4. 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 $f(4,1981)$.
(IMO -22 Problem)
4. From the given three conditions, we get $f(1,0)=f(0,1)=2, f(1,1)=f[0, f(1,0)]=f(0,2)=3, f(1,2)=f[0$, $f(1,1)]=f(0,3)=4$. This shows that: when $x=1, y=0,1,2$, we have $f(1, y)=y+2$. (1) Assuming that (1) holds when $y=k$, then when $y=k+1$, $f(1, k+1)=f[0, f(1, k)]=f(0, k+2)=(k+1)+1=(k+1)+$ 2. By the principle of induction, (1) is generally true.

From conditions (2), (3), and (1), we get $f(2,0)=f(1,1)=3, f(2,1)=f[1, f(2,0)]=f(1,3)=5, f(2,2)=$ $f[1, f(2,1)]=f(1,5)=7$. This shows that when $x=2, y=0,1,2$, we have $f(2, y)=2 y+3$. (2) Assuming that (2) holds when $y=$ $k$, then when $y=k+1$, $f(2, k+1)=f[1, f(2, k)]=f(1,2 k+3)=(2 k+3)+2=2$. $(k+1)+3$. Therefore, by the principle of induction, (2) is also generally true.

From conditions (2), (3), and (2), we get $f(3,0)=f(2,1)=5, f(3,1)=f[2, f(3,0)]=f(2,5)=13, f(3,2)=$ $f[2, f(3,1)]=f(2,13)=29$. Since $5=2^{3}-3,13=2^{4}-3,29=2^{5}-3$, we know that when $x=3, y=0,1,2$, we have $f(3, y)=2^{y+3}-3$. (3) Assuming that (3) holds when $y=k$, then when $y=k+1$, $f(3, k+1)=f[2, f(3$, $k)]=f\left(2,2^{k+3}-3\right)=2\left(2^{k+3}-3\right)+3=2^{(k+1)+3}-3$. By the principle of induction, (3) is also generally true.

Finally, from conditions (2), (3), and (3), we get $f(4,0)=f(3,1)=2^{2^{2}}-3, f(4,1)=f[3, f(4,0)]=f\left(3,2^{2^{2}}-3\right)=$ $2^{2^{2^{2}}}-3$. Using $2^{(s)}$ to represent $2^{2^{\cdots-2}}$, with $S$ layers of 2's powers, then $f(4,0)=2^{(3)}-3, f(4,1)=2^{(4)}-3, f(4,2)=f[3$, $f(4,1)]=f\left[3,2^{(4)}-3\right]=2^{2^{(4)}}-3=2^{(5)}-3$, assuming $f(4, k)=2^{(k+3)}-3$ holds, then $f(4, k+1)=f[3$, $f(4, k)]=f\left[3,2^{(k+3)}-3\right]=2^{2^{(k+3)}}-3=2^{(k+4)}-3$, by the principle of induction, we know that for any positive integer $y_{1}$, $f(4, y)=2^{(y+3)}-3$. Therefore, $f(4,1981)=2^{(1984)}-3$ (where $2^{1984}$ represents 1984 layers of 2's powers).