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

A: First of all, two sets $A$ and $B$ are equal if and only if $A\subset B$ and $B\subset A$. 
So, just use the fact that $x\in B\cup C$ if and only if $x\in B$ or $x\in C$ (or both).
Also, just use the fact that $x\in B\cap C$ if and only if $x\in B$ and $x\in C$.
Then, show that those two inclusions hold.
A: You have to show that a point is in the left side if and only if it is in the right side.
Take $x\in A\cap(B\cup C)$. It means that 
$$ x\in A\wedge(x\in B\vee x\in C)$$
To show that it is contained in the right side, assume that $x$ is not in $A\cap B$, so it is not in both $A$ and $B$. But we know that it is in $A$, so what does this imply for $B$?. Can you show that in this case $x$ must be in $A\cap C$?
For the reverse inclusion you can use the equivalences
$$X\cup Y\subseteq Z \text{ if and only if }X\subseteq Z\text{ and }Y\subseteq Z$$
as well as
$$X\subseteq Y\cap Z \text{ if and only if } X\subseteq Y\text{ and }X\subseteq Z$$
for aarbitrary sets $X,Y,Z$. It would be a good exercise to try the reverse inclusion only using these equivalences and the properties $X\cap Y\subseteq Y$ and $X\subseteq X\cup Y$ and without considering elements.
A: 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 $B \subseteq  A \implies A=B$.
