please check this

[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

[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)

[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

[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