### 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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