Question

I found some exercises where you combine n-bit 2’s complement values in different ways and simplify the output where possible. (Their practice exercises use 16-bit, but that’s irrelevant).

Eg:

!(!x&!y) == x|y

0 & y, negate the output == -1

I’m having no problem applying De Morgan’s laws with the examples using AND, OR, and NOT but I am having difficulty using NOT with + and –

Eg:

!(!x+y) == x-y

!(y-1) == -y

How does NOT distribute?

Edit: responding to comments: I realize this is a bitwise NOT. My question is: in algebraic terms, how does it distribute as per algebra? Example on Wikipedia

Answer 1

With 2’s complement numbers when you bitwise NOT them it is the same as saying the negative of the number minus 1, so !x is equivalent to -x - 1 where x can be a single variable or an expression.

Starting with !(!x+y), well !x is going to be -x - 1 so then it is !(-x - 1 + y) which becomes -(-x - 1 + y) - 1 which simplifies to x - y.

And for !(y-1), that becomes -(y - 1) - 1 = -y + 1 - 1 = -y.