1. Mar 25, 2005

### Kamataat

Verify $A\cap(B-C)=(A\cap B)-(A\cap C)$.

This is how I did it:

From $x\in[A\cap(B-C)]$ (lhs) we have that
1.) $x\in A$
2.) $x\in B$
3.) $x\notin C$
From #1 and #2 we have that
4.) $x\in(A\cap B)$
Now, since $x\in(A\cap C)$ means that $x\in A$ and $x\in C$, but we have $x\notin C$, it follows that
5.) $x\notin(A\cap C)$
Remembering that $x\in(A-B)$ means that $x\in A$ and $x\notin B$ it follows from #4 and #5 that
$x\in[(A\cap B)-(A\cap C)]$ (rhs)

Is this right?

- Kamataat

2. Mar 25, 2005

### Data

Yes, but it is only half of the proof. You have shown

$$(A \cap B) - (A \cap C) \supseteq A \cap (B-C)$$

To finish the proof you must also go in "reverse" and show

$$(A \cap B) - (A \cap C) \subseteq A \cap (B-C)$$

3. Mar 25, 2005

### Kamataat

I see. My mistake was to assume the thing I was trying to prove... and I forgot, that $A=B$ iff $A\subseteq B\wedge B\subseteq A$.

I'll post the other half in the morning. Got to sleep now. Thanks!

- Kamataat

4. Mar 26, 2005

### Kamataat

Is this it?:

From $x\in[(A\cap B)-(A\cap C)]$ (rhs) we have
1.) $x\in(A\cap B)$
2.) $x\notin(A\cap C)$
From #1 it follows that
3.) $x\in A$ and $x\in B$
From #2 and #3 it follows that
4.) $x\in A$ and $x\notin C$ (not the other way around because #3 shows that $x\in A$)
Now we have
5.) $x\in A$ and $x\in B$ and $x\notin C$
From #5 it follows that
6.) $x\in A$ and $x\in(B-C)$ from which we get
7.) $x\in[A\cap(B-C)]$ (lhs)

The one question I have is about #5-->#6. As far as I understand it, #5 could also result in "$x\in B$ and $x\in(A-C)$", or could it not because of steps #3 and #4?

- Kamataat