+Note [Inline candidates]
+~~~~~~~~~~~~~~~~~~~~~~~~
+At one point I gave is_con_app a higher score than inline-candidate,
+on the grounds that "it's *really* helpful if dictionaries get inlined fast".
+However a nofib run revealed no change if they were swapped so that
+inline-candidate has the higher score. And it's important that it does,
+else you can get a bad worker-wrapper split thus:
+ rec {
+ $wfoo x = ....foo x....
+
+ {-loop brk-} foo x = ...$wfoo x...
+ }
+But we *want* the wrapper to be inlined! If it isn't, the interface
+file sees the unfolding for $wfoo, and sees that foo is strict (and
+hence it gets an auto-generated wrapper. Result: an infinite inlining
+in the importing scope. So be a bit careful if you change this. A
+good example is Tree.repTree in nofib/spectral/minimax. If is_con_app
+has the higher score, then compiling Game.hs goes into an infinite loop.
+
+Note [Recursive rules]
+~~~~~~~~~~~~~~~~~~~~~~
+Consider this group, which is typical of what SpecConstr builds:
+
+ fs a = ....f (C a)....
+ f x = ....f (C a)....
+ {-# RULE f (C a) = fs a #-}
+
+So 'f' and 'fs' are mutually recursive. If we choose 'fs' as the loop breaker,
+all is well; the RULE is applied, and 'fs' becomes self-recursive.
+
+But if we choose 'f' as the loop breaker, we may get an infinite loop:
+ - the RULE is applied in f's RHS (see Note [Self-recursive rules] in Simplify
+ - fs is inlined (say it's small)
+ - now there's another opportunity to apply the RULE
+
+So it's very important to choose the RULE-variable as the loop breaker.
+This showed up when compiling Control.Concurrent.Chan.getChanContents.
+
+Note [Closure conversion]
+~~~~~~~~~~~~~~~~~~~~~~~~~
+We treat (\x. C p q) as a high-score candidate in the letrec scoring algorithm.
+The immediate motivation came from the result of a closure-conversion transformation
+which generated code like this:
+
+ data Clo a b = forall c. Clo (c -> a -> b) c
+
+ ($:) :: Clo a b -> a -> b
+ Clo f env $: x = f env x
+
+ rec { plus = Clo plus1 ()
+
+ ; plus1 _ n = Clo plus2 n
+
+ ; plus2 Zero n = n
+ ; plus2 (Succ m) n = Succ (plus $: m $: n) }
+
+If we inline 'plus' and 'plus1', everything unravels nicely. But if
+we choose 'plus1' as the loop breaker (which is entirely possible
+otherwise), the loop does not unravel nicely.
+
+