Indicator function properites

I just wanted to make sure my proofs are correct.
Given the indicator function.

$1_A = \left\{ \begin{array}{rcl} 1 & \mbox{for} & x \in A \\ 0 & \mbox{for} & x \notin A \end{array}\right.$

show that:
i) $1_{A \cap B} = 1_A \cdot 1_B$ :

Proof:
$1_{A \cap B} = \left\{ \begin{array}{rcl} 1 & \mbox{for} & x \in A\cap B \\ 0 & \mbox{for} & x \notin A\cap B \end{array}\right.$ which implies that $1_{A \cap B} = 1_A \cdot \left\{ \begin{array}{rcl} 1 & \mbox{for} & x \in B \\ 0 & \mbox{for} & x \notin B \end{array}\right. = 1_A \cdot 1_B$

ii) $1_{A \cup B} = 1_A + 1_B - 1_A \cdot 1_B$

Proof:
$1_{A \cup B} = \left\{ \begin{array}{rcl} 1 & \mbox{for} & x \in A\cup B \\ 0 & \mbox{for} & x \notin A\cup B \end{array}\right.$ now $A\cup B$ can be written as two disjoint sets so we have

$A\cup B \Rightarrow A\cup(B-A\cap B)$ so we have :

$1_{A \cup B} = \left\{ \begin{array}{rcl} 1 & \mbox{for} & x \in A\cup(B-A\cap B) \\ 0 & \mbox{for} & x \notin A\cup(B-A\cap B) \end{array}\right. = 1_A +\left\{ \begin{array}{rcl} 1 & \mbox{for} & x \in B -A\cap B \\ 0 & \mbox{for} & x \notin B -A\cap B \end{array}\right. = 1_A+1_B-1_A\cdot 1_B$ where the terms are expanded out.

iii) $1_{A \cap B} =\min(1_A,1_B)$

I have no idea