Binary 2s Compliment Applied Twice Gives Original - How?
First, if we wouldn't get the same number after negating it twice, it wouldn't make much sense, right?

So we just need to prove that the "complement and add 1" has indeed the effect of negation, i.e., taking $x$ into $-x$  (and thus, $-x$ to $x$). (well, maybe except for an edge case I will mention below.)

A signed number of $n$ bits, $b_{n-1}, \ldots, b_1,b_0$ where each $b_i\in\{0,1\}$,  is to be interpreted in the following way: the $i$-th bit amount $2^i$ except for the last bit (MSB). The MSB $b_{n-1}$ is what makes the number negative, and can be seen as having the value $-2^{n-1}$.


  Sanity check.
  4 bits. The number 5 is 0101 = 2^2+ 2^0 = 4+1.
  The number -5 is 1011 = -2^{3}+ 2^1+2^0 = -8+2+1 = -5.


With this in mind, given $x = -2^{n-1}b_{n-1} + \sum_{i=0}^{n-2}2^ib_i$ if we complement each bit ($b_i \to (1-b_i)$) we will get
$$\begin{align}NOT(x) &amp;= -2^{n-1}(1-b_{n-1}) + \sum_{i=0}^{n-2}2^i(1-b_i) \\
&amp; = -2^{n-1} + \sum_{i=0}^{n-2}2^i - [-2^{n-1}b_{n-1} + \sum_{i=0}^{n-2}2^ib_i] \\
&amp;= -1 -x
\end{align}$$
therefore, $NOT(x)+1$ equals $-x$.

Note the edge case(s).If $x=0$ then negating it should give 0. Yet complementing each bit gives $111111 \ne 0$ and after +1 we will get back 0  (ignoring the carry).
The other edge case is the maximal negative number $1000$ for 4-bit numbers.
If we negate it we get $0111$ after complementing and $1000$ after adding 1 to the complemented number. In this case (and this case only) the 2-complement system "doesn't work" and gives that the negation of $x$ is $x$.