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(*
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* Copyright (c) 2018-present, Facebook, Inc.
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*
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* This source code is licensed under the MIT license found in the
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* LICENSE file in the root directory of this source tree.
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*)
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(** Equality over uninterpreted functions and linear rational arithmetic *)
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type 'a exp_map = 'a Map.M(Exp).t [@@deriving compare, equal, sexp]
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let empty_map = Map.empty (module Exp)
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type subst = Exp.t exp_map [@@deriving compare, equal, sexp]
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(** see also [invariant] *)
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type t =
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{ sat: bool (** [false] only if constraints are inconsistent *)
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; rep: subst
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(** functional set of oriented equations: map [a] to [a'],
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indicating that [a = a'] holds, and that [a'] is the
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'rep(resentative)' of [a] *) }
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[@@deriving compare, equal, sexp]
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(** Pretty-printing *)
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let pp fs {sat; rep} =
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let pp_alist pp_k pp_v fs alist =
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let pp_assoc fs (k, v) =
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Format.fprintf fs "[@[%a@ @<2>↦ %a@]]" pp_k k pp_v (k, v)
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in
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Format.fprintf fs "[@[<hv>%a@]]" (List.pp ";@ " pp_assoc) alist
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in
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let pp_exp_v fs (k, v) = if not (Exp.equal k v) then Exp.pp fs v in
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Format.fprintf fs "@[{@[<hv>sat= %b;@ rep= %a@]}@]" sat
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(pp_alist Exp.pp pp_exp_v)
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(Map.to_alist rep)
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let pp_classes fs r =
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let cls =
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Map.fold r.rep ~init:empty_map ~f:(fun ~key ~data cls ->
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if Exp.equal key data then cls
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else Map.add_multi cls ~key:data ~data:key )
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in
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List.pp "@ @<2>∧ "
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(fun fs (key, data) ->
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Format.fprintf fs "@[%a@ = %a@]" Exp.pp key (List.pp "@ = " Exp.pp)
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(List.sort ~compare:Exp.compare data) )
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fs (Map.to_alist cls)
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let pp_diff fs (r, s) =
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let pp_sdiff_map pp_elt_diff equal nam fs x y =
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let sd = Sequence.to_list (Map.symmetric_diff ~data_equal:equal x y) in
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if not (List.is_empty sd) then
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Format.fprintf fs "%s= [@[<hv>%a@]];@ " nam
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(List.pp ";@ " pp_elt_diff)
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sd
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in
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let pp_sdiff_elt pp_key pp_val pp_sdiff_val fs = function
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| k, `Left v ->
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Format.fprintf fs "-- [@[%a@ @<2>↦ %a@]]" pp_key k pp_val v
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| k, `Right v ->
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Format.fprintf fs "++ [@[%a@ @<2>↦ %a@]]" pp_key k pp_val v
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| k, `Unequal vv ->
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Format.fprintf fs "[@[%a@ @<2>↦ %a@]]" pp_key k pp_sdiff_val vv
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in
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let pp_sdiff_exp_map =
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let pp_sdiff_exp fs (u, v) =
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Format.fprintf fs "-- %a ++ %a" Exp.pp u Exp.pp v
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in
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pp_sdiff_map (pp_sdiff_elt Exp.pp Exp.pp pp_sdiff_exp) Exp.equal
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in
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let pp_sat fs =
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if not (Bool.equal r.sat s.sat) then
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Format.fprintf fs "sat= @[-- %b@ ++ %b@];@ " r.sat s.sat
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in
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let pp_rep fs = pp_sdiff_exp_map "rep" fs r.rep s.rep in
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Format.fprintf fs "@[{@[<hv>%t%t@]}@]" pp_sat pp_rep
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(** Invariant *)
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(** test membership in carrier *)
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let in_car r e = Map.mem r.rep e
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let rec iter_max_solvables e ~f =
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match Exp.classify e with
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| `Interpreted -> Exp.iter ~f:(iter_max_solvables ~f) e
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| _ -> f e
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let invariant r =
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Invariant.invariant [%here] r [%sexp_of: t]
