foldr k z = foldr' k z
where
foldr' k z [] = z
foldr' k z (y:ys) = y `k` foldr' k z ys
That's obviously not sufficient, so it must have something to do with the nature of the closure. In this case I presume that it's because the closure captures k and z, although if you have any evidence to the contrary that would be interesting to see.When this happens, note that it's actually no longer constructing a closure at runtime. It has essentially closed over the values at compile time, using some very trivial transformations. If you use a definition that is too complex for those trivial transformations, you're getting in the way of the compiler doing its job. I always prefer to write my code with sympathy for the compiler. The less magic it needs to do, the better it does its job.