1. ## Proof By Contradiction

Hello,

(A-C)∩(C-B) = 0.

Do I start off by....

suppose: ~[(A-C)∩(C-B) = 0.] ?

Also, does this (0 ∩ B') equal the empty set?

Thanks...

PS. ' indicates the complement.

2. Why not start by the mere definition? Suppose $x\in\left(A-C\right)\cap\left(C-B\right)\Longrightarrow x\in A-C\,\,\,and\,\,\,also\,\,\,x\in C-B$ ...from here the way to get a contradiction is straightforward and thus no such element can exist...

Tonio

3. Thats exactly what I did, i'm not sure if I did it correctly though.

This is what I got :

(A ∩ 0) ∩ B'

(0 ∩ B')

0

is that correct?

4. Originally Posted by l flipboi l
Thats exactly what I did, i'm not sure if I did it correctly though.

This is what I got :

(A ∩ 0) ∩ B'

(0 ∩ B')

0

is that correct?

I've no idea what you had in mind, even assuming that $0=\emptyset$...how did you come up with the above? Besides this, $X\cap \emptyset =\emptyset$ for any set X, so from

the first line you get the empty set...

Tonio

5. Originally Posted by tonio
I've no idea what you had in mind, even assuming that $0=\emptyset$...how did you come up with the above? Besides this, $X\cap \emptyset =\emptyset$ for any set X, so from

the first line you get the empty set...

Tonio
I excluded alot of the work, but I used the commutative and associative property to get c and c' to cancel out.

which brought me to my answer

6. Originally Posted by l flipboi l
I excluded alot of the work, but I used the commutative and associative property to get c and c' to cancel out.

which brought me to my answer

Good, but then why did you say you "did exactly the same" as I did?!!

Tonio

7. Thanks for the help! I meant, I set it up like how you did.