This error occurs when the type checker is unable to deduce that two types are 'the same'.

For example, take this expression:

[] == [unit]

[] has type forall a. [a], but [unit] has type [Unit]. These types are not the same. But the type checker is able to determine that the type Unit may be chosen for the type variable a, so the types do become the same. We say that the type checker successfully unifies forall a. [a] with [Unit] in this case.

Another example:

f :: Number -> Number
f x = x + 1
g :: Boolean -> Boolean
g x = x || true

h = g <<< f

The type of (<<<) (that is, function composition) is forall a b c. (b -> c) -> (a -> b) -> (a -> c). For the right hand side of h to type-check, we need to find types a, b, and c such that the types match up. That is, we need to find a choice of a, b, and c such that:

• b = Boolean (from the argument type of g)
• c = Boolean (from the return type of g)
• a = Number (from the argument type of f)
• b = Number (from the return type of f).

b can not be Boolean and Number at the same time, so this system of equations is not satisfiable, and the type checker rejects the program.