Example 1-5 The number of $n$-ary Boolean functions $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$.
For $n$ Boolean variables $x_{1}, x_{2}, \cdots, x_{n}, x_{i}=(0,1), i=1,2, \cdots, n$, according to the multiplication rule, there can be $\underbrace{2 \cdot 2 \cdot \cdots \cdot 2}_{n \text { times }}=2^{n}$ different assignments, let these $2^{n}$ different assignments be
$$
a_{1}, a_{2}, \cdots, a_{2^{n}}
$$

For each assignment, the Boolean function can take a value of $(0,1)$, so the number of different Boolean functions is
$$
\underbrace{2 \cdot 2 \cdot \cdots \cdot 2}_{2^{n}}=2^{2^{n}} .
$$

For example, when $n=2$, $\left(x_{1}, x_{2}\right)$ has $(0,0),(0,1),(1,0),(1,1)$, four assignments, and $f\left(x_{1}, x_{2}\right)$ has $2^{4}=16$ possibilities. The same $\left(x_{1}, x_{2}\right)$ but with different function values represents different Boolean functions, as shown in Table 1-1.

Where $\overline{1}=0, \overline{0}=1$, and it satisfies:
For example, $\left(\bar{x}_{1} \vee x_{2}\right) \wedge\left(x_{1} \vee \bar{x}_{2}\right)$ is computed in the table as shown in Table 1-2.