### 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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