I have to show that
$1\cdot (A \cup B) \cap C = (A\cap C) \cup (B\cap C)$
$2\cdot (A\cap B) \cup C = (A \cup C) \cap (B \cup C)$
How can I show that the left hand side is equal to right hand side ? I feel like I have to use associativity or distributivity but I'm not sure how to make proof exactly.
Thank you.
On
$$ (A \cap B) \cup C = ( A \cup C) \cap (B \cup C) $$
Suppose that, $$ x \in (A \cap B) \cup C $$
Then, $$ x \in A \cap B $$ or $$ x \in C $$ // I assume that the reason why we have a difference in this step as ( AND - OR) because there exist a Union rather than an Intersection.
Therefore, $$ x \in A$$
And we can do the almost same(copy the steps) for $$ x \in B$$
Now we have to show that $$ (A \cap B) \cup C = ( A \cup C) \cap (B \cup C) $$
Suppose $$ x \in (A \cup C) \cap (B \cup C)$$
Then,$$ x \in (A \cup C) $$ and $$ x \in (B \cup C)$$
So, $$ x \in A $$ or $$ x \in C $$
Thus, $$ x \in (A \cap B)$$
Did I got right ?
I'm going to just prove $1)$, and hopefully that will give you an idea for $2)$.
We show that $ (A \cup B) \cap C \subseteq(A\cap C) \cup (B\cap C)$ and vice versa. Suppose $x\in (A \cup B) \cap C $. Then $x\in C$ and $x\in A\cup B$. WLOG, suppose $x\in A$. Then clearly $x\in A\cap C$, and we're done. The same follows for $x\in B$.
We now show $(A \cup B) \cap C \supseteq (A\cap C) \cup (B\cap C)$. Suppose $x\in(A\cap C) \cup (B\cap C)$. Then $x\in A\cap C$ or $x\in B\cap C$. WLOG, suppose the former holds. Then $x\in A$ and $x\in C$. Thus, $x\in A\cup B$, and we're done. The case where the latter holds is almost exactly the same.
Can you follow my logic to show 2)?