A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _≤3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done

A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve : Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done

A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve : Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done

A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _≤3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done

A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _≤3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done

A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve : Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done

A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve : Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done

A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _≤3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done

A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve : Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x 3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done

A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Here the Eq A argument to _==_ is not resolved since there are two conflicting candidates: Ord.eqA OrdA and Num.eqA NumA. To solve Nat → A overlap {{eqA}} : Eq A open Num {{...}} hiding (eqA) _≤3 : {A : Set} {{OrdA : Ord A}} {{NumA : Num A}} → A → Bool x ≤3 = (x == fromNat 3) || (x < fromNat 3) Whenever there are multiple valid library3 are the libraries you commonly use. While it is safe to list all your libraries in library, be aware that listing libraries with name clashes in defaults can lead to difficulties, and should be done