ah, sign should also work

@Leo so I can also work as a higher order function, confusing everyone in the room

@Razetime - 'I can also work as a higher order function' - Ok, now you've said that I can understand what Leo said, if that's correct.

@Leo - Thanks for explaining: I can understand the problem with I (especially if Razetime's comment is correct). And thanks for demonstrating how to solve it.

@Leo - But I'm still having trouble parsing the inference. If id is being applied to revnum, then why aren't they grouped together with parentheses in the inference explanation (like: ((id)(revnum)))? Instead, it looks like id is grouped with com((predN)(any)) which I can't figure out, and it isn't clear what function any is using at all...

@DominicvanEssen I relaized this cause many of my ternaries with I have failed to infer

@DominicvanEssen So, app f x means just f x, com f x means \y -> f x y, com2 f x means \y,z->f x y z

Sorry have to run, will explain more tonorrow

The current inferencer always groups things left to right, adding combinators where needed

`com2(app(com(com2(predN)(any))(id))(revnum))(base) = (\a,b->(\c->(\d,e->predN (any d e)) (id c)) revnum) base

Sorry, my earlier message was wrong, com f x means \y -> f (x y), com2 f x means \y,z->f (x y z)

To put it simply, predN gets combined with any, then they get combined with id, and the result gets applied to revnum