Commit ac73d1a7 by Simon Peyton Jones

### Revise flattening-notes

parent 1d44261c
 ... ... @@ -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 (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 (IG1) 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} } (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: ... ... @@ -73,8 +74,8 @@ The main theorem. a -fw-> t 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. is an inert generalised substitution. 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 ... ... @@ -134,6 +138,10 @@ RAE: I don't understand this lemma statement -- fs seems out of scope here. 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 ~~~~~~~~~~~~~ ... ... @@ -161,6 +169,14 @@ But if we kicked-out the inert item, we'd get 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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