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@@ fun () ->
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Map.iteri r.rep ~f:(fun ~key:a ~data:_ ->
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(* no interpreted exps in carrier *)
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assert (Poly.(Exp.classify a <> `Interpreted)) ;
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(* carrier is closed under sub-expressions *)
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iter_max_solvables a ~f:(fun b ->
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assert (
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in_car r b
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|| Trace.fail "@[subexp %a of %a not in carrier of@ %a@]" Exp.pp
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b Exp.pp a pp r ) ) )
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(** Core operations *)
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let true_ = {sat= true; rep= empty_map} |> check invariant
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(** apply a subst to an exp *)
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let apply s a = try Map.find_exn s a with Caml.Not_found -> a
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(** apply a subst to maximal non-interpreted subexps *)
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let rec norm s a =
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match Exp.classify a with
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| `Interpreted -> Exp.map ~f:(norm s) a
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[sledge] Classify fully-interpreted and simplified exps differently
Summary:
For some Exp forms, Exp.solve is not complete, and this is necessary
since the result of solve is a substitution that needs to encode the
input equation as a conjunction of equations each between a variable
and an exp. This is tantamount to, stronger even than, the theory
being convex. So Exp.solve is not complete for some exps, and some of
those have constructors that perform some simplification. For example,
`(1 ≠ 0)` simplifies to `-1` (i.e. true). To enable deductions such as
`((x ≠ 0) = b) && x = 1 |- b = -1` needs the equality solver to
substitute through subexps of simplifiable exps like = and ≠, as it
does for interpreted exps like + and ×. At the same time, since
Exp.solve for non-interpreted exps cannot be complete, to enable
deductions such as `((x ≠ 0) = (y ≠ 0)) && x = 1 |- y ≠ 0` needs the
equality solver to congruence-close over subexps of simplifiable exps
such as = and ≠, as it does for uninterpreted exps.
To strengthen the equality solver in these sorts of cases, this diff
adds a new class of exps for = and ≠, and revises the equality solver
to handle them in a hybrid fashion between interpreted and
uninterpreted.
I am not currently sure whether or not this breaks the termination
proof, but I have also not managed to adapt usual divergent cases to
break this. One notable point is that simplifying = and ≠ exps always
produces genuinely simpler and smaller exps, in contrast to
e.g. polynomial simplification and gaussian elimination.
Note that the real solution to this problem is likely to be to
eliminate the i1 type in favor or a genuine boolean type, and
translate all integer operations on i1 to boolean/logical ones. Then
the disjunction implicit in e.g. equations between disequations would
appear as actual disjunction, and could be dealt with as such.
Reviewed By: jvillard
Differential Revision: D15424823
fbshipit-source-id: 67d62df1f
6 years ago
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| `Simplified -> apply s (Exp.map ~f:(norm s) a)
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| `Atomic | `Uninterpreted -> apply s a
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(** exps are congruent if equal after normalizing subexps *)
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let congruent r a b =
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Exp.equal (Exp.map ~f:(norm r.rep) a) (Exp.map ~f:(norm r.rep) b)
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(** [lookup r a] is [b'] if [a ~ b = b'] for some equation [b = b'] in rep *)
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let lookup r a =
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With_return.with_return
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@@ fun {return} ->
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(* congruent specialized to assume [a] canonized and [b] non-interpreted *)
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let semi_congruent r a b = Exp.equal a (Exp.map ~f:(apply r.rep) b) in
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Map.iteri r.rep ~f:(fun ~key ~data ->
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if semi_congruent r a key then return data ) ;
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a
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(** rewrite an exp into canonical form using rep and, for uninterpreted
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exps, congruence composed with rep *)
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let rec canon r a =
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match Exp.classify a with
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| `Interpreted -> Exp.map ~f:(canon r) a
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[sledge] Classify fully-interpreted and simplified exps differently
Summary:
For some Exp forms, Exp.solve is not complete, and this is necessary
since the result of solve is a substitution that needs to encode the
input equation as a conjunction of equations each between a variable
and an exp. This is tantamount to, stronger even than, the theory
being convex. So Exp.solve is not complete for some exps, and some of
those have constructors that perform some simplification. For example,
`(1 ≠ 0)` simplifies to `-1` (i.e. true). To enable deductions such as
`((x ≠ 0) = b) && x = 1 |- b = -1` needs the equality solver to
substitute through subexps of simplifiable exps like = and ≠, as it
does for interpreted exps like + and ×. At the same time, since
Exp.solve for non-interpreted exps cannot be complete, to enable
deductions such as `((x ≠ 0) = (y ≠ 0)) && x = 1 |- y ≠ 0` needs the
equality solver to congruence-close over subexps of simplifiable exps
such as = and ≠, as it does for uninterpreted exps.
To strengthen the equality solver in these sorts of cases, this diff
adds a new class of exps for = and ≠, and revises the equality solver
to handle them in a hybrid fashion between interpreted and
uninterpreted.
