Proving $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
Prove the distributive property for sets:
$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
I'm not good with proofs but my understanding is that I have to prove 2 things:
(1) $A \cup (B \cap C) \subset (A \cup B) \cap (A \cap C)$
(2) $A \cap (B \cap C) \supset (A \cup B) \cap (A \cup C)$
This is what I have done so far:
Part (1)
If $x\in A$, then $x \in (A \cup B)$ and $x \in (A \cup C)$.
$\therefore x \in (A \cup B) \cap (A \cap C)$
If $x \in (B \cap C)$ then $x \in (A \cup B)$ and $x \in (A \cup C)$ because $x \in B$ and $x \in C$.
$\therefore x \in (A \cup B) \cap (A \cap C)$
$\therefore A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)$
Part (2)
Now we have to prove the reverse inequality: $(A \cup B) \cap (A \cap C)$. Then $x \in A \cup B$ and $x \in (A \cup C)$
If $x \in A$, then $x \in A \cup (B \cap C)$
This is where I am up to. I wanted to know whether my approach is correct and if I did part (1) correctly. I'm stuck on part (2) and don't know how to proceed. I'd appreciate any help.
Thank you!!
Solution 1:
You must first prove 2 cases:
(1) $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$
(2) $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$
Note that in mathematics we use the following symbols:
$\cap=$ AND = $\land$
$\cup=$ OR = $\lor$
Case 1: $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$
Let $x \in A \cap (B \cup C) \implies x \in A \land x \in (B \cup C)$
$\implies x \in A \land \{ x \in B \lor x \in C \}$
$\implies \{ x \in A \land x \in B \} \lor\{ x \in A \land x \in C \} $
$\implies x \in (A \cap B) \lor x \in (A \cap C)$
$\implies x \in (A \cap B) \cup (A \cap C)$
$\therefore x \in A \cap (B \cup C) \implies x \in (A \cap B) \cup (A \cap C)$
$\therefore A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$
Case 2: $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$
Let $x \in (A \cap B) \cup (A \cap C) \implies x \in (A \cap B) \lor x \in (A \cap C)$
$\implies \{x \in A \land x \in B \} \lor \{ x \in A \land x \in C \}$
$\implies x \in A \land \{ x \in B \lor x \in C\}$
$\implies x \in A \land \{B \cup C \}$
$\implies x \in A \cap (B \cup C)$
$\therefore x \in (A \cap B) \cup (A \cap C) \implies x \in A \cap (B \cup C)$
$\therefore (A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$
$\therefore A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
Solution 2:
Let $X = A \cap (B \cup C)$ and $Y = (A \cap B) \cup (A \cap C)$
To show that $X=Y$, we must show that:
- $X \subseteq Y$
- $Y \subseteq X$
Case 1: $X \subseteq Y$
If $x \in X$, then $x \in A$ and $x \in (B \cup C)$
The latter implies that $x$ is a member of at least one of $B$ or $C$.
We will proceed from here in 3 cases:
Case A: $x \in B$ and $x \notin C$
We know from above that $x \in A$.
If $x \in B$ and $x \in A$, then $x \in (A \cap B)$.
If $x \in (A \cap B)$, then $x \in (A \cap B) \cup (A \cap C)$.
$(A \cap B) \cup (A \cap C) = Y$, therefore $x \in Y$.
Case B: $x \notin B$ and $x \in C$
This is symmetric to Case A.
Case C: $x \in B$ and $x \in C$
This is simply an extension of Case A and Case B.
Therefore, if $x \in X$, then $x \in Y$, which implies $X \subseteq Y$.
Case 2: $Y \subseteq X$
If $y \in Y$, then $y$ is a member of at least one of $(A \cap B)$ or $(A \cap C)$.
Again, we have 3 cases:
Case A: $y \in (A \cap B)$ and $y \notin (A \cap C)$.
The former implies $y \in A$ and $y \in B$ by the definition of intersection.
If $y \in B$, then $y \in (B \cup C)$.
If $y \in A$ and $y \in (B \cup C)$, then $y \in (A \cap (B \cup C))$.
$(A \cap (B \cup C)) = X$, therefore $y \in X$.
Case B: $y \notin (A \cap B)$ and $y \in (A \cap C)$.
This is symmetric to Case A.
Case C: $y \in (A \cap B)$ and $y \in (A \cap C)$.
This is simply an extension of Case A and Case B.
Therefore, if $y \in Y$, then $y \in X$, which implies $Y \subseteq X$.
From Case 1 and Case 2, we have:
$X \subseteq Y$ and $Y \subseteq X$, therefore $X = Y$, which implies $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. This completes the proof.