So I am learning about proving intersection and union statements of sets, but the problem is I am never confident about my proofs, I never know when I am right. So if you could check my attempt, and maybe offer some help that would be great:
Prove: $A\cup \!\, (B\cap \!\ C)=(A\cup \!\, B)\cap \!\ (A\cup \!\, C)$
Here's my attempt:
$A\cup \!\, (B\cap \!\ C)=A\cup \!\,${$x:x\in \!\, B \ \text{and}\ x\in \!\, C $}
$=${$x:x\in \!\, A \ \text{or}\ x\in \!\, C\ \text{and} \ x\in \!\, B $}
(I think I need a step in between here, right?)
$=(A\cup \!\, B)\cap \!\ (A\cup \!\, C)$
Best Answer
There is no magic, just work.
One way is as follows, basically case analysis:
First show that $A \cup ( B \cap C) \subset (A\cup B) \cap (A\cup C)$.
Suppose $x \in A \cup ( B \cap C)$. Then either (i) $x \in A$ or (ii) $x \in B \cap C$.
In the first case, we see that $x \in A \subset A\cup B$ and $x \in A \subset A\cup C$, and so $x \in (A\cup B) \cap (A\cup C)$.
In the second case, we have $x \in B \cap C$, so $x \in B$ and $x \in C$. Then $x \in B \subset A \cup B$ and $x \in C \subset A \cup C$, so $x \in (A\cup B) \cap (A\cup C)$.
Now show $(A\cup B) \cap (A\cup C) \subset A \cup ( B \cap C)$.
Suppose $x \in (A\cup B) \cap (A\cup C)$, then we must have $x \in A\cup B$ and $x \in A\cup C$. Now we need to make judicious choices. I will split into (i) $x \in A$ or (ii) $x \notin A$.
In the first case, we see that $x \in A \subset A \cup ( B \cap C)$.
In the second case, since $x \in A\cup B$ and $x \notin A$ we must have $x \in B$. Similarly, since $x \in A\cup C$ and $x \notin A$ we must have $x \in C$. Hence $x \in B \cap C$ and then $x \in B \cap C \subset A \cup ( B \cap C)$.