Equality proof

**OReilly** · May 7th, 2006, 11:39 AM

Can someone help me to solve this.

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

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

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

**CaptainBlack** · May 7th, 2006, 11:55 AM

Originally Posted by OReilly

Can someone help me to solve this.

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

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

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

Isn't your first line all you require?

$ x \in [A\backslash (A\backslash B)] \Leftrightarrow [x \in A] \wedge [x \notin (A\backslash B)] \Leftrightarrow$ $ [x \in A] \wedge [x \in B] \Leftrightarrow x \in (A \cap B) $

RonL

**OReilly** · May 7th, 2006, 12:21 PM

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

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

**rgep** · May 7th, 2006, 01:26 PM

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

**OReilly** · May 7th, 2006, 04:04 PM

rgep's answer has confused me a little.

I have one question.

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

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

Then it would be
$\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} $