4. Define an operation $*$ on the set of real numbers $\mathbf{R}$ satisfying
(i) $x * 0=1$
(for any $x \in \mathbf{R}$ ).
(ii) $(x * y) * z=(z * x y)+z$ (for any $x, y, z \in \mathbf{R}$ ).
Find $31 * 32$.
4. Taking $y=0$ in (ii) and using (i) we get $1 * z=1+z$.
(1)
Thus, $(1*y)*1$ can be calculated in two ways: using (ii),
$$
(1 * y) * 1=(1 *(1 \cdot y))+1=(1 * y)+1=1+y+1=y+2;
$$

while by (1), we have $(1 * y) * 1=(1+y) * 1$.
Combining the above two equations, we get
$(1+y) * 1=y+2$, which means $x * 1=x+1$.
Therefore, $(x * y) * 1$ can be calculated in two ways: by (2),
we have $(x * y) * 1=(x * y)+1$;
and by (ii), $(x * y) * 1=(1 * x y)+1=(x y+1)+1=x y+2$.
Combining the above two equations, we get $x * y=x y+1$.
In particular, $31 * 32=31 \times 32+1=993$.