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

Question

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!

Answer 1

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.: $$[x \in A \lor x \in B] \Leftrightarrow [(x \in A \setminus B) \oplus (x \in B \setminus A) \oplus (x \in A \cap B)]$$