Proof of $A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$

If $x\in A\cap(B\cup C)$, then $x\in A$ and $x\in B\cup C$.

$x\in B\cup C\implies (x\in B$ or $x\in C)$.

So, $x\in A\cap(B\cup C)\implies x\in (A\cap B)$ or $ x\in (A\cap C)$

$\implies x\in (A\cap B)\cup(A\cap C)$

$\implies A\cap(B\cup C)\subseteq (A\cap B)\cup(A\cap C)$.

Similarly,

if $y\in (A\cap B)\cup(A\cap C),$

$\implies y\in (A\cap B)$ or $y\in (A\cap C),$

$\implies y\in A$ and $y\in (B$ or $C)$

$\implies y\in A$ and $y\in (B \cup C)$

$\implies y\in A\cap (B \cup C)$.

Now, $A \subseteq B$ and $B \subseteq A \implies A=B$.

This can be done by algebra or by a truth table. In some cases there might be reasons why it would be necessary to use algebra. But not in all. Here's a truth table: $$ \begin{array}{|c|c|c|c|c|} \hline x\in A & x\in B & x\in C & x\in A\cap(B\cup C) & x\in (A\cap B)\cup(A\cap C) \\ \hline T & T & T & ? & ? \\ T & T & f & ? & ? \\ T & f & T & ? & ? \\ f & T & T & ? & ? \\ T & f & f & ? & ? \\ f & T & f & ? & ? \\ f & f & T & ? & ? \\ f & f & f & ? & ? \\ \hline \end{array} $$ You need (1) to make sure all eight possible rows are there, (2) to fill in the blanks, and (3) to check carefully that the last two columns are identical.

Hint: proofs by characteristic function :

for any $A \subset X$ define $1_A : X \to \{ 0,1\}$ by

$$1_A(x) = \left\{ \begin{eqnarray} \begin{split} 1 & \mbox{if } x \in A \\ 0 & \mbox{if } x \notin A \\ \end{split} \end{eqnarray}\right.$$ then we have

• $A= B \Leftrightarrow 1_A = 1_B$
• $1_{A \cap B}=1_A \cdot 1_B$
• $1_{A \cup B}=1_A+1_B-1_A \cdot 1_B$

here we want prove$$ A\cap(B\cup C) = (A\cap B)\cup(A\cap C) \iff 1_{A\cap(B\cup C)} = 1_{(A\cap B)\cup(A\cap C)}$$ $\color{red}{proof:}$

1.$$ \large{1_{A\cap(B\cup C)}= 1_{A} 1_{B\cup C}=(1_{A} (1_B+1_C-1_B \cdot 1_C))=1_{A} 1_B+1_{A} 1_C-1_{A} \cdot1_B \cdot 1_C}$$ 2.$$\large{1_{(A\cap B)\cup(A\cap C)}=1_{(A\cap B)}+1_{(A\cap C)}-1_{(A\cap B)}\cdot1_{(A\cap C)}=1_{A} 1_B+1_{A} 1_C-1_{A} \cdot1_B \cdot 1_C}$$