Example 4 Let sequence $A$ be $1,3,2,5,4,3,5,7,6,11$. Property $P_{1}$ is that the element is divisible by 3, property $P_{2}$ is that the element leaves a remainder of 1 when divided by 3, and property $P_{3}$ is that the element is divisible by 2. Try to verify whether Theorem 7 holds.
$A(1)$ is $3,3,6; A(2)$ is $1,4,7; A(3)$ is $2,4,6. A(1,2)$ has no elements; $A(1,3)$ is $6; A(2,3)$ is $4. A(1,2,3)$ has no elements.$
$$B^{(0)}=B(0) \text { is } 5,5,11; B(1) \text { is } 3,3; B(2) \text { is } 1,7; B(3) \text { is } 2 \text {. }$$
$B(1,2)$ has no elements; $B(1,3)$ is $6; B(2,3)$ is $4. B(1,2,3)$ has no elements.$
$$\begin{aligned}
3 & =B^{(0)}=B(0)=A^{(0)}-A^{(1)}+A^{(2)}-A^{(3)} \\
& =10-(3+3+3)+(0+1+1)-0=3
\end{aligned}$$

This is consistent with the conclusion of Theorem 7. Similarly, the conclusion of Theorem 9 can be verified (left to the reader).