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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!

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  • $\begingroup$ Usually $\oplus$ is direct sum - what you understand under it? $\endgroup$
    – zkutch
    Jul 16, 2020 at 18:07
  • $\begingroup$ @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. $\endgroup$ Jul 16, 2020 at 18:09
  • $\begingroup$ 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. $\endgroup$
    – zkutch
    Jul 16, 2020 at 18:24

1 Answer 1

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

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