I am not currently sure whether or not this breaks the termination
proof, but I have also not managed to adapt usual divergent cases to
break this. One notable point is that simplifying = and ≠ exps always
produces genuinely simpler and smaller exps, in contrast to
e.g. polynomial simplification and gaussian elimination.
Note that the real solution to this problem is likely to be to
eliminate the i1 type in favor or a genuine boolean type, and
translate all integer operations on i1 to boolean/logical ones. Then
the disjunction implicit in e.g. equations between disequations would
appear as actual disjunction, and could be dealt with as such.
Reviewed By: jvillard
Differential Revision: D15424823
fbshipit-source-id: 67d62df1f
6 years ago
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| `Simplified | `Uninterpreted -> lookup r (Exp.map ~f:(canon r) a)
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| `Atomic -> apply r.rep a
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(** add an exp to the carrier *)
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let rec extend a r =
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match Exp.classify a with
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[sledge] Classify fully-interpreted and simplified exps differently
Summary:
For some Exp forms, Exp.solve is not complete, and this is necessary
since the result of solve is a substitution that needs to encode the
input equation as a conjunction of equations each between a variable
and an exp. This is tantamount to, stronger even than, the theory
being convex. So Exp.solve is not complete for some exps, and some of
those have constructors that perform some simplification. For example,
`(1 ≠ 0)` simplifies to `-1` (i.e. true). To enable deductions such as
`((x ≠ 0) = b) && x = 1 |- b = -1` needs the equality solver to
substitute through subexps of simplifiable exps like = and ≠, as it
does for interpreted exps like + and ×. At the same time, since
Exp.solve for non-interpreted exps cannot be complete, to enable
deductions such as `((x ≠ 0) = (y ≠ 0)) && x = 1 |- y ≠ 0` needs the
equality solver to congruence-close over subexps of simplifiable exps
such as = and ≠, as it does for uninterpreted exps.
To strengthen the equality solver in these sorts of cases, this diff
adds a new class of exps for = and ≠, and revises the equality solver
to handle them in a hybrid fashion between interpreted and
uninterpreted.
I am not currently sure whether or not this breaks the termination
proof, but I have also not managed to adapt usual divergent cases to
break this. One notable point is that simplifying = and ≠ exps always
produces genuinely simpler and smaller exps, in contrast to
e.g. polynomial simplification and gaussian elimination.
Note that the real solution to this problem is likely to be to
eliminate the i1 type in favor or a genuine boolean type, and
translate all integer operations on i1 to boolean/logical ones. Then
the disjunction implicit in e.g. equations between disequations would
appear as actual disjunction, and could be dealt with as such.
Reviewed By: jvillard
Differential Revision: D15424823
fbshipit-source-id: 67d62df1f
6 years ago
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| `Interpreted | `Simplified -> Exp.fold ~f:extend a ~init:r
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| `Atomic | `Uninterpreted ->
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Map.find_or_add r.rep a
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~if_found:(fun _ -> r)
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~default:a
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~if_added:(fun rep -> Exp.fold ~f:extend a ~init:{r with rep})
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let extend a r = extend a r |> check invariant
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let compose r s =
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let rep = Map.map ~f:(norm s) r.rep in
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let rep =
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Map.merge_skewed rep s ~combine:(fun ~key v1 v2 ->
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if Exp.equal v1 v2 then v1
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else fail "domains intersect: %a" Exp.pp key () )
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in
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[sledge] Classify fully-interpreted and simplified exps differently
Summary:
For some Exp forms, Exp.solve is not complete, and this is necessary
since the result of solve is a substitution that needs to encode the
input equation as a conjunction of equations each between a variable
and an exp. This is tantamount to, stronger even than, the theory
being convex. So Exp.solve is not complete for some exps, and some of
those have constructors that perform some simplification. For example,
`(1 ≠ 0)` simplifies to `-1` (i.e. true). To enable deductions such as
`((x ≠ 0) = b) && x = 1 |- b = -1` needs the equality solver to
substitute through subexps of simplifiable exps like = and ≠, as it
does for interpreted exps like + and ×. At the same time, since
Exp.solve for non-interpreted exps cannot be complete, to enable
deductions such as `((x ≠ 0) = (y ≠ 0)) && x = 1 |- y ≠ 0` needs the
equality solver to congruence-close over subexps of simplifiable exps
such as = and ≠, as it does for uninterpreted exps.
To strengthen the equality solver in these sorts of cases, this diff
adds a new class of exps for = and ≠, and revises the equality solver
to handle them in a hybrid fashion between interpreted and
uninterpreted.
