Calling out to idiot America

I would just like to make the following point.

If P \left( A \right) = \dfrac{ e^r }{1+e^r}, then, obviously, P \left( \neg A \right) = \dfrac{ 1 }{1+e^r}.

Last night I couldn't figure this out (yeah, I'm awesome), so I wrote a fancy function which has the result of outputting exactly the above. In somewhat simplified form, when A is the event that Y=1 and Y is a binary variable,

 P \left( Y = y \right) = \dfrac{ e^{y r} }{1 + e^r }

It takes a really special sort of blindness to write a fancy function to output an obvious fact when that fact is not itself understood.