# Relations between sets proofs

1. Mar 1, 2005

### mathrocks

I'm trying to prove the following by contradiction: [(A^B)-(B^C)]-(A^C)'=0. A, B, C are sets. All I know is in order to prove by contradiction you simply set the above not equal to zero. But I don't know where to go from there.

"^" means the intersection symbol.

2. Mar 1, 2005

### Muzza

Suppose

$$(A \cap B - B \cap C) - (A \cap C)' \neq \emptyset$$.

Then there is some element $x \in ((A \cap B - B \cap C) - (A \cap C)')$. But then

$$x \in A \cap B - B \cap C$$
$$x \notin (A \cap C)'$$

<=>

$$x \in A \cap B - B \cap C$$
$$x \in A \cap C$$

<=>

$$x \in A \cap B$$
$$x \notin B \cap C$$
$$x \in A \cap C$$

<=>

$$x \in A$$
$$x \in B$$
$$x \notin B \cap C$$
$$x \in A$$
$$x \in C$$

<=>

$$x \in A$$
$$x \in B$$
$$x \in C$$
$$x \notin B \cap C$$.

3. Mar 1, 2005

### Galileo

Suppose it is not zero. Then there is some element x in the set on the left side.
From the last term we infer that x is not an element of $(A \cap C)'$, (does that ' mean the complement?) so:

$$x \in A, \quad \mbox{and } x \in C$$,

take it from there.

Last edited: Mar 1, 2005
4. Mar 30, 2005

### kastarov

Ok, let's prove it by contradiction:
let's suppose that
(A^B-B^C)-(A^C)'=!0, where =! means "different to". Then, there is a element namely x such that x E(belongs to) (A^B-B^C)-(A^C)', then, xE(A^B-B^C) and xE!( not belongs to)(A^C)', then, xEA^B and xE!B^C and xEA^C, then
xEA, xEB, xEC and( xE!B or xE!C). Therefore, we have a contradiction.