Commit ac73d1a7 authored by Simon Peyton Jones's avatar Simon Peyton Jones
Browse files

Revise flattening-notes

parent 1d44261c
......@@ -14,12 +14,12 @@ ToDo:
The inert equalities
~~~~~~~~~~~~~~~~~~~~
Definition: can-rewrite relation.
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,
R2. If f1 >= f, and f2 >= f,
then either f1 >= f2 or f2 >= f1
Lemma. If f1 >= f then f1 >= f1
......@@ -31,9 +31,10 @@ A "generalised substitution" S is a set of triples (a -f-> t), where
t is a type
f is a flavour
such that
(WF) if (a -f1-> t1) in S
(a -f2-> t2) in S
(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 subsitution
......@@ -41,7 +42,7 @@ If S is a generalised subsitution
= a, otherwise
Application extends naturally to types S(f,t)
Theorem: S(f,a) is a function.
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)
......@@ -52,12 +53,12 @@ Notation: repeated application.
Definition: inert generalised substitution
A generalised substitution S is "inert" iff
there is an n such that
for every f,t, S^n(f,t) = S^(n+1)(f,t)
Flavours. In GHC currently drawn from {G,W,D}, but with the coercion
solver the flavours become pairs
{ (k,l) | k <- {G,W,D}, l <- {Nom,Rep} }
(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:
......@@ -68,13 +69,13 @@ 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"
The main theorem.
Suppose we have a "work item"
a -fw-> t
and an inert generalised substitution S,
and an inert generalised substitution S,
such that
(T1) S(fw,a) = a -- LHS is a fixpoint of S
(T2) S(fw,t) = t -- RHS is a fixpoint of S
(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)
......@@ -83,13 +84,14 @@ The main theorem.
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 genrealised substitution.
then the extended substition T = S+(a -fw-> t)
is an inert generalised substitution.
The idea is that
The idea is that
* (T1-2) are guaranteed by exhaustively rewriting the work-item
with S.
with S(fw,_).
* T3 is guaranteed by a simple occurs-check on the work item.
......@@ -102,17 +104,19 @@ The idea is that
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 dos not require
us to kick out wanteds that mention a, because of (K2b).
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 not (fs >= fw).
* 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, since fs>=fw we must have fs>=fs, and hence S
is not inert.
RAE: I don't understand this lemma statement -- fs seems out of scope here.
have (a -fs-> a) in S, which contradicts (WF2).
* (K1) plus (L1) guarantee that the extended substiution satisfies (WF).
* 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
......@@ -121,7 +125,7 @@ RAE: I don't understand this lemma statement -- fs seems out of scope here.
- (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.
It is always safe to extend S with such a triple.
(NB: we could strengten K1) in this way too, but see K3.
......@@ -129,11 +133,15 @@ RAE: I don't understand this lemma statement -- fs seems out of scope here.
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
- (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
~~~~~~~~~~~~~
......@@ -149,18 +157,26 @@ Suppose we have
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
......@@ -179,6 +195,9 @@ w.r.t. representational equality. Accordingly, we would to change (K3) thus:
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
......@@ -188,3 +207,21 @@ 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
solver the flavours become pairs
{ (k,l) | k <- {G,W,D}, l <- {Nom,Rep} }
But can
a -(G,R)-> Int
rewrite
b -(G,R)-> T a
?
Well, it depends on the roles at which T uses its arguments :-(.
So it may not be enough just to look at (flavour,role) pairs?
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