# 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?

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.

• Can you explain why we'd deduct c∉A∨c∉B∨c∉C from c∈A∩B∩C? I am having difficulty following the logic. – OFlyJr4 Aug 17 '17 at 2:49
• @OFlyJr4: Here's the explanation for my "logic": I didn't get much sleep last night... Of course, you're right and I am wrong there. Sorry about that. I'm going to update my post now. Once again, my apologies. – zipirovich Aug 17 '17 at 3:26
• @OFlyJr4: Gosh, I am way too tired today. On the second thought, I have to take back my taking it back. I was right and you weren't. Sorry again... One way to explain it: De Morgan's laws for negation. If the original statement is $c\in A\cap B\cap C$, which is equivalent to $c\in A\wedge c\in B\wedge c\in C$, then you need its negation to say that $c$ is not in that intersection. So you need the negation of "$c\in A\wedge c\in B\wedge c\in C$", which is $\neg(c\in A)\vee\neg(c\in B)\vee\neg(c\in C)$, which is the same as $c\notin A\vee c\notin B\vee c\notin C$. – zipirovich Aug 17 '17 at 4:01
• @OFlyJr4: Another explanation is just to think about intersection of sets. For $c$ to be an element of the intersection of several sets, $c$ must be an element of each one of them. So if $c$ is not in just one of them (any one), then it's already not in their intersection. – zipirovich Aug 17 '17 at 4:07
• DeMorgan's law. jeez. I was thinking that as I typed out my response. Thank you so much. – OFlyJr4 Aug 18 '17 at 0:02