## Proof that the decrement operator can be achieved [exclusively with binary operations]

A friend of mine was recently telling me how he’d done an electronics assignment where for extra credit they had to create an in-place decrement operator (by that I mean not taking the approach of taking the 2s complement of -1 and adding it to a number). This sounded cool and seemed like the way the ALU would do it to save allocation overhead and space.

Here’s the identity he told me about:

\[x - 1 == \texttt{~}(-x)\]where:

“~” represents the binary complement

“-“ represents the binary negative, aka 2s complement of the number.

Looks really bizarre but feels like it should make sense. I rooted about on Google search to find the proof, but either couldn’t find it, or more likely didn’t have my search terms fine tuned! (See http://www.woodmann.com/searchlores/)

Here’s the forward proof I came to. (using ‘=’ to mean comparison not assignment here)

\[x = x\] \[x = (x+1) -1\]due to 2s complement we know that \(\texttt{~}x +1 = -x\), therefore \(x+1 = -(\texttt{~}x)\), substituting this gives

\[x = -(\texttt{~} x) - 1\]as we know the negation is 2s complement of the brackets, therefore

\[x = \texttt{~}(\texttt{~}x +1) +1\]now sub back our other formula from above

\[x = \texttt{~}(-x) +1\]and here we are

\[x -1 = \texttt{~}(-x)\]*If you notice any errors above, or if you just need to “Throw some knowledgeballs down the webhills!”, please flame me on the appropriate public channels.*