F6 (22-6, Finland) The function $f(x, y)$ satisfies the following conditions 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) and (1), we get $f(1,0)=f(0,1)=2$.
In (3), let $x=0, y=n-1$, and use (1), we have
$$
\begin{aligned}
f(1, n) & =f(0, f(1, n-1)) \\
& =f(1, n-1)+1 \\
& =n+f(1,0) \\
& =n+2,
\end{aligned}
$$

From (3) and (1), we get
$$
\begin{aligned}
f(2, n) & =f(1, f(2, n-1)) \\
& =f(2, n-1)+2 \\
& =2 n+f(2,0) .
\end{aligned}
$$

Also,
$$
f(2,0)=f(1,1)=1+2=3 \text {, }
$$

Therefore,
$$
f(2, n)=2 n+3 \text {. }
$$

From (3) and (2), we get
$$
\begin{aligned}
f(3, n)+3 & =f(2, f(3, n-1))+3 \\
& =2 f(3, n-1)+6 \\
& =2(f(3, n-1)+3) \\
& =\cdots=2^{n+3}
\end{aligned}
$$

Therefore,
$$
f(3, n)=2^{n+3}-3 \text {. }
$$

From (3) and (3), we get
$$
\begin{aligned}
f(4, n)+3 & =f(3, f(4, n-1))+3 \\
& =2^{f(4, n-1)+3}=\cdots \\
& =2^{2.2^{2(4,0)+3}} \text { (a total of } n \text { 2's), }
\end{aligned}
$$

Since
$$
\begin{array}{c}
f(4,0)+3=f(3,1)+3=2^{4}, \\
f(4, n)=-3+2^{2.2} \text { (a total of } n+3 \text { 2's), }
\end{array}
$$

Therefore,
$$
f(4,1981)=-3+2^{2 \cdot 2}(1984 \text { 2's) }.
$$