1.7

证明：从本原函数出发，经复合和算子 $\prod\limits_{i=n}^{m}[\cdot]$ 可以生成所有的初等函数，这里 $\prod\limits_{i=n}^{m}[f(\vec{i})] = \left\{ \begin{array}{ll} f(n) \cdot f(n-1) \cdot\ \cdots \ \cdot f(m), & \text{if } m\geq n\\ 1, & \text{if } m < n \end{array} \right.$

证明

只需证 $\sum\limits_{i=0}^n$ ， $\prod\limits_{i=0}^n$ 和函数 $\ddot{-}$ 可表示出。

1. $\prod\limits_{i=0}^n$ 为 $\prod\limits_{i=n}^{m}$ 的特例，取 $n=0$ 即可；

2. $x^y = \prod\limits_{i=1}^{y}P_1^1(x)$，从而 $2^y = \prod\limits_{i=1}^{y}SSZ(i)$；

3. $N(x) = \prod\limits_{i=1}^{x}Z(i)$ ；

4. $\text{leq}(x,y) = \prod\limits_{i=x}^{y}Z(i) = \left\{\begin{array}{ll}0, & \text{if } x \leq y \\ 1, & \text{if } x > y\end{array}\right.$

5. $\text{geq}(x, y) = \prod\limits_{i=y}^{x} Z(i)= \left\{\begin{array}{ll}0, & \text{if } x \geq y \\ 1, & \text{if } x < y\end{array}\right.$

6. $\text{eq}(x,y) = \text{leq}(x,y)^{N\text{geq}(x,y)}=\left\{\begin{array}{ll} 0, & \text{if } x = y \\ 1, & \text{if } x \neq y\end{array}\right.$

7. $\log(x) = \prod\limits_{i=0}^x i^{N \text{ eq}(2^i,x)}$

   注意：$\log(2^y) = \prod\limits_{i=0}^{2^y}i^{N\text{ eq}(2^i, 2^y)}$

8. $\sum\limits_{i=n}^{m} f(i, \vec{x}) = \log(2^{\sum_{i=n}^mf(i, \vec{x})}) = \log(\prod\limits_{i=n}^{m}2^{f(i, \vec{x})})$

9. $x \cdot y = \sum\limits_{i=1}^{x}P_1^1(y)$

10. $x+y = \log(2^x \cdot 2^y)$

11. $x \ddot{-} y = \left(\sum\limits_{i=y+1}^{x}SZ(i)\right) + \left(\sum\limits_{i=x+1}^{y}SZ(i)\right)$​