Commit 2c261344 authored by Robbert Krebbers's avatar Robbert Krebbers

Make f_equiv stronger.

It no longer requires the functions on both sides of the relation
to be syntactically the same.
parent 63f64743
Pipeline #3823 passed with stage
in 8 minutes and 4 seconds
......@@ -281,24 +281,23 @@ Ltac f_equiv :=
| H : ?R ?x ?y |- ?R2 (match ?x with _ => _ end) (match ?y with _ => _ end) =>
destruct H
(* First assume that the arguments need the same relation as the result *)
| |- ?R (?f ?x) (?f _) => apply (_ : Proper (R ==> R) f)
| |- ?R (?f ?x) _ => apply (_ : Proper (R ==> R) f)
(* For the case in which R is polymorphic, or an operational type class,
like equiv. *)
| |- (?R _) (?f ?x) (?f _) => apply (_ : Proper (R _ ==> _) f)
| |- (?R _ _) (?f ?x) (?f _) => apply (_ : Proper (R _ _ ==> _) f)
| |- (?R _ _ _) (?f ?x) (?f _) => apply (_ : Proper (R _ _ _ ==> _) f)
| |- (?R _) (?f ?x ?y) (?f _ _) => apply (_ : Proper (R _ ==> R _ ==> _) f)
| |- (?R _ _) (?f ?x ?y) (?f _ _) => apply (_ : Proper (R _ _ ==> R _ _ ==> _) f)
| |- (?R _ _ _) (?f ?x ?y) (?f _ _) => apply (_ : Proper (R _ _ _ ==> R _ _ _ ==> _) f)
| |- (?R _ _ _ _) (?f ?x ?y) (?f _ _) => apply (_ : Proper (R _ _ _ _ ==> R _ _ _ _ ==> _) f)
| |- (?R _) (?f ?x) _ => apply (_ : Proper (R _ ==> _) f)
| |- (?R _ _) (?f ?x) _ => apply (_ : Proper (R _ _ ==> _) f)
| |- (?R _ _ _) (?f ?x) _ => apply (_ : Proper (R _ _ _ ==> _) f)
| |- (?R _) (?f ?x ?y) _ => apply (_ : Proper (R _ ==> R _ ==> _) f)
| |- (?R _ _) (?f ?x ?y) _ => apply (_ : Proper (R _ _ ==> R _ _ ==> _) f)
| |- (?R _ _ _) (?f ?x ?y) _ => apply (_ : Proper (R _ _ _ ==> R _ _ _ ==> _) f)
(* Next, try to infer the relation. Unfortunately, there is an instance
of Proper for (eq ==> _), which will always be matched. *)
(* TODO: Can we exclude that instance? *)
(* TODO: If some of the arguments are the same, we could also
query for "pointwise_relation"'s. But that leads to a combinatorial
explosion about which arguments are and which are not the same. *)
| |- ?R (?f ?x) (?f _) => apply (_ : Proper (_ ==> R) f)
| |- ?R (?f ?x ?y) (?f _ _) => apply (_ : Proper (_ ==> _ ==> R) f)
| |- ?R (?f ?x) _ => apply (_ : Proper (_ ==> R) f)
| |- ?R (?f ?x ?y) _ => apply (_ : Proper (_ ==> _ ==> R) f)
(* In case the function symbol differs, but the arguments are the same,
maybe we have a pointwise_relation in our context. *)
| H : pointwise_relation _ ?R ?f ?g |- ?R (?f ?x) (?g ?x) => apply H
......
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