Test $\mathbf{E}$ 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 $f(4,1981)$.
Solve: From $f(1, n)=f(0, f(1, n-1))=f(1, n-1)+1$ and $f(1,0)=f(0,1)=2$, we get
$$
f(1, n)=n+f(1,0)=n+2 .
$$

Also, from
$$
\begin{aligned}
f(2, n) & =f(1, f(2, n-1)) \\
& =f(2, n-1)+2 \\
& =2 n+f(2,0), \\
f(2,0) & =f(1,1)=3,
\end{aligned}
$$

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

Next, we have
$$
\begin{aligned}
f(3, n) & =f(2, f(3, n-1)) \\
& =2 f(3, n-1)+3 \\
& =2[f(3, n-1)+3]-3,
\end{aligned}
$$

thus
$$
f(3, n)+3=2^{n}(f(3,0)+3) \text {. }
$$

Since
$$
f(3,0)=f(2,1)=5,
$$

we get
$$
f(3, n)=2^{n+3}-3 .
$$

Finally, we calculate $f(4, n)$, from
$$
f(4, n)=f(3, f(4, n-1))=2^{f(4, n-1)+3}-3,
$$

which means
$$
\begin{array}{c}
f(4, n)+3=2^{f(4, n-1)+3} . \\
t_{n}=f(4, n)+3, \varphi(x)=2^{x},
\end{array}
$$

Let
then we have $t_{n}=\varphi\left(t_{n-1}\right)$. Therefore, $t_{n}=\varphi^{[n]}\left(t_{0}\right)=\varphi\left(\varphi \cdots\left(\varphi\left(t_{0}\right)\right)\right)$ is an $n$-fold iteration.
Since
$$
t_{0}=f(4,0)+3=f(3,1)+3=2^{4} \text {, }
$$

we have
$$
f(4, n)=\varphi^{[n]}(16)-3 .
$$

Thus,
$$
f(4,1981)=\varphi^{[1981]}(16)-3=2^{2^{\cdot 2}}-3 \text {, }
$$

where the exponent is iterated 1984 times.