1. ## Equality proof

Can someone help me to solve this.

Prove equality $\displaystyle A\backslash (A\backslash B) = A \cap B$

I have tried to solve it something like this, but I really don't know.

$\displaystyle x \in A \wedge x \notin (A\backslash B) \Leftrightarrow x \in A \wedge x \in B$
$\displaystyle x \in A \wedge x \notin (x \in A \wedge x \notin B) \Leftrightarrow x \in A \wedge x \in B$
$\displaystyle x \in A \wedge (x \notin A \wedge x \in B) \Leftrightarrow x \in A \wedge x \in B$

2. Originally Posted by OReilly
Can someone help me to solve this.

Prove equality $\displaystyle A\backslash (A\backslash B) = A \cap B$

I have tried to solve it something like this, but I really don't know.

$\displaystyle x \in A \wedge x \notin (A\backslash B) \Leftrightarrow x \in A \wedge x \in B$
$\displaystyle x \in A \wedge x \notin (x \in A \wedge x \notin B) \Leftrightarrow x \in A \wedge x \in B$
$\displaystyle x \in A \wedge (x \notin A \wedge x \in B) \Leftrightarrow x \in A \wedge x \in B$
Isn't your first line all you require?

$\displaystyle x \in [A\backslash (A\backslash B)] \Leftrightarrow [x \in A] \wedge [x \notin (A\backslash B)] \Leftrightarrow $$\displaystyle [x \in A] \wedge [x \in B] \Leftrightarrow x \in (A \cap B) RonL 3. This is what is confusing me. What happened to \displaystyle x \notin A in \displaystyle [x \in A] \wedge [x \notin (A\backslash B)] \Leftrightarrow$$\displaystyle [x \in A] \wedge [x \in B]$

Shouldn't be $\displaystyle x \notin (A\backslash B)$ equal to $\displaystyle x \notin A\wedge x \in B$?

4. $\displaystyle x \in A \setminus (A \setminus B)$ iff $\displaystyle x \in A \wedge x \not\in (A \setminus B)$ iff $\displaystyle x \in A \wedge {\sim}(x \in A \setminus B)$ iff $\displaystyle x \in A \wedge {\sim}(x \in A \wedge x \not\in B)$ iff $\displaystyle x \in A \wedge (x \not\in A \vee x \in B)$ iff $\displaystyle (x \in A \wedge x \not\in A) \vee (x \in A \wedge x \in B)$ iff $\displaystyle F \vee x \in A \cap B$ iff $\displaystyle x \in A \cap B$.

5. rgep's answer has confused me a little.

I have one question.

Does expression $\displaystyle \neg (x \notin A \wedge x \in B)$ means that beside negation of $\displaystyle \in$ or $\displaystyle \notin$ means that even $\displaystyle \wedge$ turns into $\displaystyle \vee$?

If that is true than answer is understable to me, because I can form tautology then.

Then it would be
$\displaystyle \begin{array}{l} x \in A \wedge x \notin (A\backslash B) \Leftrightarrow x \in A \wedge x \in B \\ x \in A \wedge \neg (x \in A \wedge \notin B) \Leftrightarrow x \in A \wedge x \in B \\ x \in A \wedge (x \notin A \vee \in B) \Leftrightarrow x \in A \wedge x \in B \\ \end{array}$

6. Does expression $\displaystyle \neg (x \notin A \wedge x \in B)$ means that beside negation of $\displaystyle \in$ or $\displaystyle \notin$ means that even $\displaystyle \wedge$ turns into $\displaystyle \vee$?
Yes: those are de Morgan's laws: $\displaystyle \neg(A \vee B) \equiv (\neg A) \wedge (\neg B)$ and $\displaystyle \neg(A \wedge B) \equiv (\neg A) \vee (\neg B)$.

7. Thats all I needed. Thanks!