6-11 The 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 $f(4,1981)$.
Let $x=0$, from (2), (1) we get
$$
f(1,0)=f(0,1)=2 \text {. }
$$

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

i.e., $f(1, y+1)=2+(y+1)$
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 $f(2, y+1)=2(y+1)+f(1,1)=2(y+1)+2+1$,
i.e., $f(2, y+1)=2(y+1)+3$
Using (3), (5) to deduce
$$
\begin{aligned}
f(3, y+1) & =f[2, f(3, y)]=2 f(3, y)+3 \\
& =2 f[2, f(3, y-1)]+3 \\
& =2^{2} f(3, y-1)+2 \times 3+3 \\
& =2^{2} f[2, f(3, y-2)]+2 \times 3+3 \\
& =2^{3} f(3, y-2)+2^{2} \times 3+2 \times 3+3 \\
& =\cdots \\
& =2^{y+1} f(3,0)+2^{y} \times 3+\cdots+2^{2} \times 3+2 \times 3+3 \\
& =2^{y+1} f(2,1)+3\left(2^{y}+\cdots+2^{2}+2+1\right) \\
& =2^{y+1} \cdot 5+3 \cdot \frac{2^{y+1}-1}{2-1} \\
& =5 \cdot 2^{y+1}+3 \cdot 2^{y+1}-3 \\
& =8 \cdot 2^{y+1}-3,
\end{aligned}
$$

i.e.,
$$
f(3, y+1)=2^{3+(y+1)}-3 \text {. }
$$

Using (3), (6) to deduce
$$
f(4, y+1)=f[3, f(4, y)]=2^{3+f(4, y)}-3
$$

i.e.,
$$
\begin{aligned}
f(4, y+1)+3 & =2^{3+f(4, y)} \\
& =2^{2^{3+f(4, y-1)}} \\
& =\cdots
\end{aligned}
$$
$$
=2^{2^{2}:^{\cdot 2^{3+f(4.0)}}} \quad y+1 \text { times. }
$$

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

Therefore
$$
\begin{array}{c}
\left.f(4, y+1)+3=2^{2^{22^{2}}}\right\} y+4 \text { times, } \\
\left.f(4,1981)+3=2^{2^{2 \cdot}}\right\} 1984 \text { times, }
\end{array}
$$

Hence $\left.f(4,1981)=-3+2^{2^{2 \cdot}}\right\} 1984$ times.