How to prove that the C statement -x, ~x+1, and ~(x-1) yield the same results?

Question

I want to know the logic behind this statement, the proof. The C expression -x, ~x+1, and ~(x-1) all yield the same results for any x. I can show this is true for specific

Answer 1

~x+1 is equivalent to 2's complement + 1 (i.e. negative number) representations of -x, ~(x-1) also represents the same (consider the case where last bit is 1, ~(x-1) = ~(b1b2.b(n-1)1 - 0) = b1'b2'...b(n-1)'1 = b1'b2'...b(n-1)'0 + 1 = ~x+1. Similar case hold for last bit is 0. ~(x-1) = ~(b1b2..bi100..00 - 1) = ~b1b2..bi011..11 = b1'b2'..bi'100..00 = b1'b2'..bi'011..11 + 1 = ~x + 1.