I am not currently sure whether or not this breaks the termination
proof, but I have also not managed to adapt usual divergent cases to
break this. One notable point is that simplifying = and ≠ exps always
produces genuinely simpler and smaller exps, in contrast to
e.g. polynomial simplification and gaussian elimination.
Note that the real solution to this problem is likely to be to
eliminate the i1 type in favor or a genuine boolean type, and
translate all integer operations on i1 to boolean/logical ones. Then
the disjunction implicit in e.g. equations between disequations would
appear as actual disjunction, and could be dealt with as such.
Reviewed By: jvillard
Differential Revision: D15424823
fbshipit-source-id: 67d62df1f
6 years ago
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{r with rep}
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let merge a b r =
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[%Trace.call fun {pf} -> pf "%a@ %a@ %a" Exp.pp a Exp.pp b pp r]
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;
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( match Exp.solve a b with
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| Some s -> compose r s
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| None -> {r with sat= false} )
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|>
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[%Trace.retn fun {pf} r' ->
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pf "%a" pp_diff (r, r') ;
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invariant r']
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(** find an unproved equation between congruent exps *)
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let find_missing r =
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With_return.with_return
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@@ fun {return} ->
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Map.iteri r.rep ~f:(fun ~key:a ~data:a' ->
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Map.iteri r.rep ~f:(fun ~key:b ~data:b' ->
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if
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Exp.compare a b < 0
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&& (not (Exp.equal a' b'))
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&& congruent r a b
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then return (Some (a', b')) ) ) ;
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None
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let rec close r =
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if not r.sat then r
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else
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match find_missing r with
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| Some (a', b') -> close (merge a' b' r)
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| None -> r
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[sledge] Classify fully-interpreted and simplified exps differently
Summary:
For some Exp forms, Exp.solve is not complete, and this is necessary
since the result of solve is a substitution that needs to encode the
input equation as a conjunction of equations each between a variable
and an exp. This is tantamount to, stronger even than, the theory
being convex. So Exp.solve is not complete for some exps, and some of
those have constructors that perform some simplification. For example,
`(1 ≠ 0)` simplifies to `-1` (i.e. true). To enable deductions such as
`((x ≠ 0) = b) && x = 1 |- b = -1` needs the equality solver to
substitute through subexps of simplifiable exps like = and ≠, as it
does for interpreted exps like + and ×. At the same time, since
Exp.solve for non-interpreted exps cannot be complete, to enable
deductions such as `((x ≠ 0) = (y ≠ 0)) && x = 1 |- y ≠ 0` needs the
equality solver to congruence-close over subexps of simplifiable exps
such as = and ≠, as it does for uninterpreted exps.
To strengthen the equality solver in these sorts of cases, this diff
adds a new class of exps for = and ≠, and revises the equality solver
to handle them in a hybrid fashion between interpreted and
uninterpreted.
I am not currently sure whether or not this breaks the termination
proof, but I have also not managed to adapt usual divergent cases to
break this. One notable point is that simplifying = and ≠ exps always
produces genuinely simpler and smaller exps, in contrast to
e.g. polynomial simplification and gaussian elimination.
Note that the real solution to this problem is likely to be to
eliminate the i1 type in favor or a genuine boolean type, and
translate all integer operations on i1 to boolean/logical ones. Then
the disjunction implicit in e.g. equations between disequations would
appear as actual disjunction, and could be dealt with as such.
