1. ## Simple set proofs

I find these problems easy enough to understand, I just have trouble with proofs that make sense.

For example, if I wanted to prove:

$(C \cup D) \setminus (C\cap D)=(C\setminus D)\cup(D\setminus C)$

Can I write:

$(C \cup D)-(C\cap D)=\{x:x\in C$ or $x\in D\}\wedge \{x\in C:x\not\in D\}\wedge \{x\in D:x\not\in C\}$

$\Rightarrow ( \forall x \in C)(x \in (C\cup D^c ))$ and $( \forall x \in D)(x \in (D\cup C^c ))$

Hence,

$(C \cup D)\setminus (C\cap D)= (C\cup D^c ) \cup (D\cup C^c ) = (C\setminus D)\cup(D\setminus C)$

2. If you can use a non 'pick-a-point' proof, here is one.
$\begin{array}{rcl}
{\left( {C \cup D} \right)\backslash \left( {C \cap D} \right)} & \equiv & {\left( {C \cup D} \right) \cap \left( {C \cap D} \right)^c } \\
{} & \equiv & {\left( {C \cup D} \right) \cap \left( {C^c \cup D^c } \right)} \\
{} & \equiv & {\left( {C \cap D} \right) \cup \left( {D \cap C^c } \right)} \\
{} & \equiv & {\left( {C\backslash D} \right) \cup \left( {D\backslash C} \right)} \\

\end{array}$

3. Yeah, I can do that much, but I think I would be accused of only "showing" it. Is the attempt I made above even an actual proof?

4. Originally Posted by MathSucker
Yeah, I can do that much, but I think I would be accused of only "showing" it. Is the attempt I made above even an actual proof?
If $x\in(C\cup D)\setminus (C\cap D)$ means that $x\in C\text{ or }x\in D\text{ and }x\notin(C\cap D).$
So the last implies $x\notin C\text{ or }x\notin D$.
$(C \cup D)-(C\cap D)=\{x:x\in C$ or $x\in D\}\wedge \{x\in C:x\not\in D\}\wedge \{x\in D:x\not\in C\}$
First, $\land$ usually denotes conjunction, which joins propositions, not sets. (Speaking of this, it's not good to use both $\land$ and "and", as well as - and $\setminus$.) Second, I don't see how the right-hand side immediately follows from the definition of the left-hand side.