Teacher $D$ thought of two positive integers $x, y (y \geq x \geq 1)$ in his mind, then he told $A$ the value of $x+y$, and told $B$ the value of $x \times y$. Next, $A$ and $B$ had the following conversation ($A$ and $B$ both know that $y \geq x \geq 1$):
$B$ said: I don't know the value of $x+y$.
$A$ said: I'll give you a hint, the value of $x+y$ does not exceed 20. Once you can determine the value of $x+y$ through this hint, then I will also know the value of $x \times y$.
$B$ said: I know the value of $x+y$ now.
Find: the values of $x$ and $y$.
Logical Reasoning
(1) From “$B$ does not know the value of $x+y$,” we can infer

The product $x \times y$ can be factored into the product of two positive integers in more than one way.
That is, $x \times y$ is neither 1 nor a prime number; $x \times y$ is a composite number.
(2) From “$B$ knows the value of $x+y$ given the hint $x+y \leq 20$,” we can infer

The composite number $x \times y$ can be factored into the product of two positive integers whose sum is no greater than 20 in a unique way.
$$
\begin{array}{l}
21=3 \times 7,22=2 \times 11,25=5 \times 5,26=2 \times 13,27=3 \times 9,33=3 \times 11,34=2 \times 17,35=5 \times 7, \\
39=3 \times 13,44=4 \times 11,49=7 \times 7,50=5 \times 10,51=3 \times 17,52=4 \times 13,54=6 \times 9,55=5 \times 11, \\
63=7 \times 9,65=5 \times 13,66=6 \times 11,75=5 \times 15,77=7 \times 11,80=8 \times 10,81=9 \times 9,88=8 \times 11
\end{array}
$$
(3) Finally, from “$A$ also knows the value of $x \times y$ after $B$ knows it,” we can infer

In the cases listed in (2), the sum of the two factors is unique.
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline $21=3 \times 7$ & $22=2 \times 11$ & $25=5 \times 5$ & $26=2 \times 13$ & $27=3 \times 9$ & $33=3 \times 11$ & $34=2 \times 17$ & $35=5 \times 7$ \\
\hline $3+7=10$ & $2+11=13$ & $5+5=10$ & $2+13=15$ & $3+9=12$ & $3+11=14$ & $2+17=19$ & $5+7=12$ \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline $39=3 \times 13$ & $44=4 \times 11$ & $49=7 \times 7$ & $50=5 \times 10$ & $51=3 \times 17$ & $52=4 \times 13$ & $54=6 \times 9$ & $55=5 \times 11$ \\
\hline $3+13=16$ & $4+11=15$ & $7+7=14$ & $5+10=15$ & $3+17=20$ & $4+13=17$ & $6+9=15$ & $5+11=16$ \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline $63=7 \times 9$ & $65=5 \times 13$ & $66=6 \times 11$ & $75=5 \times 15$ & $77=7 \times 11$ & $80=8 \times 10$ & $81=9 \times 9$ & $88=8 \times 11$ \\
\hline $7+9=16$ & $5+13=18$ & $6+11=17$ & $5+15=20$ & $7+11=18$ & $8+10=18$ & $9+9=18$ & $8+11=19$ \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline $90=9 \times 10$ & $91=7 \times 13$ & $96=8 \times 12$ & $99=9 \times 11$ & $100=10 \times 10$ & & & \\
\hline $9+10=19$ & $7+13=20$ & $8+12=20$ & $9+11=20$ & $10+10=20$ & & & \\
\hline
\end{tabular}

Among these, only “sum $=13$” appears once.
Therefore, $x=2, y=11$