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

I have to resit a calculus exam and for some reason set proofs were never my best friend...

Anyway, on a practice exam I encountered the following proof:

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

When I draw a Venn-diagram it seems quite obvious but I couldn't manage to write the proof down properly.

If someone could help me, that'd be great!

Solution 1:

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 $B \subseteq A \implies A=B$.

Solution 2:

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.