Reviewed By: jvillard
Differential Revision: D15424823
fbshipit-source-id: 67d62df1f
6 years ago
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let close r =
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[%Trace.call fun {pf} -> pf "%a" pp r]
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;
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close r
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|>
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[%Trace.retn fun {pf} r' ->
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pf "%a" pp_diff (r, r') ;
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invariant r']
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let and_eq a b r =
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if not r.sat then r
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else
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let a' = canon r a in
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let b' = canon r b in
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let r = extend a' r in
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let r = extend b' r in
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if Exp.equal a' b' then r else close (merge a' b' r)
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(** Exposed interface *)
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let is_true {sat; rep} =
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sat && Map.for_alli rep ~f:(fun ~key:a ~data:a' -> Exp.equal a a')
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let is_false {sat} = not sat
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let entails_eq r d e = Exp.equal (canon r d) (canon r e)
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let entails r s =
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Map.for_alli s.rep ~f:(fun ~key:e ~data:e' -> entails_eq r e e')
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let normalize = canon
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let difference r a b =
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[%Trace.call fun {pf} -> pf "%a@ %a@ %a" Exp.pp a Exp.pp b pp r]
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;
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let a = canon r a in
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let b = canon r b in
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( if Exp.equal a b then Some Z.zero
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else
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match (Exp.typ a, Exp.typ b) with
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| Some typ, _ | _, Some typ -> (
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match normalize r (Exp.sub typ a b) with
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| Integer {data} -> Some data
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| _ -> None )
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| _ -> None )
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|>
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[%Trace.retn fun {pf} ->
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function Some d -> pf "%a" Z.pp_print d | None -> pf ""]
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let and_ r s =
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if not r.sat then r
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else if not s.sat then s
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else
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let s, r =
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if Map.length s.rep <= Map.length r.rep then (s, r) else (r, s)
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in
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Map.fold s.rep ~init:r ~f:(fun ~key:e ~data:e' r -> and_eq e e' r)
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let or_ r s =
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if not s.sat then r
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else if not r.sat then s
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else
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let merge_mems rs r s =
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Map.fold s.rep ~init:rs ~f:(fun ~key:e ~data:e' rs ->
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if entails_eq r e e' then and_eq e e' rs else rs )
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in
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let rs = true_ in
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let rs = merge_mems rs r s in
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let rs = merge_mems rs s r in
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rs
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(* assumes that f is injective and for any set of exps E, f[E] is disjoint
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from E *)
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let map_exps ({sat= _; rep} as r) ~f =
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[%Trace.call fun {pf} -> pf "%a" pp r]
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;
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let map ~equal_key ~equal_data ~f_key ~f_data m =
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Map.fold m ~init:m ~f:(fun ~key ~data m ->
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let key' = f_key key in
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let data' = f_data data in
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if equal_key key' key then
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if equal_data data' data then m else Map.set m ~key ~data:data'
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else Map.remove m key |> Map.add_exn ~key:key' ~data:data' )
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in
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let rep' =
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map rep ~equal_key:Exp.equal ~equal_data:Exp.equal ~f_key:f ~f_data:f
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in
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(if rep' == rep then r else {r with rep= rep'})
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|>
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[sledge] Classify fully-interpreted and simplified exps differently
Summary:
For some Exp forms, Exp.solve is not complete, and this is necessary
since the result of solve is a substitution that needs to encode the
input equation as a conjunction of equations each between a variable
and an exp. This is tantamount to, stronger even than, the theory
being convex. So Exp.solve is not complete for some exps, and some of
those have constructors that perform some simplification. For example,
`(1 ≠ 0)` simplifies to `-1` (i.e. true). To enable deductions such as
`((x ≠ 0) = b) && x = 1 |- b = -1` needs the equality solver to
substitute through subexps of simplifiable exps like = and ≠, as it
does for interpreted exps like + and ×. At the same time, since
Exp.solve for non-interpreted exps cannot be complete, to enable
deductions such as `((x ≠ 0) = (y ≠ 0)) && x = 1 |- y ≠ 0` needs the
equality solver to congruence-close over subexps of simplifiable exps
such as = and ≠, as it does for uninterpreted exps.
To strengthen the equality solver in these sorts of cases, this diff
adds a new class of exps for = and ≠, and revises the equality solver
to handle them in a hybrid fashion between interpreted and
uninterpreted.
I am not currently sure whether or not this breaks the termination
proof, but I have also not managed to adapt usual divergent cases to
break this. One notable point is that simplifying = and ≠ exps always
produces genuinely simpler and smaller exps, in contrast to
e.g. polynomial simplification and gaussian elimination.
Note that the real solution to this problem is likely to be to
eliminate the i1 type in favor or a genuine boolean type, and
translate all integer operations on i1 to boolean/logical ones. Then
the disjunction implicit in e.g. equations between disequations would
appear as actual disjunction, and could be dealt with as such.
Reviewed By: jvillard
Differential Revision: D15424823
fbshipit-source-id: 67d62df1f
6 years ago
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[%Trace.retn fun {pf} r' ->
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pf "%a" pp_diff (r, r') ;
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invariant r']
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let rename r sub = map_exps r ~f:(fun e -> Exp.rename e sub)
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let fold_exps r ~init ~f =
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Map.fold r.rep ~f:(fun ~key ~data z -> f (f z data) key) ~init
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let fold_vars r ~init ~f =
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fold_exps r ~init ~f:(fun init -> Exp.fold_vars ~f ~init)
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let fv e = fold_vars e ~f:Set.add ~init:Var.Set.empty
|