Example 5 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, Property $P_{3}$ is that the element is divisible by 2, Property $P_{4}$ is that the element is divisible by 3. Try to verify whether Theorem 7 holds.
Solve $A(1)$ is $3,3,6 ; A(2)$ is $1,4,7 ; A(3)$ is $2,4,6 ; A(4)$ is $3,3,6. A(1,2)$ has no elements; $A(1,3)$ is $6 ; A(1,4)$ is $3,3,6 ; A(2,3)$ is $4 ; A(2,4)$ has no elements; $A(3,4)$ is $6. A(1,2,3)$ has no elements; $A(1,2,4)$ has no elements; $A(1,3,4)$ is $6 ; A(2,3,4)$ has no elements; $A(1,2,3,4)$ has no elements. \square$
$B^{(0)}=B(0)$ is $5,5,11 ; B(1)$ has no elements (because property $P_{1}$ is the same as property $P_{4}$); $B(2)$ is $1,7 ; B(3)$ is $2 ; B(4)$ has no elements. $B(1,2)$ has no elements; $B(1,3)$ has no elements; $B(1,4)$ is $3,3 ; B(2,3)$ is $4 ; B(2,4)$ has no elements; $B(3,4)$ has no elements. $B(1,2,3)$ has no elements; $B(1,2,4)$ has no elements; $B(1,3,4)$ is $6 ; B(2,3,4)$ has no elements. $B(1,2,3,4)$ has no elements.
$$\begin{aligned}
3= & B^{(0)}=B(0)=A^{(0)}-A^{(1)}+A^{(2)}-A^{(3)}+A^{(4)} \\
= & 10-(3+3+3+3)+(0+1+3+1+0+1) \\
& -(0+0+1+0)+0=3
\end{aligned}$$

This is consistent with the conclusion of Theorem 7. Similarly, the conclusion of Lemma 8 can be verified (left to the reader).