Explain why x == ~(~x + 1) + 1 (two's complement and back!)

Question

As we all know usually negative numbers in memory represents as two\'s complement numbers like that

from x to ~x + 1

and to get back we don

Answer 1

This is because if you increment ~x (assuming no overflow). Then converting it to back to x, you've incremented relative to ~x, but decremented relative to x. Same thing applies vice versa. Assuming your variable x has a specific value, every time you increment it, relative to ~x you'll notice it decrements.

From a programmer's point of view, this is what you'd essentially witness.

Let short int x = 1         (0x0001)
then ~x = 65534             (0xFFFE)
~x + 1 =  65534 + 1         (0xFFFF)
~(~x+1) = 0                 (0x0000)
~(~x+1) + 1 = 0 + 1         (0x0001)

Answer 2

That's the same thing anyway. That is, ~x + 1 == ~(x - 1). But let's put that aside for now.

f(x) = ~x + 1 is its own inverse. Proof:

~(~x + 1) + 1 =
(definition of subtraction: a - b = ~(~a + b))
x - 1 + 1 =
(you know this step)
x

Also, ~x + 1 == ~(x - 1). Why? Well,

~(x - 1) =
(definition of subtraction: a - b = ~(~a + b))
~(~(~x + 1)) =
(remove double negation)
~x + 1

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And that (slightly unusual) definition of subtraction, a - b = ~(~a + b)?

~(~a + b) =
(use definition of two's complement, ~x = -x - 1)
-(~a + b) - 1 =
(move the 1)
-(~a + b + 1) =
(use definition of two's complement, ~x = -x - 1)
-(-a + b) =
(you know this step)
a - b