Comments only: move flattening notes to TcFlatten

1 files changed, 1 insertions, 201 deletions

diff --git a/compiler/typecheck/Flattening-notes b/compiler/typecheck/Flattening-notes
index ffe39ab5e5..2aa9243743 100644
--- a/compiler/typecheck/Flattening-notes
+++ b/compiler/typecheck/Flattening-notes
@@ -9,207 +9,6 @@ ToDo:
   * RAE: I think it would be better to split off CNonCanonical into its own
     type, and remove it completely from Ct. Then, we would keep CIrredCan
 
-===========================
-
-The inert equalities
-~~~~~~~~~~~~~~~~~~~~
-
-Definition: can-rewrite relation.
-A "can-rewrite" relation between flavours, written f1 >= f2, is a
-binary relation with the following properties
-
-  R1.  >= is transitive
-  R2.  If f1 >= f, and f2 >= f,
-       then either f1 >= f2 or f2 >= f1
-
-Lemma.  If f1 >= f then f1 >= f1
-Proof.  By property (R2), with f1=f2
-
-Definition: generalised substitution.
-A "generalised substitution" S is a set of triples (a -f-> t), where
-  a is a type variable
-  t is a type
-  f is a flavour
-such that
-  (WF1) if (a -f1-> t1) in S
-            (a -f2-> t2) in S
-        then neither (f1 >= f2) nor (f2 >= f1) hold
-  (WF2) if (a -f-> t) is in S, then t /= a
-
-Definition: applying a generalised substitution.
-If S is a generalised substitution
-   S(f,a) = t,  if (a -fs-> t) in S, and fs >= f
-          = a,  otherwise
-Application extends naturally to types S(f,t)
-
-Theorem: S(f,a) is well defined as a function.
-Proof: Suppose (a -f1-> t1) and (a -f2-> t2) are both in S,
-               and  f1 >= f and f2 >= f
-       Then by (R2) f1 >= f2 or f2 >= f1, which contradicts (WF)
-
-Notation: repeated application.
-  S^0(f,t)     = t
-  S^(n+1)(f,t) = S(f, S^n(t))
-
-Definition: inert generalised substitution
-A generalised substitution S is "inert" iff
-
-  (IG1) there is an n such that 
-        for every f,t, S^n(f,t) = S^(n+1)(f,t)
-
-  (IG2) if (b -f-> t) in S, and f >= f, then S(f,t) = t
-        that is, each individual binding is "self-stable"
-
-----------------------------------------------------------------
-Our main invariant: 
-   the inert CTyEqCans should be an inert generalised substitution
-----------------------------------------------------------------
-
-Note that inertness is not the same as idempotence.  To apply S to a
-type, you may have to apply it recursive.  But inertness does
-guarantee that this recursive use will terminate.
-
-The main theorem.
-   Suppose we have a "work item"
-       a -fw-> t
-   and an inert generalised substitution S,
-   such that
-      (T1) S(fw,a) = a     -- LHS of work-item is a fixpoint of S(fw,_)
-      (T2) S(fw,t) = t     -- RHS of work-item is a fixpoint of S(fw,_)
-      (T3) a not in t      -- No occurs check in the work item
-
-      (K1) if (a -fs-> s) is in S then not (fw >= fs)
-      (K2) if (b -fs-> s) is in S, where b /= a, then
-              (K2a) not (fs >= fs)
-           or (K2b) not (fw >= fs)
-           or (K2c) a not in s
-     or (K3) if (b -fs-> a) is in S then not (fw >= fs)
-             (a stronger version of (K2))
-
-   then the extended substition T = S+(a -fw-> t)
-   is an inert generalised substitution.
-
-The idea is that
-* (T1-2) are guaranteed by exhaustively rewriting the work-item
-  with S(fw,_).
-
-* T3 is guaranteed by a simple occurs-check on the work item.
-
-* (K1-3) are the "kick-out" criteria.  (As stated, they are really the
-  "keep" criteria.) If the current inert S contains a triple that does
-  not satisfy (K1-3), then we remove it from S by "kicking it out",
-  and re-processing it.
-
-* Note that kicking out is a Bad Thing, because it means we have to
-  re-process a constraint.  The less we kick out, the better.
-
-* Assume we have  G>=G, G>=W, D>=D, and that's all.  Then, when performing
-  a unification we add a new given  a -G-> ty.  But doing so does NOT require
-  us to kick out an inert wanted that mentions a, because of (K2a).  This
-  is a common case, hence good not to kick out.
-
-* Lemma (L1): The conditions of the Main Theorem imply that there is no
-              (a fs-> t) in S, s.t.  (fs >= fw).
-  Proof. Suppose the contrary (fs >= fw).  Then because of (T1),
-  S(fw,a)=a.  But since fs>=fw, S(fw,a) = s, hence s=a.  But now we
-  have (a -fs-> a) in S, which contradicts (WF2).
-
-* The extended substitution satisfies (WF1) and (WF2)
