Check if for all sets $A, B, C$ the equality $(A \setminus B) \cap C = (A \cap B) \setminus (B \cap C)$ holds

I tried to solve this, I tested a variety of examples and they all made me think that this equality is valid. Therefore, I attempted to manipulate the definition of these operations but these sets simply don't want to be the equal.
From the Axiom of Extensionality, $$x \in (A \setminus B) \cap C \iff (x\in A \land x \notin B) \land x\in C \iff (x\in A \land x\notin B) \land (x \notin B \land x\in C) \Rightarrow x \in (A\cap C) \setminus (B \cup C) $$ But this is impossible to show this "the other way around" and so this equality seems not to be true. However, I cannot find a counterexample. Any hints?

  • $\begingroup$ Have you tried a Venn diagram? $\endgroup$ – Stephen Meskin Nov 23 '17 at 21:11
  • $\begingroup$ Yeah, a lot of them in fact $\endgroup$ – Aemilius Nov 23 '17 at 21:11
  • 1
    $\begingroup$ You only need two. One for the LHS and one for the RHS. Label them identically. I think you will be surprised. $\endgroup$ – Stephen Meskin Nov 23 '17 at 21:14
  • $\begingroup$ There is an error in the last implication of your work. $\endgroup$ – Stephen Meskin Nov 23 '17 at 21:17
  • $\begingroup$ @Aemilius I have re-corrected my answer; when I originally added to it, I accidentally edited the wrong equation - my answer is now correct - sorry for any confusion caused! $\endgroup$ – John Don Nov 23 '17 at 21:29

The assertion is in fact false.

Take, for example, $A = \{0, 1\}, B = \{1\}, C = \{0\}$.

Then, $(A\setminus B)\cap C = \{0\}$, but $(A \cap B)\setminus (B \cap C) = \{1\}$

However, the following assertion is true:

$$(A\setminus B)\cap C = (A \cap C)\setminus (B \cap C) $$

You should have more luck proving this (just by elementary methods).

[Intuitively, this is saying that if you find "A minus B", then "restrict this" to C, this is the same as starting by restricting everything to C, and then finding the set 'difference'.]


First I think your second equivalence relation is wrong. Additionally this whole claim is wrong. There could be an element in $A \cap C$ that is element of the set on the left side, but it does not have to be in the set on the right side, because it does not have to be an element of $B$.


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