Suppose $A, B$, and C are sets. Prove that $C\subset A\Delta B \Leftrightarrow C \subset A \cup B$ and $A \cap B \cap C = \emptyset $ The problem statement is in the title.
I'm proving a problem in class and I need to show the above containment. I've drawn some Venn diagrams to make sure the containment makes sense, and it does to me, but I am still having problems proving this statement.
I know that $A \Delta B = (A \cup B) – ( A \cap B)$, and I can see through the Venn diagrams I have drawn how [$C \subset A \Delta B$ iff $C \subset A \cup B$ and $A \cap B \cap C =\emptyset $], but I am having trouble explaining this to someone else. 
 A: Let's try necessity. Let $$c \in C \subseteq A \Delta B = (A \cup B) - (A \cap B).$$ Then, $c \in A \cup B$ and $c \not \in A \cap B$. Thus,


*

*every $c \in C$ also satisfies $c \in A \cup B$, hence, $C \subseteq A \cup B$;

*every $c \in C$ also satisfies $c \not \in A \cap B$, hence $C \cap A \cap B = \emptyset$.


Can you go the other way?
A: I would approach this as a simplification problem, which is straightforwardly solved by expanding the definitions from set theory and then simplifying using the rules of logic.
So I would start on the most complex side, and calculate as follows:
\begin{align}
& C \subseteq A \cup B \;\land\; A \cap B \cap C = \varnothing \\
\equiv & \qquad \text{"definitions of $\;\subseteq\;$; basic property of $\;\varnothing\;$"} \\
&
  \langle \forall x :: x \in C \;\Rightarrow\; x \in A \cup B \rangle
  \;\land\;
  \langle \forall x :: \lnot(x \in A \cap B \cap C) \rangle
  \\
\equiv & \qquad \text{"definition of $\;\cup\;$, and of $\;\cap\;$ twice; merge quantifications"} \\
& \langle \forall x ::
    (x \in C \;\Rightarrow\; x \in A \lor x \in B)
    \;\land\;
    \lnot(x \in A \land x \in B \land x \in C)
  \rangle \\
\equiv & \qquad \text{"logic: write $\;P \Rightarrow Q\;$ as $\;\lnot P \lor Q\;$ in left part;} \\
& \qquad \phantom{\text{"}}\text{DeMorgan for right part} \\
& \qquad \phantom{\text{"}}\text{-- to get everything in similar shape"} \\
& \langle \forall x ::
    (x \not\in C \lor x \in A \lor x \in B)
    \;\land\;
    (x \not\in A \lor x \not\in B \lor x \not\in C)
  \rangle \\
\equiv & \qquad \text{"logic: extract common disjunct $\;x \not\in C\;$"} \\
& \langle \forall x :: x \not\in C \lor (
    (x \in A \lor x \in B)
    \;\land\;
    (x \not\in A \lor x \not\in B)
  ) \rangle \\
\equiv & \qquad \text{"logic: simplify $\;(P \lor Q) \land (\lnot P \lor \lnot Q)\;$ to $\;P \not\equiv Q\;$"} \\
& \langle \forall x :: x \in C \;\Rightarrow\; (
    x \in A \not\equiv x \in B
  ) \rangle \\
\equiv & \qquad \text{"definition of $\;\triangle\;$; definition of $\;\subseteq\;$"} \\
& C \subseteq A \triangle B \\
\end{align}
This completes the proof.
In the last step I used a not-too-well-known definition of set difference:
$$
x \in A \triangle B \;\equiv\; x \in A \not\equiv x \in B
$$
