# How to prove this distributive law for sets : $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

Now, when I first proved that $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$, I did it as follows : $$\text{Let } x \in A \cap (B \cup C)$$ $$\therefore~x \in A \wedge x \in (B \cup C)$$ Now, this means that $$x$$ is a member of $$A$$ and either $$B$$ or $$C$$, which would imply that either it is a member of $$A$$ and $$B$$ or a member of $$A$$ and $$C$$. Using this : $$x \in (A \cap B) \cup (A \cap C) \implies A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$$ Using a similar method, we can prove that $$(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$$ and using the fact that if $$X \subseteq Y$$ and $$Y \subseteq X$$, it implies that $$X=Y$$, we get that : $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$.

But recently, I realised that I completely missed out the fact that if $$y \in (M \cup N)$$, it implies that either $$y \in A$$ or $$y \in B$$ or $$y \in (A \cap B)$$. This fact gives us the following definition of $$A \cup B$$: $$\forall~x \in (A \cup B) \implies x \in A \lor x \in B$$ $$\implies x \in (A-B) \oplus x \in (B-A) \oplus x \in (A \cap B)$$ Now, how do I make the proof compatible for this definition?

Thanks!

• Usually $\oplus$ is direct sum - what you understand under it? Jul 16, 2020 at 18:07
• @zkutch The XOR (exclusive or) operator. I want to say that if $x \in (A \cup B)$, then either $x \in (A-B)$ or $x \in (B-A)$ or $x \in (A \cap B)$ but no two of these statements can be true at the same time. Jul 16, 2020 at 18:09
• May be more easy will be to show $A \cup B= (A \setminus B) \cup (B \setminus C) \cup (A \cap B)$ and then show, that these 3 sets are disjoint. Jul 16, 2020 at 18:24

Firstly we can show, that if $$C, D$$ are disjoint, i.e. $$C \cap D = \emptyset$$, then $$(x \in C \lor x \in D) \Leftrightarrow (x \in C \oplus x \in D)$$
Then using it in respect to $$A \cup B= (A \setminus B) \cup (B \setminus C) \cup (A \cap B)$$ gives desired equivalence i.e.: $$[x \in A \lor x \in B] \Leftrightarrow [(x \in A \setminus B) \oplus (x \in B \setminus A) \oplus (x \in A \cap B)]$$