \cap B \cap C = \emptyset$.
$c \in C  \implies c \in A \cup B \wedge c \notin A \cap B \cap C$.
$(c \in C \implies c \in A \vee  c \in B) \wedge (c \notin A \wedge c \notin B \wedge c \notin C)$.
From this we can say, $c \in C \implies c \in A \vee c \in B \implies c \in (A \cup B) \wedge c \notin (A \cap B)$.
Also, $(c \notin A \wedge c \notin B \wedge c \notin C) \implies c \notin (A\cap B \cap C) = (A\cap B \cap C) = \emptyset$.
$\therefore C \subseteq A \Delta B \iff C \subseteq A \cup B \wedge A \cap B \cap C = \emptyset$.
How is my notation and can this be proved with venn diagrams?
 A: First of all, you should make it more clear where you're proving what. Your proof correctly consists of two parts. To make it better readable and understandable you really should state — for each part — what you're proving there and what your starting point is.
Case in point: the first three lines of your proof. I was able to figure our that there you're proving the "$\Longrightarrow$" direction of "iff", but it took me some effort because your first line looks too confusing. Here's how it can be improved:

Assume $C\subseteq A\bigtriangleup B$.
(Optional note: We need to show that $C\subseteq A\cup B$ and $A\cap B\cap C=\varnothing$.)
Pick an arbitrary $c\in C$. Then …

and you  can continue with your proof from there.
Similarly, you can improve the beginning of the second part, for the opposite direction of your proof. But there's a more serious problem with this second part: it's wrong. From $c\notin A\cap B\cap C$, the correct deduction is that $\color{blue}{(c\notin A\vee c\notin B\vee c\notin C)}$, not $\color{red}{(c\notin A\wedge c\notin B\wedge c\notin C)}$. So you have to fix that line, and then the rest of the proof. Also, the placement of parentheses in the same line (where you first have the red statement) is wrong too.
