# Equality proof

• May 7th 2006, 11:39 AM
OReilly
Equality proof
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.

$x \in A \wedge x \notin (A\backslash B) \Leftrightarrow x \in A \wedge x \in B$
$x \in A \wedge x \notin (x \in A \wedge x \notin B) \Leftrightarrow x \in A \wedge x \in B$
$x \in A \wedge (x \notin A \wedge x \in B) \Leftrightarrow x \in A \wedge x \in B$
• May 7th 2006, 11:55 AM
CaptainBlack
Quote:

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
• May 7th 2006, 12:21 PM
OReilly
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 $x \notin A\wedge x \in B$?
• May 7th 2006, 01:26 PM
rgep
$x \in A \setminus (A \setminus B)$ 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 $x \in A \cap B$.
• May 7th 2006, 04:04 PM
OReilly
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}
$
• May 7th 2006, 10:16 PM
rgep
Quote:

Does expression $\neg (x \notin A \wedge x \in B)$ means that beside negation of $\in$ or $\notin$ means that even $\wedge$ turns into $\vee$?
Yes: those are de Morgan's laws: $\neg(A \vee B) \equiv (\neg A) \wedge (\neg B)$ and $\neg(A \wedge B) \equiv (\neg A) \vee (\neg B)$.
• May 8th 2006, 02:21 AM
OReilly
Thats all I needed. Thanks!