Simple set proofs

**MathSucker** · February 2nd 2011, 12:43 PM

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,

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

**Plato** · February 2nd 2011, 12:56 PM

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

**MathSucker** · February 2nd 2011, 01:15 PM

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?

**Plato** · February 2nd 2011, 01:41 PM

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?

Well a proof is proof. But follow your instructor’s instructions.

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 $x\notin C\text{ or }x\notin D$ .
From there takes cases.

**MoeBlee** · February 2nd 2011, 02:49 PM

delete

**emakarov** · February 3rd 2011, 04:52 AM

I am struggling to understand the following.

Originally Posted by MathSucker

$(C \cup D)-(C\cap D)=\{x:x\in C$ 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.

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