-  - (K1) plus (L1) guarantee that the extended substiution satisfies (WF1).
-  - (T3) guarantees (WF2).
-
-* (K2) is about inertness.  Intuitively, any infinite chain T^0(f,t),
-  T^1(f,t), T^2(f,T).... must pass through the new work item infnitely
-  often, since the substution without the work item is inert; and must
-  pass through at least one of the triples in S infnitely often.
-
-  - (K2a): if not(fs>=fs) then there is no f that fs can rewrite (fs>=f),
-    and hence this triple never plays a role in application S(f,a).
-    It is always safe to extend S with such a triple.
-
-    (NB: we could strengten K1) in this way too, but see K3.
-
-  - (K2b): If this holds, we can't pass through this triple infinitely
-    often, because if we did then fs>=f, fw>=f, hence fs>=fw,
-    contradicting (L1), or fw>=fs contradicting K2b.
-
-  - (K2c): if a not in s, we hae no further opportunity to apply the
-    work item.
-
-  NB: this reasoning isn't water tight.
-
-Key lemma to make it watertight.
-  Under the conditions of the Main Theorem,
-  forall f st fw >= f, a is not in S^k(f,t), for any k
-
-
-Completeness
-~~~~~~~~~~~~~
-K3: completeness.  (K3) is not ncessary for the extended substitution
-to be inert.  In fact K1 could be made stronger by saying
-   ... then (not (fw >= fs) or not (fs >= fs))
-But it's not enough for S to be inert; we also want completeness.
-That is, we want to be able to solve all soluble wanted equalities.
-Suppose we have
-
-   work-item   b -G-> a
-   inert-item  a -W-> b
-
-Assuming (G >= W) but not (W >= W), this fulfills all the conditions,
-so we could extend the inerts, thus:
-
-   inert-items   b -G-> a
-                 a -W-> b
-
-But if we kicked-out the inert item, we'd get
-
-   work-item     a -W-> b
-   inert-item    b -G-> a
-
-Then rewrite the work-item gives us (a -W-> a), which is soluble via Refl.
-So we add one more clause to the kick-out criteria
-
-Another way to understand (K3) is that we treat an inert item
-        a -f-> b
-in the same way as
-        b -f-> a
-So if we kick out one, we should kick out the other.  The orientation
-is somewhat accidental.
-
------------------------
-RAE: To prove that K3 is sufficient for completeness (as opposed to a rule that
-looked for `a` *anywhere* on the RHS, not just at the top), we need this property:
-All types in the inert set are "rigid". Here, rigid means that a type is one of
-two things: a type that can equal only itself, or a type variable. Because the
-inert set defines rewritings for type variables, a type variable can be considered
-rigid because it will be rewritten only to a rigid type.
-
-In the current world, this rigidity property is true: all type families are
-flattened away before adding equalities to the inert set. But, when we add
-representational equality, that is no longer true! Newtypes are not rigid
-w.r.t. representational equality. Accordingly, we would to change (K3) thus:
-
-(K3) If (b -fs-> s) is in S with (fw >= fs), then
-  (K3a) If the role of fs is nominal: s /= a
-  (K3b) If the role of fs is representational: EITHER
-          a not in s, OR
-          the path from the top of s to a includes at least one non-newtype
-
-SPJ/DV: this looks important... follow up
-
------------------------
-RAE: Do we have evidence to support our belief that kicking out is bad? I can
-imagine scenarios where kicking out *more* equalities is more efficient, in that
-kicking out a Given, say, might then discover that the Given is reflexive and
-thus can be dropped. Once this happens, then the Given is no longer used in
-rewriting, making later flattenings faster. I tend to thing that, probably,
-kicking out is something to avoid, but it would be nice to have data to support
-this conclusion. And, that data is not terribly hard to produce: we can just
-twiddle some settings and then time the testsuite in some sort of controlled
-environment.
-
-SPJ: yes it would be good to do that. 
-
 The coercion solver
 ~~~~~~~~~~~~~~~~~~~~
 Our hope. In GHC currently drawn from {G,W,D}, but with the coercion
@@ -230,3 +29,4 @@ flattening algorithm. Flattening (T a) looks at the roles of
 T's parameters, and chooses the role for flattening `a` appropriately.
 This is why there must be the [Role] parameter to flattenMany.
 Of course, this non-uniform rewriting may gum up the proof works.
+