$6 \cdot 11$ function $f(x, y)$ satisfies for all non-negative integers $x, y$:
(1) $f(0, y)=y+1$;
(2) $f(x+1,0)=f(x, 1)$;
(3) $f(x+1, y+1)=f[x, f(x+1, y)]$;

Determine $f(4,1981)$.
[Solution] Let $x=0$, from (2), (1) we get
$$f(1,0)=f(0,1)=2 \text {. }$$

From (3), (1) we get
$$\begin{aligned}
f(1, y+1) & =f[0, f(1, y)] \\
& =f(1, y)+1 \\
& =f(1, y-1)+2=\cdots= \\
& =f(1,0)+(y+1)
\end{aligned}$$

That is $\square$
$$f(1, y+1)=2+(y+1)$$

Using (3), (4) to deduce
$$\begin{aligned}
& f(2, y+1) \\
= & f[1, f(2, y)]=2+f(2, y)=2+f[1, f(2, y-1)] \\
= & 2 \times 2+f(2, y-1)=\cdots=2 \times(y+1)+f(2,0)
\end{aligned}$$

From (2) we get $f(2, y+1)=2(y+1)+f(1,1)=2(y+1)+2+1$,
That is $f(2, y+1)=2(y+1)+3$
Using (3), (5) to deduce

That is $\square$
$$\begin{aligned}
f(3, y+1) & =f[2, f(3, y)]=2 f(3, y)+3 \\
& =2 f[2, f(3, y-1)]+3 \\
& =2^{2} f(3, y-1)+2 \times 3+3 \\
& =2^{2} f[2, f(3, y-2)]+2 \times 3+3 \\
& =2^{3} f(3, y-2)+2^{2} \times 3+2 \times 3+3 \\
& =\cdots \\
& =2^{y+1} f(3,0)+2^{y} \times 3+\cdots+2^{2} \times 3+2 \times 3+3 \\
& =2^{y+1} f(2,1)+3\left(2^{y}+\cdots+2^{2}+2+1\right) \\
& =2^{y+1} \cdot 5+3 \cdot \frac{2^{y+1}-1}{2-1} \\
& =5 \cdot 2^{y+1}+3 \cdot 2^{y+1}-3 \\
& =8 \cdot 2^{y+1}-3
\end{aligned}$$

That is

Because
$$\begin{array}{l}
f(4,0)=f(3,1)=2^{3+1}-3 \\
3+f(4,0)=2^{4}=2^{2^{2}}
\end{array}$$

Therefore
$$\begin{array}{c}
\left.f(4, y+1)+3=2^{2^{2} \cdot 2^{2}}\right\} y+4 \text {, } \\
\left.f(4,1981)+3=2^{2^{2 \cdot} \cdot 2^{2}}\right\} 1984 \text {, } \\
\text { hence } f(4,1981)=-3+2^{2^{2} \cdot \cdot^{2}} \mid 1984 \text {, }
\end{array}$$