I found this conceptually straightforward but hard to satisfactorily prove. Try it yourself.

1+NOT(y) ≡ −y ≡ x (mod ##2^n##).

Or even better:

T(x)=1+NOT(x) ≡ −x (mod ## 2^n##)
T(T(x)) ≡ −(−x) ≡ x (mod ## 2^n##)

For any n bit number b, NOT b changes every 0 to 1 and every 1 to 0. Numerically, that means NOT b = 2^n - 1 - b.

So 1 + NOT b = 2^n - b, which is equivalent to -b modulo 2^n. In other words, NOT b + 1 gives you the additive inverse of b.

Now apply the same operation again. Since the additive inverse of -b is b, we get 1 + NOT(1 + NOT b) = b modulo 2^n.

So taking the two's complement twice always gives you the original n bit number.

## \begin{align}
f(x)=1+\text{NOT}x&\implies f(x)-1=\text{NOT}x\nonumber\\
&\implies \text{NOT}(f(x)-1)=x\nonumber\\
&\implies f^{-1}(x)=\text{NOT}(x-1)\nonumber\\
\end{align} ##

For
$$ x=(\sum_{i=0}^{n}x_i10^i)_2 $$
where
## \begin{align}
\text{NOT}x&=(\sum_{i=0}^{n}(1-x_i)10^i)_2\nonumber\\
&=(\sum_{i=0}^{n}1\cdot10^i-\sum_{i=0}^{n}x_i10^i)_2\nonumber\\
\end{align}\\ ##

we have

## \begin{align}
f(x)&=1+\text{NOT}x\nonumber\\
&=(1+(\sum_{i=0}^{n}1\cdot10^i-\sum_{i=0}^{n}x_i10^i))_2\nonumber\\
&=(\sum_{i=0}^{n}1\cdot10^i-(\sum_{i=0}^{n}x_i10^i-1))_2\nonumber\\
&=\text{NOT}(x-1)\nonumber\\
&=f^{-1}(x)\nonumber\\
\end{align}\\ ##