# For all sets $A$ $B$ and $C$, if $A - B = C$ then $A - C = B$ proof

From looking at the diagram I think the statement is true . Proof: Assume $A$, $B$ and $C$ are sets. Assume $A - B = C$. Want to prove $A - B = C$

I am confused how I am supposed to go about trying to prove this . So if I want to prove $A-B$ is $C$ then $A$ and $\text{not } B$ is $C$. How do I prove it ?

• Is this assuming $C \subset A$ and $B \subset A$? Commented Jul 26, 2017 at 19:25
• Not necessarily the question did not say C⊂A and B⊂A Commented Jul 26, 2017 at 19:29
• You should be confused, since it is simply not true. Commented Jul 26, 2017 at 19:34
• Your statement is not correct ! Commented Jul 26, 2017 at 19:39

Take $A=C=\{1\}$ and $B=\{2\}$. Then $A\setminus B=A=C$, but $A\setminus C=\emptyset\neq B$

• thanks got it . so the statement was false ! Commented Jul 26, 2017 at 19:52
• @MahirShahriar Yes, as a general implication. Commented Jul 26, 2017 at 19:53

First you have: $$C=A - B = A \cap B^c$$

Then: $$A-C=A - (A-B) = A \cap (A \cap B^c)^c= A \cap (A^c \cup B)=A \cap B$$

So if you want the equality must happen that:

$$A-C=B \iff A \cap B =B \iff B \subset A$$

• How do you reconcile your "proof" with @Hellen 's correct counterexample? Commented Jul 26, 2017 at 19:36
• @EthanBolker I'm sorry I'm just showing what must happen to the equality be true. Commented Jul 26, 2017 at 19:43
• The edit does improve the answer. Commented Jul 26, 2017 at 19:45
• @EthanBolker The example does not have $B \subseteq A$ so it is a good counterexample to the general statement. Commented Jul 26, 2017 at 21:41

The Venn diagram clearly shows that if $C= A-B$ then $A -C = A \cap B$. This is not equal to $B$ unless when $B \subseteq A$.