( A ∩ B ) \ ( A ∩ C ) = ( A ∩ B ) \ C how to prove?

Question

I have to prove that. While I know this is true by thinking about it I'm having a lot of trouble actually writing the proof

Answer 1

The easiest and most intuitive way is probably to draw a diagram of the 3 sets. Showing the intersection areas between each pair of the sets and the intersection of all three sets. Once you have the diagram and it is clear why this formula is true, it should be easy to write a full proof from there.

Answer 2

If $x \in A \cap B \setminus A \cap C$, then $x \in A$ and $x \in B$, but $x \notin A \cap C$, whence $x \notin C$.
Therefore, $x \in (A \cap B) \setminus C$.

Answer 3

Before what's written below I'd like to make sure to explain the following: The intersection operator is "equivalent" to the logical "and", while the difference operator is likewise "equivalent" to ""and not"

$$\cap \sim \land ;\text{ } \text{ }\text{ }\text{ } \backslash \sim (\land \lnot) $$

I hope it's easy to see why

Whenever the proof is too hard (or you can't warp your head around it), bruteforce it

What I mean by using brute force when dealing with sets is writing everything down in logic

let's try to analyze $x \in (A \cap B )\backslash (A \cap C)$

This roughly translates to "$x$ is in A AND $x$ is in B, but $x$ is not in A AND C at the same time"

logically this would be $$(x \in A \land x \in B) \land \lnot(x \in A \land x \in C)$$

Now, we can transform that negation into $$(x \in A \land x \in B) \land (x \notin A \lor x \notin C)$$

Distribute around the "or" to get two cases

Case 1) $ x \in A \land x \in B \land x \notin A$

Case 2) $x \in A \land x \in B \land x \notin C$

You can quickly check that Case 1 is a contradiction, so we can only have the case 2. So the set on the left can be written conditionally-logically as $\{x \mid x \in A \land x \in B \land x \notin C\}$

Now let's look at the set on the right. $x \in (A \cap B) \backslash C$ roughly translates to "$x$ is in A AND B, but not in C"

Written logically this is $$(x \in A \land x \in B ) \land \lnot (x \in C)$$

which is just $x \in A \land x \in B \land x \notin C$ just like the set on the right. So they're equal

If you find it hard to find a way to prove a "logical identity", you can go "Berserk Force" on it, taking an extra step to check that the truth tables for both expressions are the same. This is like extra-brute force