-\r
-T6022.hs:3:9:\r
- No instance for (Eq ([a] -> a)) arising from a use of `=='\r
- Possible fix: add an instance declaration for (Eq ([a] -> a))\r
- In the expression: x == head\r
- In an equation for `f': f x = x == head\r
+
+T6022.hs:3:1: error:
+ • Non type-variable argument in the constraint: Eq ([a] -> a)
+ (Use FlexibleContexts to permit this)
+ • When checking the inferred type
+ f :: forall a. Eq ([a] -> a) => ([a] -> a) -> Bool