Commit 00a054f1 authored by Robbert Krebbers's avatar Robbert Krebbers

Simplify f_equiv a bit.

parent 8a1a8f00
Pipeline #166 passed with stage
......@@ -233,7 +233,7 @@ Ltac setoid_subst :=
If it cannot solve an equality, it will leave that to the user. *)
Ltac f_equiv :=
(* Deal with "pointwise_relation" *)
try lazymatch goal with
repeat lazymatch goal with
| |- pointwise_relation _ _ _ _ => intros ?
end;
(* Normalize away equalities. *)
......@@ -249,13 +249,9 @@ Ltac f_equiv :=
destruct x; f_equiv
(* First assume that the arguments need the same relation as the result *)
| |- ?R (?f ?x) (?f _) =>
let H := fresh "Proper" in
assert (Proper (R ==> R) f) as H by (eapply _);
apply H; clear H; f_equiv
apply (_ : Proper (R ==> R) f); f_equiv
| |- ?R (?f ?x ?y) (?f _ _) =>
let H := fresh "Proper" in
assert (Proper (R ==> R ==> R) f) as H by (eapply _);
apply H; clear H; f_equiv
apply (_ : Proper (R ==> R ==> R) f); f_equiv
(* 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? *)
......@@ -263,17 +259,9 @@ Ltac f_equiv :=
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 _) =>
let R1 := fresh "R" in let H := fresh "HProp" in
let T := type of x in evar (R1: relation T);
assert (Proper (R1 ==> R) f) as H by (subst R1; eapply _);
subst R1; apply H; clear H; f_equiv
apply (_ : Proper (_ ==> R) f); f_equiv
| |- ?R (?f ?x ?y) (?f _ _) =>
let R1 := fresh "R" in let R2 := fresh "R" in
let H := fresh "HProp" in
let T1 := type of x in evar (R1: relation T1);
let T2 := type of y in evar (R2: relation T2);
assert (Proper (R1 ==> R2 ==> R) f) as H by (subst R1 R2; eapply _);
subst R1 R2; apply H; clear H; f_equiv
apply (_ : Proper (_ ==> _ ==> R) f); f_equiv
(* 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) =>
......
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