7. (FIN $)^{1 \mathrm{MO6}}$ Assume that $f(x, y)$ is defined for all positive integers $x$ and $y$, and that the following equations are satisfied:
$$
\begin{aligned}
f(0, y) & =y+1, \\
f(x+1,0) & =f(x, 1), \\
f(x+1, y+1) & =f(x, f(x+1, y)) .
\end{aligned}
$$

Determine $f(2,2), f(3,3)$ and $f(4,4)$.
Alternative version: Determine $f(4,1981)$.
7. We immediately find that $f(1,0)=f(0,1)=2$. Then $f(1, y+1)=$ $f(0, f(1, y))=f(1, y)+1$; hence $f(1, y)=y+2$ for $y \geq 0$. Next we find that $f(2,0)=f(1,1)=3$ and $f(2, y+1)=f(1, f(2, y))=f(2, y)+2$, from which $f(2, y)=2 y+3$. Particularly, $f(2,2)=7$. Further, $f(3,0)=$ $f(2,1)=5$ and $f(3, y+1)=f(2, f(3, y))=2 f(3, y)+3$. This gives by induction $f(3, y)=2^{y+3}-3$. For $y=3, f(3,3)=61$. Finally, from $f(4,0)=f(3,1)=13$ and $f(4, y+1)=f(3, f(4, y))=2^{f(4, y)+3}-3$, we conclude that
$$
f(4, y)=2^{2}-3 \quad(y+3 \text { twos }) .
$$