1.5

设 $pg(x, y) = 2^x(2y+1)\dot{-}1$，证明：存在初等函数 $K(z)$ 和 $L(z)$ 使得 $K(pg(x,y))=x,$ $L(pg(x,y))=y,$ $pg(K(x), L(y)) = z.$

证明

$z = 2^x(2y+1)\dot{-}1$

iff $z+1=2^x(2y+1)$

iff $x = \text{ep}_0(z+1)$ 且 $2y+1=\frac{z+1}{2^x}$

代入 $x$，得 $2y+1 = \left[\frac{z+1}{2^{\text{ep}_0(z+1)}}\right]$

iff $x=K(z)$ 且 $y=L(z)$

这里 $K(z) = \text{ep}_0(z+1)$，$L(z) = \left[\frac{\left[\frac{z+1}{2^{\text{ep}_0(z+1)}}\right]\dot{-}1}{2}\right]$